├── book └── exercises │ └── ch1 │ ├── ch01_1-sol.pdf │ ├── ch01_2-sol.pdf │ ├── ch01_3-sol.pdf │ ├── ch01_1-sol.tex │ ├── ch01_3-sol.tex │ └── ch01_2-sol.tex ├── README.md ├── .gitignore ├── LICENSE_CC └── LICENSE_GNU /book/exercises/ch1/ch01_1-sol.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/lair001/rosen-discrete-math/HEAD/book/exercises/ch1/ch01_1-sol.pdf -------------------------------------------------------------------------------- /book/exercises/ch1/ch01_2-sol.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/lair001/rosen-discrete-math/HEAD/book/exercises/ch1/ch01_2-sol.pdf -------------------------------------------------------------------------------- /book/exercises/ch1/ch01_3-sol.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/lair001/rosen-discrete-math/HEAD/book/exercises/ch1/ch01_3-sol.pdf -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # Solutions for *Discrete Mathematics and Its Applications* by Kenneth H. Rosen 2 | Solutions for the 8th edition of *Discrete Mathematics and Its Applications* by Kenneth H. Rosen. 3 | 4 | ## Contributing 5 | 6 | If you have a question, found an error in a solution, or want to request a solution to a particular problem, feel free to [open an issue](https://github.com/lair001/rosen-discrete-math/issues) or [fork the lehman-mathematics-cs repository](https://github.com/lair001/rosen-discrete-math/fork) and [submit a pull request](https://github.com/lair001/rosen-discrete-math/pulls). You may also email me (Samuel Lair) at lair001@gmail.com. Copies of the book and related materials are in a separate [repo](https://github.com/lair001/rosen-discrete-math-materials). 7 | 8 | ## License 9 | 10 | All software (e.g. *.tex and *.py files) in this repo is licensed under [GPLv3](https://www.gnu.org/licenses/gpl-3.0.en.html). 11 | 12 | Shield: [![CC BY-NC-SA 4.0][cc-by-nc-sa-shield]][cc-by-nc-sa] 13 | 14 | All content (e.g. *.pdf files) in this repo are licensed under a 15 | [Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License][cc-by-nc-sa]. 16 | 17 | [![CC BY-NC-SA 4.0][cc-by-nc-sa-image]][cc-by-nc-sa] 18 | 19 | [cc-by-nc-sa]: http://creativecommons.org/licenses/by-nc-sa/4.0/ 20 | [cc-by-nc-sa-image]: https://licensebuttons.net/l/by-nc-sa/4.0/88x31.png 21 | [cc-by-nc-sa-shield]: https://img.shields.io/badge/License-CC%20BY--NC--SA%204.0-lightgrey.svg -------------------------------------------------------------------------------- /.gitignore: -------------------------------------------------------------------------------- 1 | ## Core latex/pdflatex auxiliary files: 2 | *.aux 3 | *.lof 4 | *.log 5 | *.lot 6 | *.fls 7 | *.out 8 | *.toc 9 | *.fmt 10 | *.fot 11 | *.cb 12 | *.cb2 13 | .*.lb 14 | 15 | ## Intermediate documents: 16 | *.dvi 17 | *.xdv 18 | *-converted-to.* 19 | # these rules might exclude image files for figures etc. 20 | # *.ps 21 | # *.eps 22 | # *.pdf 23 | 24 | ## Generated if empty string is given at "Please type another file name for output:" 25 | .pdf 26 | 27 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 28 | *.bbl 29 | *.bcf 30 | *.blg 31 | *-blx.aux 32 | *-blx.bib 33 | *.run.xml 34 | 35 | ## Build tool auxiliary files: 36 | *.fdb_latexmk 37 | *.synctex 38 | *.synctex(busy) 39 | *.synctex.gz 40 | *.synctex.gz(busy) 41 | *.pdfsync 42 | 43 | ## Build tool directories for auxiliary files 44 | # latexrun 45 | latex.out/ 46 | 47 | ## Auxiliary and intermediate files from other packages: 48 | # algorithms 49 | *.alg 50 | *.loa 51 | 52 | # achemso 53 | acs-*.bib 54 | 55 | # amsthm 56 | *.thm 57 | 58 | # beamer 59 | *.nav 60 | *.pre 61 | *.snm 62 | *.vrb 63 | 64 | # changes 65 | *.soc 66 | 67 | # comment 68 | *.cut 69 | 70 | # cprotect 71 | *.cpt 72 | 73 | # elsarticle (documentclass of Elsevier journals) 74 | *.spl 75 | 76 | # endnotes 77 | *.ent 78 | 79 | # fixme 80 | *.lox 81 | 82 | # feynmf/feynmp 83 | *.mf 84 | *.mp 85 | *.t[1-9] 86 | *.t[1-9][0-9] 87 | *.tfm 88 | 89 | #(r)(e)ledmac/(r)(e)ledpar 90 | *.end 91 | *.?end 92 | *.[1-9] 93 | *.[1-9][0-9] 94 | *.[1-9][0-9][0-9] 95 | *.[1-9]R 96 | *.[1-9][0-9]R 97 | *.[1-9][0-9][0-9]R 98 | *.eledsec[1-9] 99 | *.eledsec[1-9]R 100 | *.eledsec[1-9][0-9] 101 | *.eledsec[1-9][0-9]R 102 | *.eledsec[1-9][0-9][0-9] 103 | *.eledsec[1-9][0-9][0-9]R 104 | 105 | # glossaries 106 | *.acn 107 | *.acr 108 | *.glg 109 | *.glo 110 | *.gls 111 | *.glsdefs 112 | *.lzo 113 | *.lzs 114 | 115 | # uncomment this for glossaries-extra (will ignore makeindex's style files!) 116 | # *.ist 117 | 118 | # gnuplottex 119 | *-gnuplottex-* 120 | 121 | # gregoriotex 122 | *.gaux 123 | *.gtex 124 | 125 | # htlatex 126 | *.4ct 127 | *.4tc 128 | *.idv 129 | *.lg 130 | *.trc 131 | *.xref 132 | 133 | # hyperref 134 | *.brf 135 | 136 | # knitr 137 | *-concordance.tex 138 | # TODO Comment the next line if you want to keep your tikz graphics files 139 | *.tikz 140 | *-tikzDictionary 141 | 142 | # listings 143 | *.lol 144 | 145 | # luatexja-ruby 146 | *.ltjruby 147 | 148 | # makeidx 149 | *.idx 150 | *.ilg 151 | *.ind 152 | 153 | # minitoc 154 | *.maf 155 | *.mlf 156 | *.mlt 157 | *.mtc[0-9]* 158 | *.slf[0-9]* 159 | *.slt[0-9]* 160 | *.stc[0-9]* 161 | 162 | # minted 163 | _minted* 164 | *.pyg 165 | 166 | # morewrites 167 | *.mw 168 | 169 | # nomencl 170 | *.nlg 171 | *.nlo 172 | *.nls 173 | 174 | # pax 175 | *.pax 176 | 177 | # pdfpcnotes 178 | *.pdfpc 179 | 180 | # sagetex 181 | *.sagetex.sage 182 | *.sagetex.py 183 | *.sagetex.scmd 184 | 185 | # scrwfile 186 | *.wrt 187 | 188 | # sympy 189 | *.sout 190 | *.sympy 191 | sympy-plots-for-*.tex/ 192 | 193 | # pdfcomment 194 | *.upa 195 | *.upb 196 | 197 | # pythontex 198 | *.pytxcode 199 | pythontex-files-*/ 200 | 201 | # tcolorbox 202 | *.listing 203 | 204 | # thmtools 205 | *.loe 206 | 207 | # TikZ & PGF 208 | *.dpth 209 | *.md5 210 | *.auxlock 211 | 212 | # todonotes 213 | *.tdo 214 | 215 | # vhistory 216 | *.hst 217 | *.ver 218 | 219 | # easy-todo 220 | *.lod 221 | 222 | # xcolor 223 | *.xcp 224 | 225 | # xmpincl 226 | *.xmpi 227 | 228 | # xindy 229 | *.xdy 230 | 231 | # xypic precompiled matrices and outlines 232 | *.xyc 233 | *.xyd 234 | 235 | # endfloat 236 | *.ttt 237 | *.fff 238 | 239 | # Latexian 240 | TSWLatexianTemp* 241 | 242 | ## Editors: 243 | # WinEdt 244 | *.bak 245 | *.sav 246 | 247 | # Texpad 248 | .texpadtmp 249 | 250 | # LyX 251 | *.lyx~ 252 | 253 | # Kile 254 | *.backup 255 | 256 | # gummi 257 | .*.swp 258 | 259 | # KBibTeX 260 | *~[0-9]* 261 | 262 | # TeXnicCenter 263 | *.tps 264 | 265 | # auto folder when using emacs and auctex 266 | ./auto/* 267 | *.el 268 | 269 | # expex forward references with \gathertags 270 | *-tags.tex 271 | 272 | # standalone packages 273 | *.sta 274 | 275 | # Makeindex log files 276 | *.lpz 277 | 278 | # bak files 279 | *.bak* 280 | 281 | # intellij 282 | .idea/ 283 | -------------------------------------------------------------------------------- /book/exercises/ch1/ch01_1-sol.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \usepackage[utf8]{inputenc} 3 | \usepackage[T1]{fontenc} 4 | \usepackage{indentfirst, hyperref, gensymb, amsmath, amsthm, wasysym, amsfonts, mathtools, braket, amssymb} 5 | \hypersetup{colorlinks,allcolors=blue,linktoc=all} 6 | 7 | \title{Chapter 1 Section 1 Exercise Solutions} 8 | \author{Samuel Lair} 9 | \date{September 2022} 10 | 11 | \begin{document} 12 | 13 | \maketitle 14 | \tableofcontents 15 | 16 | \pagebreak 17 | 18 | \section{Exercise 24} 19 | \subsection{(a)} 20 | If you got promoted, then you washed the boss's car. 21 | \subsection{(b)} 22 | If there are winds from the south, then there will be a spring thaw. 23 | \subsection{(c)} 24 | If you bought the computer less than a year ago, then the warranty is good. 25 | \subsection{(d)} 26 | If Willy cheats, then he gets caught. 27 | \subsection{(e)} 28 | If you can access the website, then you paid a subscription fee. 29 | \subsection{(f)} 30 | If you know the right people, then you'll get elected. 31 | \subsection{(g)} 32 | If Carol is on a boat, then she gets seasick. 33 | 34 | \pagebreak 35 | 36 | \section{Exercise 25} 37 | \subsection{(a)} 38 | If the wind blows from the northeast, then it snows. 39 | \subsection{(b)} 40 | If it stays warm for a week, then the apple trees will bloom. 41 | \subsection{(c)} 42 | If the Pistons win the championship, then they beat the Lakers. 43 | \subsection{(d)} 44 | If you got to the top of Long's Peak, then you walked eight miles. 45 | \subsection{(e)} 46 | If you are world famous, then you will get tenure as a professor. 47 | \subsection{(f)} 48 | If you drive more than 400 miles, then you will need to buy gasoline. 49 | \subsection{(g)} 50 | If your guarantee is good, then you bought your CD player less than 90 days ago. 51 | \subsection{(h)} 52 | If the water isn't too cold, then Jan will go swimming. 53 | \subsection{(i)} 54 | If people believe in science, then we will have a future. 55 | 56 | \pagebreak 57 | 58 | \section{Exercise 26} 59 | \subsection{(a)} 60 | If I remembered to send you the address, then you sent me an e-mail message. 61 | \subsection{(b)} 62 | If you were born in the United States, then you are a citizen of this country. 63 | \subsection{(c)} 64 | If you keep your textbook, then it will be a useful reference in your future courses. 65 | \subsection{(d)} 66 | If their goalie plays well, then the Red Wings will win the Stanley Cup. 67 | \subsection{(e)} 68 | If you get the job, then you had the best credentials. 69 | \subsection{(f)} 70 | If there is a storm, then the beach erodes. 71 | \subsection{(g)} 72 | If you log on to the server, then you have a valid password. 73 | \subsection{(h)} 74 | If you don't begin your climb too late, then you will reach the summit. 75 | \subsection{(i)} 76 | If you are among the first 100 customers tomorrow, then you will get a free ice cream cone. 77 | 78 | \pagebreak 79 | 80 | \section{Exercise 27} 81 | \subsection{(a)} 82 | It is hot outside if and only if you buy an ice cream cone. 83 | \subsection{(b)} 84 | You win the contest if and only if you have the only winning ticket. 85 | \subsection{(c)} 86 | You get promoted if and only if you have connections. 87 | \subsection{(d)} 88 | You watch television if and only if your mind will decay. 89 | \subsection{(e)} 90 | The trains run late if and only if it is a day when I take the train. 91 | 92 | \pagebreak 93 | 94 | \section{Exercise 28} 95 | \subsection{(a)} 96 | You get an A in this course if and only if you learn how to solve discrete mathematics problems. 97 | \subsection{(b)} 98 | You will be informed if and only if you read the newspaper every day. 99 | \subsection{(c)} 100 | It rains if and only if it is a weekend day. 101 | \subsection{(d)} 102 | The wizard is not in if and only if you can see him. 103 | \subsection{(e)} 104 | My airplane flight is late if and only if I have to catch a connecting flight. 105 | 106 | \pagebreak 107 | 108 | \section{Exercise 29} 109 | \subsection{(a)} 110 | Original: If it snows today, then I will ski tomorrow. 111 | 112 | Converse: If I will ski tomorrow, then it snows today. 113 | 114 | Contrapositive: If I won't ski tomorrow, then it doesn't snow today. 115 | 116 | Inverse: If it doesn't snow today, then I won't ski tomorrow. 117 | 118 | \subsection{(b)} 119 | Original: If there is going to be a quiz, then I come to class. 120 | 121 | Converse: If I come to class, then there is going to be a quiz. 122 | 123 | Contrapositive: If I don't come to class, then there isn't going to be a quiz. 124 | 125 | Inverse: If there isn't going to be a quiz, then I don't come to class. 126 | 127 | \subsection{(c)} 128 | Original: If a positive integer is prime, then it has no divisors other than 1 and itself. 129 | 130 | Converse: If a positive integer has no divisors other than 1 and itself, then it is prime. 131 | 132 | Contrapositive: If a positive integer has divisors other than 1 and itself, then it isn't prime. 133 | 134 | Inverse: If a positive integer isn't prime, then it has divisors other than 1 and itself. 135 | 136 | \pagebreak 137 | 138 | \section{Exercise 30} 139 | 140 | \subsection{(a)} 141 | Original: If it snows tonight, then I will stay at home. 142 | 143 | Converse: If I will stay at home, then it snows tonight. 144 | 145 | Contrapositive: If I won't stay at home, then it doesn't snow tonight. 146 | 147 | Inverse: If it doesn't snow tonight, then I won't stay at home. 148 | 149 | \subsection{(b)} 150 | Original: If it is a sunny summer day, then I go to the beach. 151 | 152 | Converse: If I go to the beach, then it is a sunny summer day. 153 | 154 | Contrapositive: If I don't go the beach, then it isn't a sunny summer day. 155 | 156 | Inverse: If it isn't a sunny summer day, then I don't go to the beach. 157 | 158 | \subsection{(c)} 159 | Original: If I stay up late, then I sleep until noon. 160 | 161 | Converse: If I sleep until noon, then I stay up late. 162 | 163 | Contrapositive: If I don't sleep until noon, then I don't stay up late. 164 | 165 | Inverse: If I don't stay up late, then I don't sleep until noon. 166 | 167 | \pagebreak 168 | 169 | \section{Exercise 42} 170 | Let 171 | \begin{align} 172 | & (p \lor \neg q) \label{e42e1} \\ 173 | & (q \lor \neg r) \label{e42e2} \\ 174 | & (r \lor \neg p) \label{e42e3} \\ 175 | & \eqref{e42e1} \land \eqref{e42e2} \land \eqref{e42e3} \label{e42e4} 176 | \end{align} 177 | 178 | \eqref{e42e1} is true if and only if p is true or q is false. 179 | 180 | \eqref{e42e2} is true if and only if q is true or r is false. 181 | 182 | \eqref{e42e3} is true if and only if r is true or p is false. 183 | 184 | \eqref{e42e4} is true if and only if \eqref{e42e1}, \eqref{e42e2}, and \eqref{e42e3} are all true. 185 | 186 | Therefore, when $p,q,r$ have the same truth value, \eqref{e42e1}, \eqref{e42e2}, \eqref{e42e3}, and \eqref{e42e4} are all true. However, when one of $p,q,r$ has a different truth value from the other two, two of \eqref{e42e1}, \eqref{e42e2}, and \eqref{e42e3} are true while the other is false. \eqref{e42e4} is false in this case. 187 | 188 | Hence, \eqref{e42e4} is true when $p$, $q$, and $r$ have the same truth value and false otherwise. 189 | 190 | \pagebreak 191 | 192 | \section{Exercise 43} 193 | Let 194 | \begin{align} 195 | & (p \lor q \lor r) \label{e43e1} \\ 196 | & (\neg p \lor \neg q \lor \neg r) \label{e43e2} \\ 197 | & \eqref{e43e1} \land \eqref{e43e2} \label{e43e3} 198 | \end{align} 199 | 200 | \eqref{e43e1} is true if and only if at least one of $p, q, r$ is true. 201 | 202 | \eqref{e43e2} is true if and only if at least one of $p, q, r$ is false. 203 | 204 | Hence, \eqref{e43e3} is true if and only if at least one of $p, q, r$ is true and at least one of $p, q, r$ is false. I.e. \eqref{e43e3} is true if at least one of $p, q, r$ is true and at least one is false, but is false when all three variables have the same truth value 205 | 206 | 207 | \pagebreak 208 | 209 | \section{Exercise 44} 210 | If $p_1, p_2, ..., p_n$ are $n$ propositions explain why 211 | \begin{equation}\label{e44claim} 212 | \bigwedge_{i=1}^{n-1} \bigwedge_{j=i+1}^n (\neg p_i \lor \neg p_j) 213 | \end{equation} 214 | is true if and only if at most of $p_1, p_2, ..., p_n$ is true. 215 | 216 | \eqref{e44claim} is a conjunction of disjuncts of the form 217 | \begin{equation}\label{e44dj} 218 | \neg p_i \lor \neg p_j 219 | \end{equation} 220 | Due to the limits of the conjunctions, there is exactly one disjunct for every pair $(i,j)$ such that $1 \le i < j \le n$. Therefore, if more than one of the $p_k$'s are true, then there exists a pair $(i',j')$ such that both $p_{i'}$ and $p_{j'}$ are true and the corresponding disjunct \eqref{e44dj} is false. A single false disjunct is enough to cause \eqref{e44claim} to evaluate to false. 221 | 222 | If no more than 1 of the $p_k$'s is true, then \eqref{e44dj} is true. If \eqref{e44dj} is true, then no more 1 of the $p_k$'s is true. Our claim follows. 223 | 224 | \pagebreak 225 | 226 | \section{Exercise 45} 227 | \begin{equation} 228 | \bigvee_{h=1}^n p_h 229 | \end{equation} 230 | is true if and only if at least one of the $p_k$'s are true. Therefore, 231 | \begin{equation} 232 | \left( \bigwedge_{i=1}^{n-1} \bigwedge_{j=i+1}^n (\neg p_i \lor \neg p_j) \right) \land \left( \bigvee_{h=1}^n p_h \right) 233 | \end{equation} 234 | is true if and only if exactly one of the $p_k$'s is true. 235 | 236 | \pagebreak 237 | 238 | \section{Exercise 47} 239 | \subsection{(a)} 240 | \begin{align*} 241 | & \text{101 1110} & \\ 242 | & \underline{\text{010 0001}} & \\ 243 | & \text{111 1111} & (\text{bitwise OR}) \\ 244 | & \text{000 0000} & (\text{bitwise AND}) \\ 245 | & \text{111 1111} & (\text{bitwise XOR}) \\ 246 | \end{align*} 247 | 248 | \subsection{(b)} 249 | \begin{align*} 250 | & \text{1111 0000} & \\ 251 | & \underline{\text{1010 1010}} & \\ 252 | & \text{1111 1010} & (\text{bitwise OR}) \\ 253 | & \text{1010 0000} & (\text{bitwise AND}) \\ 254 | & \text{0101 1010} & (\text{bitwise XOR}) \\ 255 | \end{align*} 256 | 257 | \subsection{(c)} 258 | \begin{align*} 259 | & \text{00 0111 0001} & \\ 260 | & \underline{\text{10 0100 1000}} & \\ 261 | & \text{10 0111 1001} & (\text{bitwise OR}) \\ 262 | & \text{00 0100 0000} & (\text{bitwise AND}) \\ 263 | & \text{10 0011 1001} & (\text{bitwise XOR}) \\ 264 | \end{align*} 265 | 266 | \subsection{(d)} 267 | \begin{align*} 268 | & \text{11 1111 1111} & \\ 269 | & \underline{\text{00 0000 0000}} & \\ 270 | & \text{11 1111 1111} & (\text{bitwise OR}) \\ 271 | & \text{00 0000 0000} & (\text{bitwise AND}) \\ 272 | & \text{11 1111 1111} & (\text{bitwise XOR}) \\ 273 | \end{align*} 274 | 275 | \section{Exercise 48} 276 | \subsection{(a)} 277 | \begin{align*} 278 | & \text{0 1011} \\ 279 | \lor \; & \underline{\text{1 1011}} \\ 280 | & \text{1 1011} \\ 281 | & \\ 282 | & \text{1 1000} \\ 283 | \land \; & \underline{\text{1 1011}} \\ 284 | & \text{1 1000} \; \leftarrow \text{answer} 285 | \end{align*} 286 | 287 | \subsection{(b)} 288 | \begin{align*} 289 | & \text{0 1111} \\ 290 | \land \; & \underline{\text{1 0101}} \\ 291 | & \text{0 0101} \\ 292 | & \\ 293 | & \text{0 0101} \\ 294 | \lor \; & \underline{\text{0 1000}} \\ 295 | & \text{0 1101} \; \leftarrow \text{answer} 296 | \end{align*} 297 | 298 | \subsection{(c)} 299 | \begin{align*} 300 | & \text{0 1010} \\ 301 | \oplus \; & \underline{\text{1 1011}} \\ 302 | & \text{1 0001} \\ 303 | & \\ 304 | & \text{1 0001} \\ 305 | \oplus \; & \underline{\text{0 1000}} \\ 306 | & \text{1 1001} \; \leftarrow \text{answer} 307 | \end{align*} 308 | 309 | \subsection{(d)} 310 | \begin{align*} 311 | & \text{1 1011} \\ 312 | \lor \; & \underline{\text{0 1010}} \\ 313 | & \text{1 1011} \\ 314 | & \\ 315 | & \text{1 0001} \\ 316 | \lor \; & \underline{\text{1 1011}} \\ 317 | & \text{1 1011} \\ 318 | & \\ 319 | & \text{1 1011} \\ 320 | \land \; & \underline{\text{1 1011}} \\ 321 | & \text{1 1011} \; \leftarrow \text{answer} 322 | \end{align*} 323 | 324 | \pagebreak 325 | 326 | \section{Exercise 49} 327 | The truth value of "Fred is not happy" is $1 - 0.8 = 0.2$. 328 | 329 | The truth value of "John is not happy" is $1 - 0.4 = 0.6$. 330 | 331 | \pagebreak 332 | 333 | \section{Exercise 50} 334 | The truth value of "Fred and John are happy" is $min(0.8, 0.4) = 0.4$. 335 | 336 | "Neither Fred nor John is happy" is equivalent to "Fred is not happy and John is not happy". Therefore, the truth value of "Neither Fred nor John is happy" is $min(0.2, 0.6) = 0.2$. 337 | 338 | \pagebreak 339 | 340 | \section{Exercise 51} 341 | The truth value of "Fred is happy or John is happy" is $max(0.8, 0.4) = 0.8$. 342 | 343 | The truth value of "Fred is not happy or John is not happy" is $max(0.2, 0.6) = 0.6$. 344 | 345 | \pagebreak 346 | 347 | \section{Exercise 52} 348 | "This statement is false" is not a proposition. It is a declarative sentence but it doesn't have a definite truth value. It is, in fact, a logical paradox. Regardless of whether you assume it is true or false, you reach a contradiction. 349 | 350 | \pagebreak 351 | 352 | \section{Exercise 53} 353 | \subsection{(a)} 354 | If we assume that the 100th statement is true, we reach a contradiction since we must conclude that the 100th statement is false. No contradiction is reached if the 100th statement is false. Therefore, the 100th statement is false, which implies that at least one of the statements are true. 355 | 356 | If we take $1 \le n \le 98$ and assume that the $n^{th}$ statement is true, we reach a contradiction. If the $n^{th}$ statement is true, then $100 - n$ statements must true. Since $100 - n > 1$, there must be true statements other than $n^{th}$ statement. However, if more than 1 statement is true, they will contradict each other. No contradiction is reached if we assume that the $n^{th}$ statement is false. Therefore, statements $1-98$ are be false. 357 | 358 | Now if we assume that the $99^{th}$ statement is false, we reach a contradiction in that all statements are false yet exactly 1 must be true. However, no contradiction is reached if we assume that the $99^{th}$ statement is true. Therefore, the 99th statement must be true. 359 | 360 | Hence, we conclude that the $99^{th}$ statement is true and the rest are false. 361 | 362 | \subsection{(b)} 363 | First of all, note that if the $n^{th}$ statement is true then all prior statements are true as well. 364 | 365 | If we assume that the $51^{st}$ statement is true, we reach a contradiction. If the $51^{st}$ statement is true, then statements $1-50$ must also be true. This leaves only statements $52-100$ available to be false. However, this is only a total of $49$ statements, contradicting the $51^{st}$ statement. Therefore, $51^{st}$ statement is false, which implies that statements $52-100$ are also false. 366 | 367 | If we assume that the $50^{th}$ statement is false, then we reach a contradiction where at least $50$ statements are false since statements $51-100$ must also be false. We reach no contradiction if we assume that the $50^{th}$ statement is true. Therefore, the $50^{th}$ statement is true, which implies that statements $1-49$ are also true. 368 | 369 | Hence, we conclude that statements $1-50$ are true and statements $51-100$ are false. 370 | 371 | \subsection{(c)} 372 | If we assume that the $50^{th}$ statement is true, then we reach a contraction since statements $1-49$ must also be true leaving only statements $51-99$ available to be false. 373 | 374 | However, if we assume that the $50^{th}$ statement is false, then we reach a contraction where at $50$ statements are false since statements $51-99$ must also be false. 375 | 376 | Hence, we conclude that this system of statements represents an unsolvable logical paradox. 377 | 378 | \end{document} -------------------------------------------------------------------------------- /book/exercises/ch1/ch01_3-sol.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \usepackage[utf8]{inputenc} 3 | \usepackage[T1]{fontenc} 4 | \usepackage{indentfirst, hyperref, gensymb, amsmath, amsthm, wasysym, amsfonts, mathtools, braket, amssymb} 5 | \hypersetup{colorlinks,allcolors=blue,linktoc=all} 6 | 7 | \let\biconditional\leftrightarrow 8 | \let\conditional\rightarrow 9 | 10 | \title{Chapter 1 Section 3 Exercise Solutions} 11 | \author{Samuel Lair} 12 | \date{September 2022} 13 | 14 | \begin{document} 15 | 16 | \maketitle 17 | \tableofcontents 18 | 19 | \pagebreak 20 | 21 | \section{Exercise 5} 22 | \[ 23 | \begin{array}{|c|c|c|c|c|c|c|c|} 24 | p & q & r & q \lor r & p \land (q \lor r) & p \land q & p \land r & (p \land q) \lor (p \land r) \\ 25 | \hline 26 | T & T & T & T & T & T & T & T \\ 27 | T & T & F & T & T & T & F & T \\ 28 | T & F & T & T & T & F & T & T \\ 29 | T & F & F & F & F & F & F & F \\ 30 | F & T & T & T & F & F & F & F \\ 31 | F & T & F & T & F & F & F & F \\ 32 | F & F & T & T & F & F & F & F \\ 33 | F & F & F & F & F & F & F & F 34 | \end{array} 35 | \] 36 | 37 | Since the truth values of the compound propositions $p \land (q \lor r)$ and $(p \land q) \lor (p \land r)$ agree for all possible combinations of the truth values of $p$, $q$, and $r$, said compound propositions are logically equivalent. 38 | 39 | \section{Exercise 6} 40 | \[ 41 | \begin{array}{|c|c|c|c|c|c|c|} 42 | p & q & p \land q & \neg (p \land q) & \neg p & \neg q & \neg p \lor \neg q \\ 43 | \hline 44 | T & T & T & F & F & F & F \\ 45 | T & F & F & T & F & T & T \\ 46 | F & T & F & T & T & F & T \\ 47 | F & F & F & T & T & T & T 48 | \end{array} 49 | \] 50 | 51 | Since the truth values of the compound propositions $\neg (p \land q)$ and $\neg p \lor \neg q$ agree for all possible combinations of the truth values of $p$ and $q$, said compound propositions are logically equivalent. 52 | 53 | \pagebreak 54 | 55 | \section{Exercise 9} 56 | 57 | \subsection{(a)} 58 | \begin{align*} 59 | & p \implies \neg q & \equiv \\ 60 | \neg & p \lor \neg q 61 | \end{align*} 62 | 63 | \subsection{(b)} 64 | \begin{align*} 65 | & (p \implies q) \implies r & \equiv \\ 66 | \neg & (p \implies q) \lor r & \equiv \\ 67 | \neg & (\neg p \lor q) \lor r & \equiv \\ 68 | & (p \land \neg q) \lor r 69 | \end{align*} 70 | 71 | \subsection{(c)} 72 | \begin{align*} 73 | & (\neg q \implies p) \implies (p \implies \neg q) & \equiv \\ 74 | \neg & (\neg q \implies p) \lor (p \implies \neg q) & \equiv \\ 75 | \neg & (q \lor p) \lor (\neg p \lor \neg q) & \equiv \\ 76 | & (\neg q \land \neg p) \lor (\neg p \lor \neg q) & \equiv \\ 77 | & (\neg p \land \neg q) \lor (\neg p \lor \neg q) & \equiv \\ 78 | & ((\neg p \land \neg q) \lor \neg p) \lor \neg q & \equiv \\ 79 | & (\neg p \lor (\neg p \land \neg q)) \lor \neg q & \equiv \\ 80 | \neg & p \lor \neg q 81 | \end{align*} 82 | 83 | \pagebreak 84 | 85 | \section{Exercise 10} 86 | 87 | \subsection{(a)} 88 | \begin{align*} 89 | \neg & p \implies \neg q & \equiv \\ 90 | \neg (\neg & p) \lor \neg q & \equiv \\ 91 | & p \lor \neg q 92 | \end{align*} 93 | 94 | \subsection{(b)} 95 | \begin{align*} 96 | & p \lor q \implies \neg p & \equiv \\ 97 | \neg ( & p \lor q) \lor \neg p & \equiv \\ 98 | (\neg & p \land \neg q) \lor \neg p & \equiv \\ 99 | \neg & p \lor (\neg p \land \neg q) & \equiv \\ 100 | \neg & p 101 | \end{align*} 102 | 103 | \subsection{(c)} 104 | \begin{align*} 105 | ( & p \implies \neg q) \implies (\neg p \implies q) & \equiv \\ 106 | \neg ( & p \implies \neg q) \lor (\neg p \implies q) & \equiv \\ 107 | \neg (\neg & p \lor \neg q) \lor (p \lor q) & \equiv \\ 108 | ( & p \land q) \lor (p \lor q) & \equiv \\ 109 | (( & p \land q) \lor p) \lor q & \equiv \\ 110 | ( & p \lor (p \land q)) \lor q & \equiv \\ 111 | & p \lor q 112 | \end{align*} 113 | 114 | \pagebreak 115 | 116 | \section{Exercise 15} 117 | 118 | \subsection{(a)} 119 | \begin{align*} 120 | & (p \land q) \implies p & \equiv \\ 121 | \neg & (p \land q) \lor p & \equiv \\ 122 | (\neg & p \lor \neg q) \lor p & \equiv \\ 123 | \neg & q \lor (p \lor \neg p) & \equiv \\ 124 | \neg & q \lor T & \equiv \\ 125 | & T 126 | \end{align*} 127 | 128 | \subsection{(b)} 129 | \begin{align*} 130 | & p \implies (p \lor q) & \equiv \\ 131 | \neg & p \lor (p \lor q) & \equiv \\ 132 | & q \lor (p \lor \neg p) & \equiv \\ 133 | & q \lor T & \equiv \\ 134 | & T 135 | \end{align*} 136 | 137 | \subsection{(c)} 138 | \begin{align*} 139 | \neg & p \implies (p \implies q) & \equiv \\ 140 | \neg & p \implies (\neg p \lor q) & \equiv \\ 141 | & p \lor (\neg p \lor q) & \equiv \\ 142 | & q \lor (p \lor \neg p) & \equiv \\ 143 | & q \lor T & \equiv \\ 144 | & T 145 | \end{align*} 146 | 147 | \subsection{(d)} 148 | \begin{align*} 149 | ( & p \land q) \implies (p \implies q) & \equiv \\ 150 | ( & p \land q) \implies (\neg p \lor q) & \equiv \\ 151 | \neg ( & p \land q) \lor (\neg p \lor q) & \equiv \\ 152 | (\neg & p \lor \neg q) \lor (\neg p \lor q) & \equiv \\ 153 | (\neg & p \lor \neg p) \lor (q \lor \neg q) & \equiv \\ 154 | \neg & p \lor T & \equiv \\ 155 | & T 156 | \end{align*} 157 | 158 | \subsection{(e)} 159 | \begin{align*} 160 | \neg( & p \implies q) \implies p & \equiv \\ 161 | \neg(\neg & p \lor q) \implies p & \equiv \\ 162 | (\neg & p \lor q) \lor p & \equiv \\ 163 | & q \lor (p \lor \neg p) & \equiv \\ 164 | & q \lor T & \equiv \\ 165 | & T 166 | \end{align*} 167 | 168 | \subsection{(f)} 169 | \begin{align*} 170 | \neg( & p \implies q) \implies \neg q & \equiv \\ 171 | \neg(\neg & p \lor q) \implies \neg q & \equiv \\ 172 | (\neg & p \lor q) \lor \neg q & \equiv \\ 173 | \neg & p \lor (q \lor \neg q) & \equiv \\ 174 | \neg & p \lor T & \equiv \\ 175 | & T 176 | \end{align*} 177 | 178 | \pagebreak 179 | 180 | \section{Exercise 16} 181 | \subsection{(a)} 182 | \begin{align*} 183 | [\neg & p \land (p \lor q)] \implies q & \equiv \\ 184 | \neg [\neg & p \land (p \lor q)] \lor q & \equiv \\ 185 | & p \lor \neg (p \lor q) \lor q & \equiv \\ 186 | & p \lor (\neg p \land \neg q) \lor q & \equiv \\ 187 | ( & p \lor q) \lor (\neg p \land \neg q) & \equiv \\ 188 | (( & p \lor q) \lor \neg p) \land ((p \lor q) \lor \neg q) & \equiv \\ 189 | ( & q \lor (p \lor \neg p)) \land (p \lor (q \lor \neg q)) & \equiv \\ 190 | ( & q \lor T) \land (p \lor T) & \equiv \\ 191 | & T \land T & \equiv \\ 192 | & T 193 | \end{align*} 194 | 195 | \subsection{(b)} 196 | \begin{align*} 197 | [( & p \implies q) \land (q \implies r)] \implies (p \implies r) & \equiv \\ 198 | \neg [( & p \implies q) \land (q \implies r)] \lor (p \implies r) & \equiv \\ 199 | \neg [(\neg & p \lor q) \land (\neg q \lor r)] \lor (\neg p \lor r) & \equiv \\ 200 | [\neg(\neg & p \lor q) \lor \neg (\neg q \lor r)] \lor (\neg p \lor r) & \equiv \\ 201 | [( & p \land \neg q) \lor (q \land \neg r)] \lor (\neg p \lor r) & \equiv \\ 202 | [(( & p \land \neg q) \land q) \lor ((p \lor \neg q) \land \neg r)] \lor (\neg p \lor r) & \equiv \\ 203 | [( & p \land (q \land \neg q)) \lor ((p \lor \neg q) \land \neg r)] \lor (\neg p \lor r) & \equiv \\ 204 | [( & p \land F) \lor (\neg r \land (p \lor \neg q))] \lor (\neg p \lor r) & \equiv \\ 205 | [ & F \lor ((\neg r \land p) \lor (\neg r \land \neg q))] \lor (\neg p \lor r) & \equiv \\ 206 | ((\neg & r \land p) \lor (\neg r \land \neg q)) \lor (\neg p \lor r) & \equiv \\ 207 | ((\neg & p \lor r) \lor (\neg r \land p)) \lor (\neg r \land \neg q) & \equiv \\ 208 | (((\neg & p \lor r) \lor \neg r) \land ((\neg p \lor r) \lor p)) \lor (\neg r \land \neg q) & \equiv \\ 209 | ((\neg & p \lor T) \land (r \lor T)) \lor (\neg r \land \neg q) & \equiv \\ 210 | ( & T \land T) \lor (\neg r \land \neg q) & \equiv \\ 211 | (\neg & r \land \neg q) \lor T & \equiv \\ 212 | & T 213 | \end{align*} 214 | 215 | \subsection{(c)} 216 | \begin{align*} 217 | [ & p \land (p \implies q)] \implies q & \equiv \\ 218 | \neg [ & p \land (\neg p \lor q)] \lor q & \equiv \\ 219 | [\neg & p \lor \neg (\neg p \lor q)] \lor q & \equiv \\ 220 | [\neg & p \lor (p \land \neg q)] \lor q & \equiv \\ 221 | (\neg & p \lor q) \lor (p \land \neg q) & \equiv \\ 222 | [((\neg & p \lor q) \lor p) \land ((\neg p \lor q) \lor \neg q)] & \equiv \\ 223 | [( & q \lor T) \land (\neg p \lor T)] & \equiv \\ 224 | & T \land T \equiv \\ 225 | & T 226 | \end{align*} 227 | 228 | \subsection{(d)} 229 | \begin{align*} 230 | [( & p \lor q) \land (p \implies r) \land (q \implies r)] \implies r & \equiv \\ 231 | \neg [( & p \lor q) \land (\neg p \lor r) \land (\neg q \lor r)] \lor r & \equiv \\ 232 | [\neg ( & p \lor q) \lor \neg (\neg p \lor r) \lor \neg (\neg q \lor r)] \lor r & \equiv \\ 233 | [(\neg & p \land \neg q) \lor (p \land \neg r) \lor (q \land \neg r)] \lor r & \equiv \\ 234 | [(\neg & p \land \neg q) \lor (p \land \neg r)] \lor [r \lor (q \land \neg r)] & \equiv \\ 235 | [(\neg & p \land \neg q) \lor (p \land \neg r)] \lor [(r \lor q) \land (r \lor \neg r)] & \equiv \\ 236 | [(\neg & p \land \neg q) \lor (p \land \neg r)] \lor (r \lor q) & \equiv \\ 237 | [( & r \lor q) \lor (\neg p \land \neg q)] \lor (p \land \neg r) & \equiv \\ 238 | [(( & r \lor q) \lor \neg p) \land ((r \lor q) \lor \neg q)] \lor (p \land \neg r) & \equiv \\ 239 | (( & r \lor q) \lor \neg p) \lor (p \land \neg r) & \equiv \\ 240 | ((( & r \lor q) \lor \neg p) \lor p) \land (((r \lor q) \lor \neg p) \lor \neg r) & \equiv \\ 241 | (( & r \lor q) \lor (p \lor \neg p)) \land ((\neg p \lor q) \lor (r \lor \neg r)) & \equiv \\ 242 | (( & r \lor q) \lor T) \land ((\neg p \lor q) \lor T) & \equiv \\ 243 | & T \land T & \equiv \\ 244 | & T 245 | \end{align*} 246 | 247 | \pagebreak 248 | 249 | \section{Exercise 17} 250 | 251 | \subsection{(a)} 252 | \[ 253 | \begin{array}{|c|c|c|c|} 254 | p & q & p \land q & p \lor (p \land q) \\ 255 | \hline 256 | \textbf{T} & T & T & \textbf{T} \\ 257 | \textbf{T} & F & F & \textbf{T} \\ 258 | \textbf{F} & T & F & \textbf{F} \\ 259 | \textbf{F} & F & F & \textbf{F} 260 | \end{array} 261 | \] 262 | 263 | Since the truth values of $p \lor (p \land q)$ and $p$ agree for all possible combinations of truth values for $p$ and $q$, $p \lor (p \land q)$ and $p$ are logically equivalent. I.e. $p \lor (p \land q) \equiv p$ is true. 264 | 265 | \subsection{(b)} 266 | \[ 267 | \begin{array}{|c|c|c|c|} 268 | p & q & p \lor q & p \land (p \lor q) \\ 269 | \hline 270 | \textbf{T} & T & T & \textbf{T} \\ 271 | \textbf{T} & F & T & \textbf{T} \\ 272 | \textbf{F} & T & T & \textbf{F} \\ 273 | \textbf{F} & F & F & \textbf{F} 274 | \end{array} 275 | \] 276 | 277 | Since the truth values of $p \land (p \lor q)$ and $p$ agree for all possible combinations of truth values for $p$ and $q$, $p \land (p \lor q)$ and $p$ are logically equivalent. I.e. $p \land (p \lor q) \equiv p$ is true. 278 | 279 | \pagebreak 280 | 281 | \section{Exercise 18} 282 | 283 | \begin{align*} 284 | (\neg & p \land (p \conditional q)) \conditional \neg q & \equiv \\ 285 | \neg (\neg & p \land (\neg p \lor q)) \lor \neg q & \equiv \\ 286 | ( & p \lor \neg (\neg p \lor q)) \lor \neg q & \equiv \\ 287 | ( & p \lor (p \land \neg q)) \lor \neg q & \equiv \\ 288 | & p \lor \neg q & \equiv \\ 289 | \neg & p \conditional \neg q 290 | \end{align*} 291 | 292 | Hence, $(\neg p \land (p \conditional q)) \conditional \neg q$ is not a tautology. 293 | 294 | \pagebreak 295 | 296 | \section{Exercise 19} 297 | 298 | \begin{align*} 299 | (\neg & q \land (p \conditional q)) \conditional \neg q & \equiv \\ 300 | \neg (\neg & q \land (\neg p \lor q)) \lor \neg q & \equiv \\ 301 | ( & q \lor \neg (\neg p \lor q)) \lor \neg q & \equiv \\ 302 | ( & q \lor (p \land \neg q)) \lor \neg q & \equiv \\ 303 | ( & p \land \neg q) \lor (q \lor \neg q) & \equiv \\ 304 | ( & p \land \neg q) \lor T & \equiv \\ 305 | & T 306 | \end{align*} 307 | 308 | Hence, $(\neg q \land (p \conditional q)) \conditional \neg q$ is a tautology. 309 | 310 | \pagebreak 311 | 312 | \section{Exercise 20} 313 | Let 314 | \begin{align} 315 | & p \biconditional q \label{ex20eq1} \\ 316 | ( & p \land q) \lor (\neg p \land \neg q) \label{ex20eq2} 317 | \end{align} 318 | 319 | \[ 320 | \begin{array}{|c|c|c|c|c|c|} 321 | p & q & \eqref{ex20eq1} & p \land q & \neg p \land \neg q & \eqref{ex20eq2} \\ 322 | \hline 323 | T & T & \textbf{T} & T & F & \textbf{T} \\ 324 | T & F & \textbf{F} & F & F & \textbf{F} \\ 325 | F & T & \textbf{F} & F & F & \textbf{F} \\ 326 | F & F & \textbf{T} & F & T & \textbf{T} 327 | \end{array} 328 | \] 329 | 330 | Since the truth values of \eqref{ex20eq1} and \eqref{ex20eq2} agree for all possible combinations of truth values for $p$ and $q$, \eqref{ex20eq1} and \eqref{ex20eq2} are logically equivalent. 331 | 332 | \pagebreak 333 | 334 | \section{Exercise 21} 335 | Let 336 | \begin{align} 337 | \neg ( & p \biconditional q) \label{ex21eq1} \\ 338 | & p \biconditional \neg q \label{ex21eq2} 339 | \end{align} 340 | 341 | \[ 342 | \begin{array}{|c|c|c|c|c|c|} 343 | p & q & p \biconditional q & \eqref{ex21eq1} & \neg q & \eqref{ex21eq2} \\ 344 | \hline 345 | T & T & T & \textbf{F} & F & \textbf{F} \\ 346 | T & F & F & \textbf{T} & T & \textbf{T} \\ 347 | F & T & F & \textbf{T} & F & \textbf{T} \\ 348 | F & F & T & \textbf{F} & T & \textbf{F} 349 | \end{array} 350 | \] 351 | 352 | Since the truth values of \eqref{ex21eq1} and \eqref{ex21eq2} agree for all possible combinations of truth values for $p$ and $q$, \eqref{ex21eq1} and \eqref{ex21eq2} are logically equivalent. 353 | 354 | \pagebreak 355 | 356 | \section{Exercise 22} 357 | Let 358 | \begin{align} 359 | & p \conditional q \label{ex22eq1} \\ 360 | \neg & q \conditional \neg p \label{ex22eq2} 361 | \end{align} 362 | 363 | \[ 364 | \begin{array}{|c|c|c|c|c|c|} 365 | p & q & \eqref{ex22eq1} & \neg q & \neg p & \eqref{ex22eq2} \\ 366 | \hline 367 | T & T & \textbf{T} & F & F & \textbf{T} \\ 368 | T & F & \textbf{F} & T & F & \textbf{F} \\ 369 | F & T & \textbf{T} & F & T & \textbf{T} \\ 370 | F & F & \textbf{T} & T & T & \textbf{T} 371 | \end{array} 372 | \] 373 | 374 | Since the truth values of \eqref{ex22eq1} and \eqref{ex22eq2} agree for all possible combinations of truth values for $p$ and $q$, \eqref{ex22eq1} and \eqref{ex22eq2} are logically equivalent. 375 | 376 | \pagebreak 377 | \section{Exercise 23} 378 | Let 379 | \begin{align} 380 | \neg & p \biconditional q \label{ex23eq1} \\ 381 | & p \biconditional \neg q \label{ex23eq2} 382 | \end{align} 383 | 384 | \[ 385 | \begin{array}{|c|c|c|c|c|c|} 386 | p & q & \neg p & \eqref{ex23eq1} & \neg q & \eqref{ex23eq2} \\ 387 | \hline 388 | T & T & F & F & F & F \\ 389 | T & F & F & T & T & T \\ 390 | F & T & T & T & F & T \\ 391 | F & F & T & F & T & F 392 | \end{array} 393 | \] 394 | 395 | Since the truth values of \eqref{ex23eq1} and \eqref{ex23eq2} agree for all possible combinations of truth values for $p$ and $q$, \eqref{ex23eq1} and \eqref{ex23eq2} are logically equivalent. 396 | 397 | \pagebreak 398 | \section{Exercise 24} 399 | $\neg (p \oplus q)$ is true when $p \oplus q$ is false, which means that $p$ and $q$ share the same truth value. This is exactly when $p \biconditional q$ is true. Hence, $\neg (p \oplus q)$ and $p \biconditional q$ are logically equivalent. 400 | 401 | \pagebreak 402 | \section{Exercise 25} 403 | $\neg (p \biconditional q)$ is true when $p \biconditional q$ is false, which means that $p$ and $q$ have different truth values. This is exactly when $\neg p \biconditional q$ is true. Hence, $\neg (p \biconditional q)$ and $\neg p \biconditional q$ are logically equivalent. 404 | 405 | \pagebreak 406 | \section{Exercise 26} 407 | $(p \conditional q) \land (p \conditional r)$ is true when both $(p \conditional q)$ and $(p \conditional r)$ are true, which means either $p = F$ or both $q = T$ and $r = T$. This is exactly when $p \conditional (q \land r)$ is true. Hence, $(p \conditional q) \land (p \conditional r)$ and $p \conditional (q \land r)$ are logically equivalent. 408 | 409 | \pagebreak 410 | \section{Exercise 27} 411 | $(p \conditional r) \land (q \conditional r)$ is true when both $(p \conditional r)$ and $(q \conditional r)$ are true, which means either $r = T$ or both $p = F$ and $q = F$. This is exactly when $(p \lor q) \conditional r$ is true. Hence, $(p \conditional r) \land (q \conditional r)$ and $(p \lor q) \conditional r$ are logically equivalent. 412 | 413 | \pagebreak 414 | \section{Exercise 28} 415 | $(p \conditional q) \lor (p \conditional r)$ is true when either $(p \conditional q)$ or $(p \conditional r)$ is true, which means either $p = F$, $q = T$, or $r = T$. This is exactly when $p \conditional (q \lor r)$ is true. Hence, $(p \conditional q) \lor (p \conditional r)$ and $p \conditional (q \lor r)$ are logically equivalent. 416 | 417 | \pagebreak 418 | \section{Exercise 29} 419 | $(p \conditional r) \lor (q \conditional r)$ is true when either $(p \conditional r)$ or $(q \conditional r)$ is true, which means either $p = F$, $q = F$, or $r = T$. This is exactly when $(p \land q) \conditional r$ is true. Hence, $(p \conditional r) \lor (q \conditional r)$ and $(p \land q) \conditional r$ are logically equivalent. 420 | 421 | \pagebreak 422 | \section{Exercise 30} 423 | $\neg p \conditional (q \conditional r)$ is true when either $\neg p$ is false or $(q \conditional r)$ is true, which means that either $p = T$, $q = F$, or $r = T$. This is exactly when $q \conditional (p \lor r)$ is true. Hence, $\neg p \conditional (q \conditional r)$ and $q \implies (p \lor r)$ are logically equivalent. 424 | 425 | \pagebreak 426 | \section{Exercise 31} 427 | $p \biconditional q$ is true when $p$ and $q$ share the same truth value. This is exactly when $(p \conditional q) \land (q \conditional p)$ is true. Hence, $p \biconditional q$ and $(p \conditional q) \land (q \conditional p)$ are logically equivalent. 428 | 429 | \pagebreak 430 | \section{Exercise 32} 431 | $p \biconditional q$ is true when $p$ and $q$ share the same truth value. This is exactly when $\neg p \biconditional \neg q$ is true. Hence, $p \biconditional q$ and $\neg p \biconditional \neg q$ are logically equivalent. 432 | 433 | \pagebreak 434 | \section{Exercise 34} 435 | \begin{align*} 436 | ( & p \lor q) \land (\neg p \lor r) \conditional (q \lor r) & \equiv \\ 437 | \neg [( & p \lor q) \land (\neg p \lor r)] \lor (q \lor r) & \equiv \\ 438 | [\neg ( & p \lor q) \lor \neg (\neg p \lor r)] \lor (q \lor r) & \equiv \\ 439 | [(\neg & p \land \neg q) \lor (p \land \neg r)] \lor (q \lor r) & \equiv \\ 440 | (\neg & p \land \neg q) \lor [(q \lor r) \lor (p \land \neg r)] & \equiv \\ 441 | (\neg & p \land \neg q) \lor [((q \lor r) \lor p) \land ((q \lor r) \lor \neg r)] & \equiv \\ 442 | (\neg & p \land \neg q) \lor [((q \lor r) \lor p) \land T] & \equiv \\ 443 | (( & q \lor r) \lor p) \lor (\neg p \land \neg q) & \equiv \\ 444 | ((( & q \lor r) \lor p) \lor \neg p) \land (((q \lor r) \lor p) \lor \neg q) & \equiv \\ 445 | & T \land T & \equiv \\ 446 | & T 447 | \end{align*} 448 | 449 | \end{document} -------------------------------------------------------------------------------- /LICENSE_CC: -------------------------------------------------------------------------------- 1 | Attribution-NonCommercial-ShareAlike 4.0 International 2 | 3 | ======================================================================= 4 | 5 | Creative Commons Corporation ("Creative Commons") is not a law firm and 6 | does not provide legal services or legal advice. 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For 434 | the avoidance of doubt, this paragraph does not form part of the 435 | public licenses. 436 | 437 | Creative Commons may be contacted at creativecommons.org. 438 | -------------------------------------------------------------------------------- /book/exercises/ch1/ch01_2-sol.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \usepackage[utf8]{inputenc} 3 | \usepackage[T1]{fontenc} 4 | \usepackage{indentfirst, hyperref, gensymb, amsmath, amsthm, wasysym, amsfonts, mathtools, braket, amssymb} 5 | \hypersetup{colorlinks,allcolors=blue,linktoc=all} 6 | 7 | \title{Chapter 1 Section 2 Exercise Solutions} 8 | \author{Samuel Lair} 9 | \date{September 2022} 10 | 11 | \begin{document} 12 | 13 | \maketitle 14 | \tableofcontents 15 | 16 | \pagebreak 17 | 18 | \section{Exercise 1} 19 | \[ 20 | \neg a \implies \neg e 21 | \] 22 | 23 | \pagebreak 24 | 25 | \section{Exercise 2} 26 | \[ 27 | m \implies e \lor p 28 | \] 29 | 30 | \pagebreak 31 | 32 | \section{Exercise 3} 33 | \[ 34 | g \implies r \land \neg m \land \neg b 35 | \] 36 | 37 | \pagebreak 38 | 39 | \section{Exercise 4} 40 | \[ 41 | \neg s \implies (d \implies w) 42 | \] 43 | 44 | \pagebreak 45 | 46 | \section{Exercise 5} 47 | \[ 48 | e \implies a \land (b \lor p) \land r 49 | \] 50 | 51 | \pagebreak 52 | 53 | \section{Exercise 6} 54 | \[ 55 | u \implies (b_{32} \land g_1 \land r_1 \land h_{16}) \lor (b_{64} \land g_2 \land r_2 \land h_{32}) 56 | \] 57 | 58 | \pagebreak 59 | 60 | \section{Exercise 7} 61 | \subsection{(a)} 62 | \[ 63 | q \implies p 64 | \] 65 | \subsection{(b)} 66 | \[ 67 | q \land \neg p 68 | \] 69 | \subsection{(c)} 70 | \[ 71 | q \implies p 72 | \] 73 | \subsection{(d)} 74 | \[ 75 | \neg q \implies \neg p 76 | \] 77 | 78 | \pagebreak 79 | 80 | \section{Exercise 8} 81 | 82 | \subsection{(a)} 83 | \[ 84 | r \land \neg p 85 | \] 86 | 87 | \subsection{(b)} 88 | \[ 89 | r \land p \implies q 90 | \] 91 | 92 | \subsection{(c)} 93 | \[ 94 | \neg r \implies \neg q 95 | \] 96 | \subsection{(d)} 97 | \[ 98 | \neg p \land r \implies q 99 | \] 100 | 101 | \pagebreak 102 | 103 | \section{Exercise 9} 104 | Let 105 | \begin{align*} 106 | p & \Coloneqq \text{"The system is in multiuser state."} \\ 107 | q & \Coloneqq \text{"The system is operating normally."} \\ 108 | r & \Coloneqq \text{"The kernel is functioning."} \\ 109 | s & \Coloneqq \text{"The system is in interrupt mode"} 110 | \end{align*} 111 | Then our system specifications can be expressed as the following system of logical expressions: 112 | \begin{align} 113 | & p \iff q \label{ex9eq1} \\ 114 | & q \implies r \label{ex9eq2} \\ 115 | & \neg r \lor s \label{ex9eq3} \\ 116 | & \neg p \implies s \label{ex9eq4} \\ 117 | & \neg s \label{ex9eq5} 118 | \end{align} 119 | In order for \eqref{ex9eq5} to be true, $s$ must be false. Since $s$ is false, $p$ must be true in order for \eqref{ex9eq4} to be true. Since $p$ is true, $q$ must be true in order for \eqref{ex9eq1} to be true. Since $q$ is true, $r$ must be true in order in order for \eqref{ex9eq2} to be true. However, we must conclude that \eqref{ex9eq3} is false since $r$ is true and $s$ is false. 120 | 121 | Therefore, there is no assignment of truth values such that all of our logical expressions are true. Hence, our system specifications are inconsistent. 122 | 123 | \pagebreak 124 | 125 | \section{Exercise 10} 126 | Let 127 | \begin{align*} 128 | p & \Coloneqq \text{"The system software is being upgraded."} \\ 129 | q & \Coloneqq \text{"Users can access the file system."} \\ 130 | r & \Coloneqq \text{"Users can save new files"} 131 | \end{align*} 132 | Then our system specifications can be expressed as the following system of logical expressions: 133 | \begin{align} 134 | p & \implies \neg q \label{ex10eq1} \\ 135 | q & \implies r \label{ex10eq2} \\ 136 | \neg r & \implies \neg p \label{ex10eq3} 137 | \end{align} 138 | All of our logical expressions are true if we take $p = true$, $q = false$ and $r = true$. Hence, our system specifications are consistent. 139 | 140 | \pagebreak 141 | 142 | \section{Exercise 11} 143 | Let 144 | \begin{align*} 145 | p & \Coloneqq \text{"The router can send packets to the edge system."} \\ 146 | q & \Coloneqq \text{"The router supports the new address space."} \\ 147 | r & \Coloneqq \text{"The latest software release is installed."} 148 | \end{align*} 149 | Then our system specifications can be expressed as the following system of logical expressions: 150 | \begin{align} 151 | & p \implies q \label{ex11eq1} \\ 152 | & q \implies r \label{ex11eq2} \\ 153 | & r \implies p \label{ex11eq3} \\ 154 | & \neg q \label{ex11eq4} 155 | \end{align} 156 | All of our logical expressions are true if we take $p = false$, $q = false$, and $r = false$. Hence, our system specifications are consistent. 157 | 158 | \pagebreak 159 | 160 | \section{Exercise 12} 161 | Let 162 | \begin{align*} 163 | p & \Coloneqq \text{"The file system is locked."} \\ 164 | q & \Coloneqq \text{"New messages will be queued."} \\ 165 | r & \Coloneqq \text{"The system is functioning normally."} \\ 166 | s & \Coloneqq \text{"New messages will be sent to the message buffer."} 167 | \end{align*} 168 | Then our system specifications can be expressed as the following system of logical expressions: 169 | \begin{align} 170 | & \neg p \implies q \label{ex12eq1} \\ 171 | & \neg p \iff r \label{ex12eq2} \\ 172 | & \neg q \implies s \label{ex12eq3} \\ 173 | & \neg p \implies s \label{ex12eq4} \\ 174 | & \neg s \label{ex12eq5} 175 | \end{align} 176 | In order for \eqref{ex12eq5} to be true, s must be false. Since s is false, p must be true in order for \eqref{ex12eq4} to be true. Since s is false, q must be true in order for \eqref{ex12eq3} to be true. However, since p is false and q is false, we must conclude that \eqref{ex12eq1} is false. 177 | 178 | All of our logical expressions are true if we take $p = true$, $q = true$, $r = false$, and $s = false$. Hence, our system specifications are consistent. 179 | 180 | \pagebreak 181 | 182 | \section{Exercise 13} 183 | \subsection{(a)} 184 | beaches AND New AND Jersey 185 | \subsection{(b)} 186 | (beaches AND Jersey) NOT New 187 | 188 | \pagebreak 189 | 190 | \section{Exercise 14} 191 | \subsection{(a)} 192 | hiking AND West AND Virginia 193 | \subsection{(b)} 194 | (hiking AND Virginia) NOT West 195 | 196 | \pagebreak 197 | 198 | \section{Exercise 15} 199 | Ethiopian AND restaurant AND New AND (York OR Jersey) 200 | 201 | \pagebreak 202 | 203 | \section{Exercise 16} 204 | (men AND (shoes or boots)) NOT work 205 | 206 | \pagebreak 207 | 208 | \section{Exercise 17} 209 | \subsection{(a)} 210 | The statement that "All of the inscriptions are false" is equivalent to the propositional expression: 211 | \[ 212 | \neg p_3 \land \neg p_1 \land \neg (\neg p_3) \equiv \neg p_1 \land \neg p_3 \land p_3 \equiv \neg p_1 \land F \equiv F 213 | \] 214 | Therefore, the Queen who never lies cannot make this statement. 215 | 216 | \subsection{(b)} 217 | The statement that "Exactly one of the inscriptions is true" is equivalent to the propositional expression: 218 | \begin{align*} 219 | & (p_3 \land \neg p_1 \land \neg (\neg p_3)) \lor (\neg p_3 \land p_1 \land \neg (\neg p_3)) \lor (\neg p_3 \land \neg p_1 \land \neg p_3) & \equiv \\ 220 | & (p_3 \land \neg p_1) \lor (\neg p_3 \land \neg p_1) 221 | \end{align*} 222 | The Queen who never lies could make this statement if the treasure is in either Trunk 3 or Trunk 2. 223 | 224 | \subsection{(c)} 225 | The statement that "Exactly two of the inscriptions are true" is equivalent to the propositional expression: 226 | \begin{align*} 227 | & (p_3 \land p_1 \land \neg (\neg p_3)) \lor (p_3 \land \neg p_1 \land \neg p_3) \lor (\neg p_3 \land p_1 \land \neg p3) & \equiv \\ 228 | & (p_3 \land p_1) \lor (\neg p_3 \land p_1) 229 | \end{align*} 230 | The Queen who never lies could make this statement. If we assume that the treasure cannot be in multiple trunks, we can conclude that the treasure is in Trunk 1. 231 | 232 | \subsection{(d)} 233 | The statement that "All three inscriptions are true" is equivalent to the propositional expression: 234 | \begin{align*} 235 | & p_3 \land p_1 \land \neg p_3 \equiv p_1 \land p_3 \land \neg p_3 \equiv p_1 \land F \equiv F 236 | \end{align*} 237 | The Queen who never lies cannot make this statement. 238 | 239 | \pagebreak 240 | 241 | \section{Exercise 18} 242 | \subsection{(a)} 243 | The statement that "All of the inscriptions are false" is equivalent to the propositional expression: 244 | 245 | \[ 246 | \neg (\neg p_1) \land \neg p_1 \land \neg p_2 \equiv p_1 \land \neg p_1 \land \neg p_2 \equiv F \land \neg p_2 \equiv F 247 | \] 248 | Therefore, the Queen who never lies cannot make this statement. 249 | 250 | \subsection{(b)} 251 | The statement that "Exactly one of the inscriptions is true" is equivalent to the propositional expression: 252 | \begin{align*} 253 | & (\neg p_1 \land \neg p_1 \land \neg p_2) \lor (\neg (\neg p_1) \land p_1 \land \neg p_2) \lor (\neg (\neg p_1) \land \neg p_1 \land p_2) & \equiv \\ 254 | & (\neg p_1 \land p_2) \lor (p_1 \land \neg p_2) 255 | \end{align*} 256 | The Queen who never lies could make this statement if the treasure is in either Trunk 2 or Trunk 1. 257 | 258 | \subsection{(c)} 259 | The statement that "Exactly two of the inscriptions are true" is equivalent to the propositional expression: 260 | \begin{align*} 261 | & (\neg p_1 & p_1 & \neg p_2) \lor (\neg p1 \land \neg p_1 \land p_2) \lor (\neg (\neg p1) \land p_1 \land p_2) & \equiv \\ 262 | & (\neg p_1 \land p_2) \lor (p_1 \land p_2) 263 | \end{align*} 264 | The Queen who never lies could make this statement. If we assume that the treasure cannot be in multiple trunks, we can conclude that the treasure is in Trunk 2. 265 | 266 | \subsection{(d)} 267 | The statement that "All three inscriptions are true" is equivalent to the propositional expression: 268 | \[ 269 | \neg p_1 \land p_1 \land p_2 \equiv F \land p_2 \equiv F 270 | \] 271 | Therefore, the Queen who never lies cannot make this statement. 272 | 273 | \pagebreak 274 | 275 | \section{Exercise 19} 276 | Let 277 | \begin{align*} 278 | p & \Coloneqq \text{"The left fork leads to the ruins."} \\ 279 | q & \Coloneqq \text{"The villager always tells the truth."} 280 | \end{align*} 281 | 282 | Suppose we ask a straightfoward question like: 283 | \begin{equation}\label{ex19eq1} 284 | \text{"Does the left fork lead to the ruins?"} 285 | \end{equation} 286 | 287 | This question is equivalent to the following propositional expression: 288 | \begin{equation}\label{ex19eq1p} 289 | (p \land q) \lor (\neg p \land \neg q) 290 | \end{equation} 291 | 292 | In other words, the truth table for \eqref{ex19eq1} is as follows: 293 | \[ 294 | \begin{array}{|c|c|c|} 295 | p & q & \eqref{ex19eq1p} \\ 296 | \hline 297 | T & T & T \\ 298 | T & F & F \\ 299 | F & T & F \\ 300 | F & F & T 301 | \end{array} 302 | \] 303 | 304 | This is problematic since we want the truth value of our question to match the truth value of p regardless of the truth value of q. We can achieve this by making the liar lie about his answer to \eqref{ex19eq1}. Due to the double negation rule, a lie about a lie will be equal to the truth. Therefore, the question we will actually ask is: 305 | 306 | \begin{equation}\label{ex19eq2} 307 | \text{"If I were to ask you about whether the left fork leads to the ruins, would you say yes?"} 308 | \end{equation} 309 | 310 | The truth table for \eqref{ex19eq2} is as follows: 311 | \[ 312 | \begin{array}{|c|c|r c|} 313 | p & q & & \eqref{ex19eq2} \\ 314 | \hline 315 | T & T & \eqref{ex19eq1} \equiv & T \\ 316 | T & F & \neg \eqref{ex19eq1} \equiv & T \\ 317 | F & T & \eqref{ex19eq1} \equiv & F \\ 318 | F & F & \neg \eqref{ex19eq1} \equiv & F 319 | \end{array} 320 | \] 321 | 322 | The truth value of \eqref{ex19eq2} matches $p$ as desired, allowing us to infer which fork leads to the ruins. 323 | 324 | \pagebreak 325 | 326 | \section{Exercise 20} 327 | \subsection{(a)} 328 | The question "Are you a liar?" does not work because both types of cannibals will answer no. The cannibals who always tell the truth will answer no because that is the truth. The cannibals who always lie will answer no because that is the lie. 329 | 330 | \subsection{(b)} 331 | Similar to Exercise 19, the explorer needs to make the liars lie about a lie. Therefore, the explorer should ask: "If I were to ask you whether you are a liar, would you say yes?" 332 | 333 | \pagebreak 334 | 335 | \section{Exercise 21} 336 | From the first professor's response, we can infer that he wants coffee but does not know whether the other two professors want coffee. From the second professor's response, we can infer that he wants coffee but does not know whether the third professor wants coffee. From the third professor's response, we can infer that he does not want coffee. 337 | 338 | Therefore, the hostess gives coffee to the first and second professors. 339 | 340 | \pagebreak 341 | 342 | \section{Exercise 22} 343 | Let 344 | \begin{align*} 345 | j & \Coloneqq \text{"Jasmine attends."} \\ 346 | s & \Coloneqq \text{"Samir attends."} \\ 347 | k & \Coloneqq \text{"Kanti attends."} 348 | \end{align*} 349 | 350 | In order for noone to be unhappy, we must satisfy the following system specifications: 351 | \begin{align} 352 | j & \implies \neg s \label{ex22eq1} \\ 353 | s & \implies k \label{ex22eq2} \\ 354 | \neg j & \implies \neg k \label{ex22eq3} 355 | \end{align} 356 | 357 | Suppose that Jasmine attends. From \eqref{ex22eq1}, Samir cannot also attend. Our specifications are satisfied regardless of whether Kanti attends. 358 | 359 | Now suppose that Jasmine doesn't attend. From \eqref{ex22eq3}, Kanti also doesn't attend. Finally, from \eqref{ex22eq2}, Samir also doesn't attend. 360 | 361 | Hence, we can invite either Jasmine alone, both Jasmine and Kanti, or none of the three. 362 | 363 | \pagebreak 364 | 365 | \section{Exercise 23} 366 | Let 367 | \begin{align*} 368 | & p \Coloneqq \text{"A is a knight."} \\ 369 | & q \Coloneqq \text{"B is a knight."} 370 | \end{align*} 371 | Then A's statement is equivalent to the propositional expression: 372 | \[ 373 | R \Coloneqq \neg p \lor \neg q 374 | \] 375 | 376 | First, let us consider the case where A is a knave. Then $\neg p$ is true and A's statement is false. We could informally reason that the negation of A's statement is "Neither of us is knave" or $p \land q$. However, let us try a more formal approach by applying De Morgan's Law for OR: 377 | 378 | \begin{align*} 379 | & \neg R & \equiv \\ 380 | & \neg (\neg p \lor \neg q) & \equiv \\ 381 | & p \land q 382 | \end{align*} 383 | Therefore, we've contradicted our assumption that $\neg p$ is true. 384 | 385 | Next, let us consider the case where A is a knight. Then $p$ is true and A's statement is true. In order for $R$ to evaluate to true, $\neg q$ must be true. I.e. B is a knave. 386 | 387 | Hence, we conclude that A is a knight and B is a knave. 388 | 389 | \pagebreak 390 | 391 | \section{Exercise 24} 392 | The case where A is a knight is equivalent to the following system specifications: 393 | \begin{align*} 394 | & p \\ 395 | & p \land q \\ 396 | & (\neg p \land q) \lor (p \land \neg q) 397 | \end{align*} 398 | These specifications are inconsistent. In order for the first two expressions to be true, $p$ must be true and $q$ must be true. However, these truth value assignments result in the third expression evaluating to false. 399 | 400 | The case where A is a knave is equivalent to the following system specifications: 401 | \begin{align*} 402 | & \neg p \\ 403 | & \neg (p \land q) \equiv \neg p \lor \neg q \\ 404 | & (\neg p \land q) \lor (p \land \neg q) 405 | \end{align*} 406 | These specifications are all satisfied for $p = F$ and $q = T$. 407 | 408 | Hence, we conclude that A is a knave and B is a knight. 409 | 410 | \pagebreak 411 | 412 | \section{Exercise 25} 413 | The case where A is a knight is equivalent to the following system specifications: 414 | \begin{align*} 415 | & p \\ 416 | & \neg p \lor q 417 | \end{align*} 418 | These specifications are all satisfied for $p = T$ and $q = T$. 419 | 420 | The case where A is a knave is equivalent to the following system specifications: 421 | \begin{align*} 422 | & \neg p \\ 423 | & \neg (\neg p \lor q) \equiv p \land \neg q 424 | \end{align*} 425 | These specifications are inconsistent. In order for the first expression to be true, $p$ must be false. However, this truth value assignment results in the second expression evaluating to false regardless of the value for $q$. 426 | 427 | Hence, we conclude that A is a knight and B is a knight. 428 | 429 | \pagebreak 430 | 431 | \section{Exercise 26} 432 | Let $R$ be the logical expression equivalent to A's statement: 433 | \[ 434 | R \Coloneqq p 435 | \] 436 | 437 | Let $S$ be the logical expression equivalent to B's statement: 438 | \[ 439 | S \Coloneqq q 440 | \] 441 | 442 | This scenario is equivalent to the system specifications: 443 | \begin{align*} 444 | & (R \land p) \lor (\neg R \land \neg p) \equiv p \lor \neg p \equiv T \\ 445 | & (S \land q) \lor (\neg S \land \neg q) \equiv q \lor \neg q \equiv T 446 | \end{align*} 447 | 448 | Any truth value assignment will satisfy all of the specifications. 449 | 450 | Hence, we conclude that we cannot draw any conclusions in this scenario. $A$ can be either a knight or a knave independent of the status of $B$. $B$ can be either a knight or a knave independent of the status of $A$. 451 | 452 | \pagebreak 453 | 454 | \section{Exercise 27} 455 | Let $R$ be the logical expression equivalent to A's statement: 456 | \[ 457 | \neg p \land \neg q 458 | \] 459 | 460 | This scenario is equivalent to the logical expression: 461 | \begin{align*} 462 | & (R \land p) \lor (\neg R \land \neg p) & \equiv \\ 463 | & (\neg p \land \neg q \land p) \lor (\neg (\neg p \land \neg q) \land \neg p) & \equiv \\ 464 | & (p \land \neg p \land \neg q) \lor ((p \lor q) \land \neg p) & \equiv \\ 465 | & (F \land \neg q) \lor ((\neg p \land p) \lor (\neg p \land q)) & \equiv \\ 466 | & F \lor (\neg p \land q) & \equiv \\ 467 | & \neg p \land q 468 | \end{align*} 469 | 470 | Hence, we conclude that $A$ is a knave and $B$ is a knight. 471 | 472 | \pagebreak 473 | 474 | \section{Exercise 28} 475 | Brute forcing knight, knave, and spy problems using logical formalism can get quite complicated as we need at least 4 propositional variables to uniquely express who is the knight, who is the knave, and who is the spy. Therefore, for exercises 28-35, we'll try to simplify the problem by applying the constraints before possibly resorting to logical expressions. 476 | 477 | $B$ cannot be the knight since then we would have two knights. If $C$ is the knight, then $C$'s statement would result in a contradiction. Therefore, $C$ isn't the knight either. $A$ must the knight. 478 | 479 | Hence, we conclude that $A$ is the knight, $B$ is the spy, and $C$ is the knave. 480 | 481 | \pagebreak 482 | 483 | \section{Exercise 29} 484 | $C$ cannot be the knight since then we would have two knights. If $B$ is the knight, then $B$'s statement would result in a contradiction. Therefore, $B$ isn't the knight either. $A$ must be the knight. 485 | Now suppose that $B$ is the knave. Then $B$'s statement would be false, resulting in a contradiction. Therefore, $B$ must be the spy. 486 | 487 | Hence, we conclude that $A$ is the knight, $B$ is the spy, and $C$ is the knave. 488 | 489 | \pagebreak 490 | 491 | \section{Exercise 30} 492 | This scenario is a paradox since it requires the knight to lie. I.e. there is no solution. 493 | 494 | \pagebreak 495 | 496 | \section{Exercise 31} 497 | If either $B$ or $C$ is the knight, then their respective statements results in a contradiction. $A$ must be the knight. Since this means that $B$'s statement is also true, $B$ must be the spy. 498 | 499 | Hence, we conclude that $A$ is the knight, $B$ is the spy, and $C$ is the knave. 500 | 501 | \pagebreak 502 | 503 | \section{Exercise 32} 504 | Suppose that $A$ is the knight. Then $B$'s statement is also true so $B$ is the spy. However, this also makes $C$'s statement true and we reach a contradiction since there is no knave. Therefore, $A$ isn't the knight. 505 | 506 | Suppose that $A$ is spy. Then B's statement is true so $B$ is the knight. However, this also makes $C$'s statement true and we reach a contradiction since there is no knave. Therefore, $A$ isn't the spy. 507 | 508 | Suppose that $A$ is the knave. Then B's statement is false so $B$ is the spy. $C$'s statement is true so $C$ is the knight. 509 | 510 | Hence, we conclude that $A$ is the knave, $B$ is the spy, and $C$ is the knight. 511 | 512 | \pagebreak 513 | 514 | \section{Exercise 33} 515 | These statements don't allow us to eliminate any solutions. I.e. all 6 permutations of knight, knave, and spy are possible in this scenario. 516 | 517 | \pagebreak 518 | 519 | \section{Exercise 34} 520 | Suppose that $C$ is the knight. Then $A$ is the spy and $B$ is the knave. 521 | 522 | Suppose that $C$ is the knave. Then C's statement is true, resulting in a contradiction. 523 | 524 | Suppose that $C$ is the spy. Then either $A$ or $B$ must be the knave. This results in a contradiction since the knave's statement is true. 525 | 526 | Hence, we conclude that $A$ is the spy, $B$ is the knave, and $C$ is the knight. 527 | 528 | \pagebreak 529 | 530 | \section{Exercise 35} 531 | This scenario is a paradox since it requires the knave to tell the truth. I.e. there is no solution. 532 | 533 | \pagebreak 534 | 535 | \section{Exercise 36} 536 | 537 | \subsection{(a)} 538 | 539 | If we assume that Jones is innocent, then his statement is true. Jones's statement implies that Smith's statement is false. Therefore, Smith is guilty. We can assume that Williams's statement is true without contradicting Jones's statement or the conclusion that Smith is guilty. In this case, we conclude that Smith is guilty and Jones and Williams are innocent. 540 | 541 | If we assume that Jones is guilty, then both Smith's and Williams's statements can be true without contradicting each other. In this case, we conclude that Jones is guilty and Smith and Williams are innocent. 542 | 543 | Hence, we cannot determine who the murderer is. We need additional information. For example, if one cannot both be with Cooper and out of town on the day of the killing, then we can conclude that Jones is the murderer since his statement would now contradicts those of both Smith and Williams. 544 | 545 | (In all honesty, this a probably a sloppily worded question where the author intended for Jones's statement to contradict Williams's statement, leading to the conclusion that Jones is guilty.) 546 | 547 | \subsection{(b)} 548 | We have even less information than in part (a). We cannot determine who the murderer is. Even if one cannot both be with Cooper and out of town on the day of the killing, we still cannot determine who the murderer(s) is/are since there could be multiple murderers. 549 | 550 | \pagebreak 551 | 552 | \section{Exercise 37} 553 | Let 554 | \begin{align*} 555 | f & \Coloneqq \text{"Fred is the highest paid."} \\ 556 | j & \Coloneqq \text{"Janice is the highest paid."} \\ 557 | k & \Coloneqq \text{"Janice is the lowest paid."} \\ 558 | m & \Coloneqq \text{"Maggie is the highest paid."} 559 | \end{align*} 560 | 561 | Then the known facts are equivalent to the following system specifications: 562 | \begin{align} 563 | j & \implies \neg k \label{ex37eq1} \\ 564 | k & \implies \neg j \label{ex37eq2} \\ 565 | f & \implies \neg j \label{ex37eq3} \\ 566 | f & \implies \neg m \label{ex37eq4} \\ 567 | j & \implies \neg f \label{ex37eq5} \\ 568 | j & \implies \neg m \label{ex37eq6} \\ 569 | m & \implies \neg f \label{ex37eq7} \\ 570 | m & \implies \neg j \label{ex37eq8} \\ 571 | \neg f & \implies j \label{ex37eq9} \\ 572 | \neg k & \implies m \label{ex37eq10} 573 | \end{align} 574 | 575 | Suppose that $f = T$. From \eqref{ex37eq3}, $j = F$. From \eqref{ex37eq4}, $m = F$. From \eqref{ex37eq10}, $k = T$. Therefore, Fred > Maggie > Janice in this case. 576 | 577 | Suppose that $f = F$. From \eqref{ex37eq9}, $j = T$. From \eqref{ex37eq1}, $k = F$. From \eqref{ex37eq10}, $m = T$. However, this contradicts both \eqref{ex37eq6} and \eqref{ex37eq8}. I.e. Janice and Maggie cannot both be the highest paid. 578 | 579 | Hence, we conclude that Fred is the highest paid, Janice is the lowest paid, and Maggie's pay is between that of Fred and Janice. 580 | 581 | \pagebreak 582 | 583 | \section{Exercise 38} 584 | Let 585 | \begin{align*} 586 | k & \Coloneqq \text{"Kevin is chatting."} \\ 587 | h & \Coloneqq \text{"Heather is chatting."} \\ 588 | r & \Coloneqq \text{"Randy is chatting."} \\ 589 | v & \Coloneqq \text{"Vijay is chatting."} \\ 590 | a & \Coloneqq \text{"Abby is chatting."} 591 | \end{align*} 592 | 593 | Our scenario is equivalent to the following system specifications: 594 | 595 | \begin{align} 596 | & k \lor h \label{ex38eq1} \\ 597 | & r \oplus v \label{ex38eq2} \\ 598 | & a \implies r \label{ex38eq3} \\ 599 | & v \iff k \label{ex38eq4} \\ 600 | & h \implies a \land k \label{ex38eq5} 601 | \end{align} 602 | 603 | Suppose that $v = T$. From \eqref{ex38eq4}, $k = T$. From \eqref{ex38eq2}, $r = F$. From \eqref{ex38eq3}, $a = F$. From \eqref{ex38eq5}, $h = F$. $\eqref{ex38eq1} = T$ since $k = T$. 604 | 605 | Suppose that $v = F$. From \eqref{ex38eq4}, $k = F$. From \eqref{ex38eq5}, $h = F$. However, this a contradiction since $\eqref{ex38eq1} = F$. 606 | 607 | Hence, we conclude that Kevin and Vijay are chatting while Heather, Randy and Abby are not chatting. 608 | 609 | \pagebreak 610 | 611 | \section{Exercise 39} 612 | Let 613 | \begin{align*} 614 | b & \Coloneqq \text{"The butler is telling the truth."} \\ 615 | c & \Coloneqq \text{"The cook is telling the truth."} \\ 616 | g & \Coloneqq \text{"The gardener is telling the truth."} \\ 617 | h & \Coloneqq \text{"The handyman is telling the truth."} 618 | \end{align*} 619 | 620 | Our facts are equivalent to the following system specifications: 621 | \begin{align} 622 | b & \implies c \label{ex39eq1} \\ 623 | c & \implies \neg g \label{ex39eq2} \\ 624 | g & \implies \neg c \label{ex39eq3} \\ 625 | \neg g & \implies h \label{ex39eq4} \\ 626 | \neg h & \implies g \label{ex39eq5} \\ 627 | h & \implies \neg c \label{ex39eq6} 628 | \end{align} 629 | 630 | Suppose that $b = T$. From \eqref{ex39eq1}, $c = T$. From \eqref{ex39eq2}, $g = F$. From \eqref{ex39eq4}, $h = T$. However, \eqref{ex39eq6} leads us to a contradiction where $c = F$. 631 | 632 | Suppose that $b = F$. We must also take $c = F$ to avoid the previous contradiction. If we take $g = T$, $h$ can be either true or false without reaching a contradiction. If we take $g = F$, then \eqref{ex39eq4} requires that $h = T$. This doesn't lead to any contradictions either. 633 | 634 | Hence, we can conclude that the butler and cook are lying. However, we cannot determine whether the gardener is telling the truth or whether the handyman is telling the truth. 635 | 636 | \pagebreak 637 | 638 | \section{Exercise 40} 639 | 640 | \subsection{(a)} 641 | 642 | Suppose Alice is telling the truth. Then we reach a contradiction where John is also telling the truth. 643 | 644 | Suppose that John is telling the truth. Then we reach a contradiction where either Carlos or Diana is also telling the truth. Since we can conclude that John's statement is false, we now know that he did it. However, let's also see if we can figure out who is telling the truth. 645 | 646 | Suppose that Carlos is telling the truth. This contradicts the fact that John's statement is false. 647 | 648 | Suppose that Diana is telling the truth. There is no contradiction in this case. 649 | 650 | Therefore, Diana is telling the truth while the other three are lying. Hence, we conclude that John accessed the computer system without authorization since his statement is false. 651 | 652 | \subsection{(b)} 653 | Alice's and Carlos's statements cannot both be true since they implicate different people. Also, if Diana's statement is true, then Carlos's statement is false. 654 | 655 | Suppose that Carlos's statement is true. Then we reach a contradiction where both Alice's and Diana's statements are false. Therefore, Carlos's statement is false while the other three are true. 656 | 657 | Hence, we conclude that Carlos accessed the computer system without authorization since Alice's statement is true. 658 | 659 | \pagebreak 660 | 661 | \section{Exercise 41} 662 | If the sign on the first door is true, then the sign on the second door is true as well. Since we know that exactly one sign is true, the sign on the first door must be false and the sign on the second door must be true. 663 | 664 | Hence, the lady is in the second room and the tiger is in the first room. 665 | 666 | \pagebreak 667 | 668 | \section{Exercise 43} 669 | If more than one senator is honest, then there exists a pair of senators such that both are honest (i.e. neither is corrupt). This contradicts the requirement that "given any two Freedonian senators, at least one is corrupt." 670 | 671 | Hence we conclude that one senator is honest and the other 49 are corrupt. 672 | 673 | \pagebreak 674 | 675 | \section{Exercise 44} 676 | \subsection{(a)} 677 | \[ 678 | \neg p \lor \neg q 679 | \] 680 | \subsection{(b)} 681 | \[ 682 | \neg (p \lor (\neg p \land q)) 683 | \] 684 | 685 | \pagebreak 686 | 687 | \section{Exercise 45} 688 | \subsection{(a)} 689 | \[ 690 | \neg (p \land (q \lor \neg r)) 691 | \] 692 | 693 | \subsection{(b)} 694 | \[ 695 | (\neg p \land \neg q) \lor (p \land r) 696 | \] 697 | 698 | \end{document} 699 | 700 | -------------------------------------------------------------------------------- /LICENSE_GNU: -------------------------------------------------------------------------------- 1 | GNU GENERAL PUBLIC LICENSE 2 | Version 3, 29 June 2007 3 | 4 | Copyright (C) 2007 Free Software Foundation, Inc. 5 | Everyone is permitted to copy and distribute verbatim copies 6 | of this license document, but changing it is not allowed. 7 | 8 | Preamble 9 | 10 | The GNU General Public License is a free, copyleft license for 11 | software and other kinds of works. 12 | 13 | The licenses for most software and other practical works are designed 14 | to take away your freedom to share and change the works. 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You may not convey a covered 525 | work if you are a party to an arrangement with a third party that is 526 | in the business of distributing software, under which you make payment 527 | to the third party based on the extent of your activity of conveying 528 | the work, and under which the third party grants, to any of the 529 | parties who would receive the covered work from you, a discriminatory 530 | patent license (a) in connection with copies of the covered work 531 | conveyed by you (or copies made from those copies), or (b) primarily 532 | for and in connection with specific products or compilations that 533 | contain the covered work, unless you entered into that arrangement, 534 | or that patent license was granted, prior to 28 March 2007. 535 | 536 | Nothing in this License shall be construed as excluding or limiting 537 | any implied license or other defenses to infringement that may 538 | otherwise be available to you under applicable patent law. 539 | 540 | 12. No Surrender of Others' Freedom. 541 | 542 | If conditions are imposed on you (whether by court order, agreement or 543 | otherwise) that contradict the conditions of this License, they do not 544 | excuse you from the conditions of this License. If you cannot convey a 545 | covered work so as to satisfy simultaneously your obligations under this 546 | License and any other pertinent obligations, then as a consequence you may 547 | not convey it at all. For example, if you agree to terms that obligate you 548 | to collect a royalty for further conveying from those to whom you convey 549 | the Program, the only way you could satisfy both those terms and this 550 | License would be to refrain entirely from conveying the Program. 551 | 552 | 13. Use with the GNU Affero General Public License. 553 | 554 | Notwithstanding any other provision of this License, you have 555 | permission to link or combine any covered work with a work licensed 556 | under version 3 of the GNU Affero General Public License into a single 557 | combined work, and to convey the resulting work. The terms of this 558 | License will continue to apply to the part which is the covered work, 559 | but the special requirements of the GNU Affero General Public License, 560 | section 13, concerning interaction through a network will apply to the 561 | combination as such. 562 | 563 | 14. Revised Versions of this License. 564 | 565 | The Free Software Foundation may publish revised and/or new versions of 566 | the GNU General Public License from time to time. Such new versions will 567 | be similar in spirit to the present version, but may differ in detail to 568 | address new problems or concerns. 569 | 570 | Each version is given a distinguishing version number. If the 571 | Program specifies that a certain numbered version of the GNU General 572 | Public License "or any later version" applies to it, you have the 573 | option of following the terms and conditions either of that numbered 574 | version or of any later version published by the Free Software 575 | Foundation. If the Program does not specify a version number of the 576 | GNU General Public License, you may choose any version ever published 577 | by the Free Software Foundation. 578 | 579 | If the Program specifies that a proxy can decide which future 580 | versions of the GNU General Public License can be used, that proxy's 581 | public statement of acceptance of a version permanently authorizes you 582 | to choose that version for the Program. 583 | 584 | Later license versions may give you additional or different 585 | permissions. However, no additional obligations are imposed on any 586 | author or copyright holder as a result of your choosing to follow a 587 | later version. 588 | 589 | 15. Disclaimer of Warranty. 590 | 591 | THERE IS NO WARRANTY FOR THE PROGRAM, TO THE EXTENT PERMITTED BY 592 | APPLICABLE LAW. EXCEPT WHEN OTHERWISE STATED IN WRITING THE COPYRIGHT 593 | HOLDERS AND/OR OTHER PARTIES PROVIDE THE PROGRAM "AS IS" WITHOUT WARRANTY 594 | OF ANY KIND, EITHER EXPRESSED OR IMPLIED, INCLUDING, BUT NOT LIMITED TO, 595 | THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR 596 | PURPOSE. THE ENTIRE RISK AS TO THE QUALITY AND PERFORMANCE OF THE PROGRAM 597 | IS WITH YOU. SHOULD THE PROGRAM PROVE DEFECTIVE, YOU ASSUME THE COST OF 598 | ALL NECESSARY SERVICING, REPAIR OR CORRECTION. 599 | 600 | 16. Limitation of Liability. 601 | 602 | IN NO EVENT UNLESS REQUIRED BY APPLICABLE LAW OR AGREED TO IN WRITING 603 | WILL ANY COPYRIGHT HOLDER, OR ANY OTHER PARTY WHO MODIFIES AND/OR CONVEYS 604 | THE PROGRAM AS PERMITTED ABOVE, BE LIABLE TO YOU FOR DAMAGES, INCLUDING ANY 605 | GENERAL, SPECIAL, INCIDENTAL OR CONSEQUENTIAL DAMAGES ARISING OUT OF THE 606 | USE OR INABILITY TO USE THE PROGRAM (INCLUDING BUT NOT LIMITED TO LOSS OF 607 | DATA OR DATA BEING RENDERED INACCURATE OR LOSSES SUSTAINED BY YOU OR THIRD 608 | PARTIES OR A FAILURE OF THE PROGRAM TO OPERATE WITH ANY OTHER PROGRAMS), 609 | EVEN IF SUCH HOLDER OR OTHER PARTY HAS BEEN ADVISED OF THE POSSIBILITY OF 610 | SUCH DAMAGES. 611 | 612 | 17. Interpretation of Sections 15 and 16. 613 | 614 | If the disclaimer of warranty and limitation of liability provided 615 | above cannot be given local legal effect according to their terms, 616 | reviewing courts shall apply local law that most closely approximates 617 | an absolute waiver of all civil liability in connection with the 618 | Program, unless a warranty or assumption of liability accompanies a 619 | copy of the Program in return for a fee. 620 | 621 | END OF TERMS AND CONDITIONS 622 | 623 | How to Apply These Terms to Your New Programs 624 | 625 | If you develop a new program, and you want it to be of the greatest 626 | possible use to the public, the best way to achieve this is to make it 627 | free software which everyone can redistribute and change under these terms. 628 | 629 | To do so, attach the following notices to the program. It is safest 630 | to attach them to the start of each source file to most effectively 631 | state the exclusion of warranty; and each file should have at least 632 | the "copyright" line and a pointer to where the full notice is found. 633 | 634 | 635 | Copyright (C) 636 | 637 | This program is free software: you can redistribute it and/or modify 638 | it under the terms of the GNU General Public License as published by 639 | the Free Software Foundation, either version 3 of the License, or 640 | (at your option) any later version. 641 | 642 | This program is distributed in the hope that it will be useful, 643 | but WITHOUT ANY WARRANTY; without even the implied warranty of 644 | MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the 645 | GNU General Public License for more details. 646 | 647 | You should have received a copy of the GNU General Public License 648 | along with this program. If not, see . 649 | 650 | Also add information on how to contact you by electronic and paper mail. 651 | 652 | If the program does terminal interaction, make it output a short 653 | notice like this when it starts in an interactive mode: 654 | 655 | Copyright (C) 656 | This program comes with ABSOLUTELY NO WARRANTY; for details type `show w'. 657 | This is free software, and you are welcome to redistribute it 658 | under certain conditions; type `show c' for details. 659 | 660 | The hypothetical commands `show w' and `show c' should show the appropriate 661 | parts of the General Public License. Of course, your program's commands 662 | might be different; for a GUI interface, you would use an "about box". 663 | 664 | You should also get your employer (if you work as a programmer) or school, 665 | if any, to sign a "copyright disclaimer" for the program, if necessary. 666 | For more information on this, and how to apply and follow the GNU GPL, see 667 | . 668 | 669 | The GNU General Public License does not permit incorporating your program 670 | into proprietary programs. If your program is a subroutine library, you 671 | may consider it more useful to permit linking proprietary applications with 672 | the library. If this is what you want to do, use the GNU Lesser General 673 | Public License instead of this License. But first, please read 674 | . 675 | --------------------------------------------------------------------------------