├── .gitignore ├── Miscellanea ├── DualLemma.pdf ├── DualLemma.tex ├── Errata.pdf └── Errata.tex ├── README.md └── Solutions ├── Chap10 ├── Lecture10.pdf └── Lecture10.tex ├── Chap11 ├── Lecture11.pdf └── Lecture11.tex ├── Chap12 ├── Lecture12.pdf └── Lecture12.tex ├── Chap3 ├── Lecture3.pdf └── Lecture3.tex ├── Chap6 ├── Lecture6.pdf └── Lecture6.tex ├── Chap7 ├── Lecture7.pdf └── Lecture7.tex └── README.MD /.gitignore: -------------------------------------------------------------------------------- 1 | 2 | *.gz 3 | *.aux 4 | *.log 5 | *.out 6 | .DS_Store 7 | -------------------------------------------------------------------------------- /Miscellanea/DualLemma.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/sleepycoke/Mathematical_Logic_NJUCS/88e4550da4597d5343fe5a595f94a544ed475d70/Miscellanea/DualLemma.pdf -------------------------------------------------------------------------------- /Miscellanea/DualLemma.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | 3 | \usepackage{fontspec, xunicode, xltxtra} 4 | \usepackage{amsmath, amsfonts, amssymb} 5 | \usepackage[a4paper,top=30mm,bottom=30mm,left=30mm,right=30mm]{geometry} 6 | \usepackage{amsthm} 7 | \usepackage[shortlabels]{enumitem} 8 | 9 | \begin{document} 10 | \title{A Proof of Dual Replacement Lemma} 11 | \author{DING Chao} 12 | \maketitle 13 | 14 | Let $\varphi$ be a formula in basic modal and we denote $\varphi'$ as the formula obtained by replacing a ``$\lozenge$'' inside $\varphi$ with ``$\neg \square \neg$''. We claim that 15 | $$\vdash_K \varphi \leftrightarrow \varphi'. $$ 16 | \begin{proof} 17 | 18 | We shall prove this by structural induction on $\varphi$. 19 | \begin{enumerate}[{Case} 1: ] 20 | \item $\varphi = \lozenge \psi$ and we are replacing the first $\lozenge$, then $\varphi' = \neg \square \neg \varphi$. 21 | \begin{enumerate}[(1)] 22 | \item $\lozenge p \leftrightarrow \neg \square \neg p$ \hfill DUAL 23 | \item $\lozenge \varphi \leftrightarrow \neg \square \neg \varphi$ \hfill US:(1) 24 | \end{enumerate} 25 | 26 | \item $\varphi = \neg \psi$, $\varphi' = \neg \psi'$. We informally claim that the definition of $\psi'$ is clear: For this case, if $\varphi'$ replaces the $i$th $\lozenge$, $\psi'$ does the $i$th. 27 | \begin{enumerate}[(1)] 28 | \item $(p \leftrightarrow q) \rightarrow (\neg p \leftrightarrow \neg q)$ \hfill TAUT 29 | \item $(\psi \leftrightarrow \psi') \leftrightarrow (\neg \psi \leftrightarrow \neg \psi')$ \hfill US:(1) 30 | \item $\psi \leftrightarrow \psi'$ \hfill I.H. 31 | \item $\neg \psi \leftrightarrow \neg \psi'$ \hfill MP(2)(3) 32 | \end{enumerate} 33 | \item $\varphi = \psi * \rho$, $\varphi' = \psi' * \rho$, where $* \in \{\wedge, \vee, \rightarrow, \leftrightarrow\}$. 34 | \begin{enumerate}[(1)] 35 | \item $(p \leftrightarrow q) \rightarrow ((p * r )\leftrightarrow (q * r) )$ \hfill TAUT 36 | \item $(\psi \leftrightarrow \psi') \rightarrow ((\psi * \rho) \leftrightarrow (\psi' * \rho))$ \hfill US:(1) 37 | \item $\psi \leftrightarrow \psi'$ \hfill I.H. 38 | \item $(\psi * \rho) \leftrightarrow (\psi' * \rho) $ \hfill MP(2)(3) 39 | \end{enumerate} 40 | \item $\varphi = \psi * \rho$, $\varphi' = \psi * \rho'$. A similar proof as above holds. 41 | \item $\varphi = \square \psi$, $\varphi' = \square \psi'$. 42 | \begin{enumerate}[(1)] 43 | \item $\psi \rightarrow \psi'$ \hfill PL:I.H. 44 | \item $\square(\psi \rightarrow \psi')$ \hfill N(1) 45 | \item $\square(\psi \rightarrow \psi') \rightarrow (\square \psi \rightarrow \square \psi')$ \hfill US:K 46 | \item $\square \psi \rightarrow \square \psi'$ \hfill MP(3)(2) 47 | \end{enumerate} 48 | Analogously, $\square \psi' \rightarrow \square \psi$. Therefore $\square \psi \leftrightarrow \square \psi'$. 49 | \item $\varphi = \lozenge \psi$ and we are replacing a $\lozenge$ inside $\psi$, then $\varphi' = \lozenge \psi'$. 50 | \begin{enumerate}[(1)] 51 | \item $\square \neg \psi \leftrightarrow \square \neg \psi'$ \hfill I.H.(Case 5) 52 | \item $\lozenge \psi \leftrightarrow \neg \square \neg \psi$ \hfill I.H.(Case 1) 53 | \item $\lozenge \psi' \leftrightarrow \neg \square \neg \psi'$ \hfill I.H.(Case 1) 54 | \item $\lozenge \psi \leftrightarrow \lozenge \psi'$ \hfill PL:(1),(2),(3) 55 | \end{enumerate} 56 | \end{enumerate} 57 | 58 | Finally we have proved $\vdash_K \varphi \leftrightarrow \varphi'$ holds for any basic modal formula $\varphi$. 59 | \end{proof} 60 | 61 | % 62 | \end{document} 63 | -------------------------------------------------------------------------------- /Miscellanea/Errata.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/sleepycoke/Mathematical_Logic_NJUCS/88e4550da4597d5343fe5a595f94a544ed475d70/Miscellanea/Errata.pdf -------------------------------------------------------------------------------- /Miscellanea/Errata.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \usepackage{ctex} 3 | \usepackage{fontspec, xunicode, xltxtra} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsfonts} 7 | \usepackage{mathrsfs} 8 | \usepackage{enumerate} 9 | \usepackage{amsthm} 10 | \usepackage{hyperref} 11 | \usepackage[a4paper,top=30mm,bottom=30mm,left=30mm,right=30mm]{geometry} 12 | 13 | \begin{document} 14 | \title{数理逻辑讲义勘误表\footnote{v14. 此文档在 \url{ https://github.com/sleepycoke/Mathematical_Logic_NJUCS} 维护}} 15 | \author{丁超} 16 | \maketitle 17 | \begin{enumerate} 18 | \item 书后没有索引 19 | \item 没有指明引用参照文献的地方. 20 | \item 没有公式相等和逻辑符优先级的定义. 21 | \item P10 例1.6 子项的等价替换没有证明. 需要在命题1.14的基础上完成证明.或参见\\ \url{https://proofwiki.org/wiki/Substitution_for_Equivalent_Subformula_is_Equivalent} 22 | \item P12 第4行$A_n$改成$A_m$. (由吴骏给出) 23 | \item P18-13,14 P48-15,19,20 由于各公式可能永真，永假，或不独立, 导致结论不一定成立. 24 | \item P34定义3.8 “约定...”一行内的$M$都要改成$\mathbb{M}$. 25 | \item P34定义3.9(2)改为：“$\mathscr{L}$的一个模型为二元组$(\mathbb{M},\sigma)$, 这里$\mathbb{M}$为$\mathscr{L}$-结构且$\sigma$为$M$上的赋值." 26 | \item P35定义3.10 $M$都要改成$\mathbb{M}$. 27 | \item P36定义3.14 除了(3)中的最后一个$M$,其它$M$都要改成$\mathbb{M}$. 类似错误在本书内不胜枚举，不一一列出. 28 | \item P37 P56$\models \Delta$被重定义为$\vdash \Delta$有效和$\Delta$永真. 二者还不等价. (由周涛指出) 29 | \item P39 倒数第三行两处``$\mathscr{L}\ne$''改成``$\mathscr{L}-$''. (由吴骏给出) 30 | \item P40 情况1 下一行$M\models_\rho$改为$M\models_\sigma$. (由吴骏给出) 31 | \item P40 倒数第3, 4行$\frac{t_1}{t_n}$改为$t_1$. (由吴骏给出)\item P56 定义4.6 (1) 最后\((\bigvee\limits_{j=1}^m B_j)\)前面加上\(\models\).(由董杨静提交) 32 | \item P48-19(1)$\leftrightarrow$没有定义. 而且不是永真式. (由董杨静指出) 33 | \item P51 规则$\exists R$上矢列后件在$A[t/x]$前少了$\Lambda, $. (由吴骏给出) 34 | \item P56 命题4.7 1) 两个"任意"改为"某个".(由谢逸指出) 35 | \item P56 命题4.7 1) 最后的\(B_i\)改为\(B_j\).(由董杨静提交) 36 | \item P41 出现了$y\equiv x$但书上没有定义$\equiv$. 事实上我认为多余定义它, 用"="就够了. 37 | \item P41 case 6.1 若$y\equiv x$那么$x$不在$\forall xA$中自由出现, 也就有$(\forall xA)[t/x] = \forall xA$，实际上没有发生替换也就不用这么麻烦证明. 38 | \item P47-13 $I(~^{-1})$的定义没有加mod; $x_1x_2$ 中间要加 *. 39 | \item P47-14(1)就是书上例题但又没有给证明. 有同学写作业就直接当它可用了. 40 | \item P48-17 第二个$\Gamma$改为$\Gamma_n$. 41 | \item P49-21 没有解释一个公式(集)有模型是指这个模型满足它. 42 | \item P55 左上角用了元定理，不是G的规则. 43 | %By Sun Xudong 44 | \item (由孙旭东指出)P68命题6.8没有说明公式集可满足的定义. 是每个公式可以有不同的模型还是要有统一模型. 虽然从证明中可以看出默认的是后者. 45 | \item P69定义6.10(1)(2)删除开头的两个``若''. (由吴骏给出) 46 | \item P72 第三行将c换成y没有说明得到的还是证明树 47 | \item P73 定理6.19 $\Gamma$和$\Psi$不在同一个语言内，要说明$\Gamma$的模型如何由$\Psi$的模型得到. 48 | %By Shen Mingjie 49 | \item P72 (5)第二行, ``由前命题''改成``由命题6.5''; ``只需证若$Con(\Psi_n$''改为``--$Con(\Psi$; (6)$\exists x.A \in \Gamma$改为``$\in \Psi$''. (由沈明杰指出) 50 | \item P75 第六讲习题五没说$\Phi$是公式集. 51 | \item P80定义7.9前一行L-闭项没有定义，我认为是没有变元的项. 52 | \item P82第二行闭项没有定义，我认为是闭公式. 53 | \item P84 第四行``$f(t_1)$''后加一个``)''(由沈明杰指出) 54 | \item P84第4行``$f(t_1)$"后面加一个``)''. (由吴骏给出) 55 | \item P85 第一行''Sklolem"改为''Skolem''. (由吴骏给出) 56 | \item P87 公式命名不同于出现顺序, 为何不一律以第一个出现的公式为A... 57 | \item P88 $T15$不应有$\vdash$. 按照P87定义8.1(3)定理是公式. 类似问题不胜枚举,不再赘述. 58 | \item P92 $T21$是元定理, 用$T$编号是不合适的. 59 | \item P90 (9) missing ``)''. 60 | \item P95 第一行起至本章末尾$G$改为$G'$ . 61 | \item P95 倒数第6行$(T32)$改为$(T22)$. (由吴骏给出) 62 | \item 第03讲Slides第27页 有同学反映[*]没有定义. 63 | \item P102定理9.7要除去条件$x\notin FV(B)$并重新证明. (由吴骏给出) 64 | \item P103 第5行 (c为新变元) 改为 (c为新常元) 65 | \item 第十讲P108 定义10.5约定(1)A(t)表示A(t), 本页倒数第二行前件多个A. 66 | %By Chen Qinlin 67 | \item P118 引理10.18, 归纳步骤，改为``不妨设P终于度为d的Cut推理''. (由陈钦霖指出) 68 | \item P121 定义11.1(c)加上$B\subseteq E$. 69 | \item P122 命题11.2定义加上$C\subseteq \mathcal{P}(E)$. 70 | \item P134 第5题第二行$\mathbb{M}\models(\mathbf{M}, \mathbf{I})$改为$\mathbb{M}=(\mathbf{M}, \mathbf{I})$ 71 | \item P138 倒数第三行 $\exists s \in S$改成 $\exists t \in S$. 72 | \item P141第一行多个$\bot$ 倒数第二行$\psi_1 \psi_2$改成$\varphi_1 \varphi_2$. 73 | \item P136定义12.1 第二行``$\mathfrak{F}$上的关系''改成``$W$上的关系''. 74 | \item P143图12-7,$w_6$哪来的？模型中没有. 75 | \item 第十二讲框架类没有定义. 76 | 77 | 78 | \item 定义12.4最好结合一下语义不然很难区别路径公式与状态公式. 79 | \item P154 (h),(i)一样. 80 | \item P147 (P1) 第二个$x$改成$s_0$. 81 | \item 闭公式closed formula是指没有自由变元的公式. 书上可能没有定义. 82 | \item P140 ``合式公式(well-formed-formula)''没有定义. 83 | \item P148 图12-9, black与grey颜色相近. 84 | \item P149 定义12.13 ``K-证明是一个无穷的'' 改成 ``--有穷的''. 85 | \item P150 例12.9第4步前面少加了$\Box$. 每行前不要加$\vdash$. 86 | 87 | \item Hauptsatz和紧性定理的证明过程完全不看也不影响做习题. 88 | 89 | 90 | 91 | 92 | 93 | 94 | 95 | 96 | \end{enumerate} 97 | 98 | \end{document} -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # Mathematical_Logic_NJUCS 2 | 南京大学计算机系数理逻辑课程资料 3 | 4 | 仅限于学习交流，禁止用于商业用途。由于作者水平有限，错误在所难免，恳请指正。 5 | 6 | 传播此文档时请遵守 7 | https://creativecommons.org/licenses/by-nc-sa/4.0/ 8 | -------------------------------------------------------------------------------- /Solutions/Chap10/Lecture10.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/sleepycoke/Mathematical_Logic_NJUCS/88e4550da4597d5343fe5a595f94a544ed475d70/Solutions/Chap10/Lecture10.pdf -------------------------------------------------------------------------------- /Solutions/Chap10/Lecture10.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \usepackage{ctex} 3 | \usepackage{fontspec, xunicode, xltxtra} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsfonts} 7 | \usepackage{mathrsfs} 8 | \usepackage{enumerate} 9 | \usepackage{amsthm} 10 | \usepackage{hyperref} 11 | \usepackage{bussproofs} 12 | \usepackage[a4paper,top=30mm,bottom=30mm,left=30mm,right=30mm]{geometry} 13 | \EnableBpAbbreviations 14 | \newcommand{\vv}[0]{\vdash\dashv} 15 | \newcommand{\rl}[1]{\RightLabel{$#1$}} 16 | \def\fCenter{\ \vdash \ } 17 | \renewcommand{\today}{\number\year 年 \number\month 月 \number\day 日} 18 | 19 | \setlength\parindent{0pt} 20 | \vspace{-8ex} 21 | %\date{} 22 | \begin{document} 23 | 24 | \title{数理逻辑第十讲习题参考答案\footnote{Ver. 1.0. 原答案由宋方敏教授给出手稿, 由丁超同学录入并修订补充. 此文档来源为\url{https://github.com/sleepycoke/Mathematical_Logic_NJUCS} 25 | 欢迎各位同学提出意见共同维护. 26 | }} 27 | \maketitle 28 | 29 | \section*{习题10.1} 30 | 由于这两套系统的推理规则相同, 只要保证一套系统的公理在另一套系统内可证即可保证证明树存在, 从而矢列保持可证. 31 | \subsection*{``$\Rightarrow$''} 32 | $A\vdash A$为$\Gamma, A, \Delta \vdash \Pi, A, \Lambda$的特例, 故其为$LK'$中的公理, 也就在$LK'$中可证. 33 | \subsection*{``$\Leftarrow$''} 34 | 首先证明以下引理: 35 | 36 | 对于任何公式$A$, 有$A\vdash A$在$LK$中可证. (*) 37 | 38 | 下面对$A$的结构进行归纳证明. 39 | \begin{enumerate}[情况1: ] 40 | \item[Basis: ] $A$是原子公式, 从而$A\vdash A$是$LK$的公理, 故(*)成立. 41 | \item[I.H.: ]对于复合公式$A$的所有真子公式$B$都有$B\vdash B$在$LK$中可证. 42 | \item[I.S.: ]对复合公式$A$的最外层的逻辑联结词进行讨论: 43 | \item $A$形为$\neg B$, 由归纳假设, $B\vdash B$在$LK$中可证, 从而有 44 | \begin{prooftree} 45 | \Axiom$B \fCenter B$ 46 | \rl{\neg L, \neg R} 47 | \doubleLine 48 | \UI $\neg B \fCenter \neg B$ 49 | \end{prooftree} 50 | 51 | \item $A$形为$B \rightarrow C$, 由归纳假设$B \vdash B$和$C\vdash C$在$LK$中可证, 从而有 52 | \begin{prooftree} 53 | \AX$B \fCenter B$ \rl{WR} 54 | \UI$B \fCenter B, C$ \rl{ER} 55 | \UI$B \fCenter C, B$ 56 | \AX$C \fCenter C$ \rl{WL} 57 | \UI$B, C \fCenter C$ \rl{EL} 58 | \UI$C, B \fCenter C$ \rl{\rightarrow L} 59 | \BI$B\rightarrow C, B \fCenter C$ \rl{EL} 60 | \UI$B, B \rightarrow C \fCenter C$ \rl{\rightarrow R} 61 | \UI$B\rightarrow C \fCenter B\rightarrow C$ 62 | \end{prooftree} 63 | \item $A$形为$B \wedge C$或$B\vee C$, 同理可证. 64 | \item $A$形为$\forall x B(x)$, 由归纳假设, $B(a) \vdash B(a)$在$LK$中可证. 从而有: 65 | \begin{prooftree} 66 | \AX$B(a) \fCenter B(a)$ \rl{\forall L} 67 | \UI$\forall x B(x) \fCenter B(a)$ \rl{\forall R} 68 | \UI$\forall x B(x) \fCenter \forall x B(x)$ 69 | \end{prooftree} 70 | \item $A$形为$\exists x B(x)$与上同理. 71 | \end{enumerate} 72 | 综上(*)成立. 73 | \bigbreak 74 | 接下来我们证明$\Gamma, A, \Delta \vdash \Pi, A, \Lambda$在$LK$中可证. 我们已证$A\vdash A$可证, 从而有证明树: 75 | \begin{prooftree} 76 | \AX$A\fCenter A$\rl{WL, WR} 77 | \doubleLine 78 | \UI $\Gamma, \Delta, A \fCenter A, \Pi, \Lambda$ \rl{EL, ER} 79 | \doubleLine 80 | \UI $\Gamma, A, \Delta \fCenter \Pi, A, \Lambda$ 81 | \end{prooftree} 82 | \qed 83 | 84 | \section*{习题10.2} 85 | 定义: 86 | \begin{align*} 87 | R_1 &\triangleq \{(A, B)|A \in sub(B)\}\\ 88 | R_2 &\triangleq \{(A, B)|A \text{是$B$的前辈}\}\\ 89 | R_3 &\triangleq \{(A, B)|A \text{是$B$的立接前辈}\} 90 | \end{align*} 91 | 易见$R_1$满足传递性, 且$R_2$是$R_3$的传递闭包. 92 | 所以只需证 93 | \begin{center} 94 | \emph{若$C$为$D$的立接前辈, 则$C$为$D$的子公式...(*).} 95 | \end{center} 96 | 97 | 设$C$与$D$分别在规则$J$的上矢列和下矢列中. 98 | \begin{enumerate}[情况1: ] 99 | \item $J$为$Cut$, 由于$Cut$公式没有立接后辈, 从而$C$只能是旁公式. 100 | \item $C$为旁公式, 从而$D=C$, (*)成立. 101 | \item $J$为$Exchange$, $C$为$J$的辅公式, $D$为主公式, 从而$D=C$, (*)成立. 102 | \item $J$不是$Exchange$或$Cut$, $C$为$J$的辅公式, $D$为主公式；可逐个验证$C$是$D$的子公式. 103 | \end{enumerate} 104 | 综上(*)成立. 105 | \qed 106 | 107 | \section*{习题10.3} 108 | \begin{prooftree} 109 | \AX$A(a) \fCenter A(a)$\rl{WR} 110 | \UI$A(A) \fCenter B, A(a)$\rl{\rightarrow R} 111 | \UI$\fCenter A(a)\rightarrow B, A(a)$\rl{\exists R} 112 | \UI$\fCenter \exists x(A(x)\rightarrow B), A(a)$\rl{\forall R} 113 | \UI$\fCenter \exists x A(x) \rightarrow B, \forall x A(x)$ 114 | \AX$B\fCenter B$\rl{WL} 115 | \UI$A(a), B\fCenter B$\rl{\rightarrow R} 116 | \UI$B\fCenter A(a)\rightarrow B$\rl{\exists R} 117 | \UI$B\fCenter \exists x(A(x) \rightarrow B)$\rl{\rightarrow L} 118 | \BI$\forall x A(x) \rightarrow B \fCenter \exists x(A(x)\rightarrow B)$ 119 | \end{prooftree} 120 | 121 | \qed 122 | 123 | \section*{习题10.4} 124 | 对$P(a)$的深度$n$归纳证明: 125 | 126 | \begin{enumerate}[情况1: ] 127 | \item[Basis: ]$n = 0$, $P(a)$为公理, 设其呈形$A(a)\vdash A(a)$. 从而$P(t)$为$A(t)\vdash A(t)$, 也为公理, 于是$P(t)$为证明树. 128 | \item[I.H.: ]$n \le k$时, 有$P(t)$为证明树. 129 | \item[I.S.: ]往证当$n = k + 1$时, $P(t)$同样是证明树. 130 | \item $P(a)$呈形 131 | \AXC{$Q(a)$}\rl{J} 132 | \UIC{$B(a)$} 133 | \DP 134 | 其中$Q(a)$是深度为$k$的证明树, $B(a)$为$P(a)$的终矢列. 135 | 以下仅证明$J$为$\forall R$的情况, 其它情况类似. 136 | 我们设$Q(a)$的终矢列$C(a)$呈形 137 | $\Gamma(a)\vdash \Delta(a), A(a, b)$； $B(a)$呈形$\Gamma(a)\vdash \Delta(a), \forall x A(a, x)$. 由于$a$不是特征变元, $a \ne b, A(a, b)[\frac{t}{a}] = A(t, b)$, 那么$P(t)$最后一层推理呈形 138 | \begin{prooftree} 139 | \AX$\Gamma(t)\fCenter \Delta(t), A(t, b)$\rl{\forall R} 140 | \UI$\Gamma(t)\fCenter \Delta(t), \forall x A(t, x)$ 141 | \end{prooftree} 142 | 由于$t$中不自由出现特征变元, 所以$b$是新变元, 这是有效推理. 又由归纳假设$C(t)$可证, 从而$P(t)$是证明树. 143 | \item $P(a)$呈形 144 | \AXC{$Q(a)$} 145 | \AXC{$R(a)$}\rl{J} 146 | \BIC{$B(a)$} 147 | \DP 148 | 同样由归纳假设$Q(t)$, $R(t)$为证明树. 再对推理规则逐条验证可证明$P(t)$为证明树. 不再赘述. 149 | \end{enumerate} 150 | \qed 151 | 152 | 153 | \section*{习题10.5} 154 | \subsection*{$Mix\Rightarrow Cut$} 155 | \begin{prooftree} 156 | \AX$\Gamma \fCenter \Delta, A$ 157 | \AX$A, \Gamma \fCenter \Delta$\rl{Mix(A)} 158 | \BI$\Gamma, \Gamma^* \fCenter \Delta^*, \Delta$\rl{Exchange, Contraction} 159 | \doubleLine 160 | \UI$\Gamma \fCenter \Delta$ 161 | \end{prooftree} 162 | \subsection*{$Cut \Rightarrow Mix$} 163 | \begin{prooftree} 164 | \AX$\Gamma \fCenter \Delta$\rl{ER, CR} 165 | \doubleLine 166 | \UI$\Gamma \fCenter \Delta^*, A$\rl{Weakening, Exchange} 167 | \UI$\Gamma, \Pi^* \fCenter \Delta^*, \Lambda, A$ 168 | \AX$\Pi \fCenter \Lambda$\rl{EL, CL} 169 | \UI$A, \Pi^* \fCenter \Lambda$\rl{Weakening, Exchange} 170 | \doubleLine 171 | \UI$A, \Gamma, \Pi^* \fCenter \Delta^*, \Lambda$\rl{Cut} 172 | \BI$\Gamma, \Pi^* \fCenter \Delta^*, \Lambda$ 173 | \end{prooftree} 174 | \qed 175 | 176 | \section*{习题10.6} 177 | 利用习题10.2, 只需证任何公式都有在终矢列的后辈. 178 | 对除$Cut$之外的规则逐条验证上矢列中的公式都有后辈即可. 179 | %太烦了不想写, 宋老师也没写 180 | 181 | \section*{习题10.7} 182 | 反设$\vdash$可证, 那么它有一个无切证明树$T$. 设$T$的一个叶子为$A\vdash A$, 那么$A$不可能是终矢列中任何公式的子公式, 与习题10.6矛盾. 183 | 184 | 于是$\vdash$不可证. 185 | \qed 186 | 187 | \section*{习题10.8} 188 | 反设$P(a)\vdash Q(a)$可证, 那么它存在一个无切证明树$T$. 由于$P(a)$与$Q(a)$皆为原子公式, 由习题10.6, $T$中出现的公式只能是二者之一. 于时$T$中不能出现逻辑规则, 只能有三种弱规则. 189 | 190 | 注意到这三种规则都只有一个上矢列, 从而$T$只有一个叶子, 设其呈形$A\vdash A$. 若$A = P(a)$,注意到$CR$不能将$P(a)$从后件完全消除, $WR$与$ER$不会消除公式, 从而$T$中所有矢列的后件都会有$P(a)$那么, $P(a)\vdash Q(a)$不可能在$T$中. $A = Q(a)$同理. 191 | 192 | 于是, $P(a)\vdash Q(a)$不可证. 193 | \qed 194 | 195 | 196 | 197 | \section*{习题10.9} 198 | 根据习题10.1我们有以下证明树的叶子可证: 199 | \begin{prooftree} 200 | \AX$A, B \fCenter C, A$ 201 | \AX$A, B \fCenter C, B$ 202 | \AX$C, A, B \fCenter C$ \rl{\rightarrow L} 203 | \BI$B\rightarrow C, A, B \fCenter C$ \rl{\rightarrow L} 204 | \BI$A\rightarrow(B\rightarrow C), A, B \fCenter C$\rl{EL} 205 | \doubleLine 206 | \UI$A, B, A\rightarrow(B\rightarrow C) \fCenter C$\rl{\rightarrow R} 207 | \UI$B, A\rightarrow(B\rightarrow C) \fCenter A\rightarrow C$\rl{\rightarrow R} 208 | \UI$A\rightarrow(B\rightarrow C) \fCenter B\rightarrow A\rightarrow C$ 209 | 210 | 211 | \end{prooftree} 212 | \qed 213 | 214 | \end{document} 215 | -------------------------------------------------------------------------------- /Solutions/Chap11/Lecture11.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/sleepycoke/Mathematical_Logic_NJUCS/88e4550da4597d5343fe5a595f94a544ed475d70/Solutions/Chap11/Lecture11.pdf -------------------------------------------------------------------------------- /Solutions/Chap11/Lecture11.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \usepackage{ctex} 3 | \usepackage{fontspec, xunicode, xltxtra} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsfonts} 7 | \usepackage{mathrsfs} 8 | \usepackage{enumerate} 9 | \usepackage{amsthm} 10 | \usepackage{hyperref} 11 | \usepackage[a4paper,top=30mm,bottom=30mm,left=30mm,right=30mm]{geometry} 12 | 13 | \newcommand{\vv}[0]{\vdash\dashv} 14 | \renewcommand{\today}{\number\year 年 \number\month 月 \number\day 日} 15 | 16 | \setlength\parindent{0pt} 17 | \vspace{-8ex} 18 | %\date{} 19 | \begin{document} 20 | 21 | \title{数理逻辑第十一讲习题参考答案\footnote{Ver. 0.6. 原答案由宋方敏教授给出手稿, 乔羽同学录入, 最后由丁超同学修订补充. 此文档来源为\url{https://github.com/sleepycoke/Mathematical_Logic_NJUCS}由于时间紧张，丁超只更正了比较明显的笔误, 没有仔细验证, 见谅. 22 | 欢迎各位同学提出意见共同维护. 23 | }} 24 | \maketitle 25 | 26 | \section{} 27 | 28 | 设:$\Delta \subseteq \Gamma, \Delta$ 为有穷集， 29 | 30 | 从而有$k \in \mathbb{N} $使得 31 | 32 | $\Delta \subseteq \{ (x > S^nO) | n=0,1,2,..,k\}$ 33 | 34 | 设$N = \{ \mathbb{N},O,Suc , > \}$ 为算术的标准模型 35 | 36 | 其中$ Suc (x) =x+1 $, $ > $为大于关系 37 | 38 | 令$\sigma$ 为$\mathbb{N}$上赋值使$\sigma (x) = k+1$ 39 | 40 | 从而$N \vDash_\sigma (x>S^n(O))(n=0,1,2,..,k)$ 41 | 42 | 故$N \vDash_\sigma \Delta$即$\Delta$可满足 43 | 44 | 由紧致性定理知$\Gamma$可满足. 45 | 46 | \section{} 47 | 48 | 49 | 设$\Gamma \vDash \varphi$ 50 | 51 | 反设不存在$\Gamma$的有穷子集$\Delta$使$\Delta \vDash \varphi$ 52 | 53 | 从而 对任何$\Gamma$的有穷子集$\Delta$有$\Delta \nvDash \varphi$ 54 | 55 | 因此对任何$\Gamma$的有穷子集$\Delta$有$\Delta \cup \{\neg \varphi\}$可满足 56 | 57 | 故$\Gamma \cup \{\neg \varphi\}$的任何有穷子集可满足, 58 | 59 | 由紧致性定理知$\Gamma \cup \{\neg \varphi\}$可满足，设 60 | 61 | $\mathfrak{M} \vDash \Gamma \cup\{\neg \varphi\}$. 即$\mathfrak{M} \vDash \Gamma$且$\mathfrak{M} \vDash \neg\varphi$, 与$\Gamma \models \varphi$矛盾. 62 | 63 | 因此存在$\Gamma$的有穷子集$\Delta$使$\Delta \models \varphi$. 64 | 65 | \section{} 66 | 67 | 令$\varphi_n \triangleq \exists x_1,...,x_n. \bigwedge_{1\leq i < j\leq n} \neg (x_i \doteq x_j)$ 68 | 69 | 易见$\mathfrak{M} \models \varphi_n \Leftrightarrow |M|\geq n$ 70 | 71 | $\mathfrak{M} \models \{ \varphi_i | i\in \mathbb{N}^+ \} \Leftrightarrow |M|\geq \aleph_0$ 72 | 73 | 令$\Gamma \triangleq \Sigma \cup \{ \varphi_i | i \in \mathbb{N}^+ \}$，对于任何$\Gamma$的 74 | 75 | 有穷子集$\Delta \subseteq \Sigma \cup \{ \varphi_i | i \in \mathbb{N}^+ \} $，存在$k$使 76 | 77 | $\Delta \subseteq \Sigma \cup \{ \varphi_1 ,..., \varphi_k\} $,由于$\Sigma$具有论域 78 | 79 | 基数大于k的模型，故$\Delta$可满足, 80 | 81 | 由紧致性定理知$\Gamma$ 可满足，那么有，$\mathfrak{M}\vDash \Gamma$ 82 | 83 | 从而$\mathfrak{M} \vDash \{ \varphi_i | i \in \mathbb{N}^+\}$,故$|M|\geq \aleph_0$. 84 | 85 | 86 | \section{} 87 | 88 | 只需证每个有穷图可4色则无穷图可四色. 89 | 90 | 设MAP为一张无穷地图，令全体国家的集合为 91 | 92 | $\{a_i |i\in I\}$,这里$|I|\geq \aleph_0$. 93 | 94 | 设一阶语言$\mathcal{L}$由以下构成 95 | 96 | (1)常元:$\{a_i | i\in I\}$ 97 | 98 | (2)一元谓词符：$C_k (x) (k=1,2,3,4)$\quad($C_k (x)$表示$x$着$k$色) 99 | 100 | (3)二元谓词符：$g(x,y)$\quad($g(x,y)$表示$x$与$y$有大于0的公共边界). 101 | 102 | 令$Q \triangleq \{|i,j\in I\}$且在MAP中$a_i$与$a_j$有大于0的公共边界. 103 | 104 | 令$\Gamma \triangleq \{q(a_i,a_j) | \in Q\}$ 105 | 106 | $\cup \{\neg q(a_i,a_j) | \notin Q\}$ 107 | 108 | $\cup\{ \forall x(c_1(x)\vee c_2(x)\vee c_3(x)\vee c_4(x))\}$ 109 | 110 | $\cup\{ \forall x \forall y (q(x, y)\rightarrow (\neg (c_1(x)\wedge c_1(y)) \wedge \neg(c_2(x)\wedge c_2(y))\wedge \neg(c_3(x)\wedge c_3(y))\wedge \neg(c_4(x)\wedge c_4(y))))\}$ 111 | 112 | 设$S\subseteq \Gamma$ 为$\Gamma$的任何有穷子集，不妨设$\{a_0,...,a_n\}$ 113 | 114 | 为出现在$S$中的全体常元，令$M=\{a_0,...,a_n\},MAP[s]$为$\{a_0,\dots,a_n\}$的生成子图. 从而$MAP[s]$可着4色. 115 | 116 | 令$(C_{k})_\frak{M} \triangleq \{a_i|a_i$着$k$色$i\leq n\}k=1,...,4$ 117 | 118 | $q_\frak{M} =\{|\in Q\}$从而$\mathfrak{M} \vDash S$ 119 | 120 | 由compactness知有$\mathfrak{M}$使$\mathfrak{M}\vDash \Gamma $即$MAP$可4染色. 121 | 122 | \section{} 123 | 反设，对任何$m\in \mathbb{N}$ 都存在结构$\mathfrak{M} \triangleq (M,I)$使 124 | 125 | $m <|M| < \aleph_0$ 且$\mathfrak{M}\vDash \neg \varphi $ 126 | 127 | 令 $\varphi_n$ 为 $\exists x_1 \exists x_2 ... \exists x_n (\bigwedge_{0a_1>a_2...$为无穷下降链，从而 161 | 162 | $\{a_n | n \in \mathbb{N}\}$ 没有最小元. 163 | 164 | 故$R_\frak{M}$并非$M$上的良序.矛盾 165 | 166 | 167 | \end{document} -------------------------------------------------------------------------------- /Solutions/Chap12/Lecture12.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/sleepycoke/Mathematical_Logic_NJUCS/88e4550da4597d5343fe5a595f94a544ed475d70/Solutions/Chap12/Lecture12.pdf -------------------------------------------------------------------------------- /Solutions/Chap12/Lecture12.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \usepackage{ctex} 3 | \usepackage{fontspec, xunicode, xltxtra} 4 | \usepackage{amsmath, amsfonts, amssymb} 5 | \usepackage[a4paper,top=30mm,bottom=30mm,left=30mm,right=30mm]{geometry} 6 | \usepackage[shortlabels]{enumitem} 7 | \usepackage{bussproofs} 8 | 9 | \setmainfont{Times New Roman} 10 | \newcommand{\combine}[2]{{C_{#1}^{#2}}} 11 | 12 | \setlist{nolistsep} 13 | 14 | \begin{document} 15 | \title{数理逻辑作业} 16 | \author{陈劭源 \and 161240004} 17 | \maketitle 18 | 19 | \section*{3} 20 | \begin{enumerate}[(a)] 21 | \item $\mathfrak{F} = (W, R)$，其中$W = \{a, b\}$, $R = \{(a, b)\}$，则对于状态$a$，$\square \bot$不成立（无论标记函数是什么），从而$\square \bot$ 在该框架下非有效。 22 | 23 | 该公式在满足$R = \varnothing$的框架$\mathfrak{F} = (W, R)$上均有效。 24 | 25 | \item $\mathfrak{F} = (W, R)$，其中$W = \{a, b, c\}$, $R = \{(a, b), (a, c)\}$，则对于标记函数$L = \{(a, \varnothing), (b, \{p\}), (c, \varnothing)\}$，在模型$(\mathfrak{F}, L)$ 下，公式$\lozenge p \rightarrow \square p$ 在状态$a$上非真，从而该公式在上述框架下非有效。 26 | 27 | 该公式在满足对于任意$u \in W$，存在至多一个$v \in W$使得$Ruv$的框架$\mathfrak{F} = (W, R)$上均有效。 28 | \item $\mathfrak{F} = (W, R)$，其中$W = \{a, b\}$, $R = \{(a, b)\}$，则对于标记函数$L = \{(a, \{p\}), (b,\varnothing)\}$，在模型$(\mathfrak{F}, L)$ 下，公式$p \rightarrow \square \lozenge p$ 在状态$a$上非真，从而该公式在前述框架下非有效。 29 | 30 | 该公式在所有对称框架下均有效。 31 | \item $\mathfrak{F} = (W, R)$，其中$W = \{a, b, c, d\}$, $R = \{(a, b), (b, c), (b, d)\}$，则对标记函数$L = \{(a, \varnothing), (b,\varnothing), $ $(c, \{p\}), (d,\varnothing) \}$，在模型$(\mathfrak{F}, L)$ 下，公式$\square \lozenge p \rightarrow \lozenge \square p$ 在状态$a$上非真，从而该公式在前述框架下非有效。 32 | 33 | 该公式在状态转换图呈一个环形的所有框架下均有效。 34 | \end{enumerate} 35 | 36 | \section*{4} 37 | \begin{enumerate}[(a)] 38 | \item[(a)] 1, 2, 3, 4 39 | \item[(b)] 无 40 | \item[(c)] 1, 2, 3, 4 41 | \item[(h)] 1, 2, 3, 4 42 | \item[(k)] 1\footnote{丁超注: $\forall(y \mathcal{U} r)$ 只在状态1上有效, 因为其它结点都有不经过标记$r$的全路径. } 43 | \end{enumerate} 44 | 45 | \section*{6} 46 | \subsection*{$(\square p \wedge \lozenge q) \rightarrow \lozenge (p \wedge q)$的$\mathbf{K}$-证明} 47 | \begin{enumerate} 48 | \item $\square (p \rightarrow q) \rightarrow (\square p \rightarrow \square q)$ \hfill $\mathbf{K}$ 49 | \item $\square (p \rightarrow \neg q) \rightarrow (\square p \rightarrow \square \neg q)$ \hfill $\mathbf{US}: 1$ 50 | \item $(p \rightarrow q) \rightarrow \neg(p \wedge \neg q)$ \hfill $\mathbf{TAUT}$ 51 | \item $(\square p \rightarrow \square \neg q) \rightarrow \neg(\square p \wedge \neg \square \neg q)$ \hfill $\mathbf{US}: 3$ 52 | \item $\square (p \rightarrow \neg q) \rightarrow \neg(\square p \wedge \neg \square \neg q)$ $\hfill \mathbf{PL}: 2, 3$ 53 | 54 | \item $\square(\neg (p \wedge q) \rightarrow (p \rightarrow \neg q)) $ \hfill $\mathbf{N: TAUT}$ 55 | \item $\square(\neg (p \wedge q) \rightarrow (p \rightarrow \neg q)) \rightarrow (\square \neg (p \wedge q)) \rightarrow \square (p \rightarrow \neg q))$ \hfill $\mathbf{US}: 1$ 56 | \item $(\square\neg (p \wedge q)) \rightarrow \square (p \rightarrow \neg q)$ \hfill $\mathbf{MP}: 6, 7$ 57 | \item $(\square\neg (p \wedge q)) \rightarrow \neg(\square p \wedge \neg \square \neg q)$ \hfill $\mathbf{PL}: 8, 5$ 58 | \item $(\square p \wedge \neg \square \neg q) \rightarrow (\neg \square\neg (p \wedge q))$ \hfill $\mathbf{PL}: 9$ 59 | 60 | \item $(\square p \wedge \lozenge q) \rightarrow \lozenge (p \wedge q)$ \hfill $\mathbf{PL}: 10, \mathbf{DUAL}, (\mathbf{US: DUAL})$ 61 | \end{enumerate} 62 | 63 | \subsection*{$\lozenge(p \vee q) \leftrightarrow (\lozenge p \vee \lozenge q)$的$\mathbf{K}$-证明} 64 | \begin{enumerate} 65 | \item $\square (p \wedge q \rightarrow p)$ \hfill $\mathbf{N: TAUT}$ 66 | \item $\square (p \wedge q \rightarrow p) \rightarrow (\square (p \wedge q) \rightarrow \square p)$ \hfill $\mathbf{US: K}$ 67 | \item $\square (p \wedge q) \rightarrow \square p$ \hfill $\mathbf{MP}: 1, 2$ 68 | \item $\square (p \wedge q) \rightarrow \square q$ \hfill 同理 69 | \item $\square (p \wedge q) \rightarrow (\square p \wedge \square q)$ \hfill $\mathbf{PL}: 3, 4$ 70 | 71 | \item $(\square p \wedge \square q) \rightarrow \square (p \wedge q)$ \hfill 课本例12.9 72 | \item $(\square p \wedge \square q) \leftrightarrow \square (p \wedge q)$ \hfill $\mathbf{PL}: 5, 6$ 73 | \item $(\square \neg p \wedge \square \neg q) \leftrightarrow \square (\neg p \wedge \neg q)$ \hfill $\mathbf{US}: 7$ 74 | \item $(\neg \square \neg (p \vee q)) \leftrightarrow ((\neg \square \neg p) \vee (\neg \square \neg q))$ \hfill $\mathbf{PL}: 8$ 75 | \item $\lozenge(p \vee q) \leftrightarrow (\lozenge p \vee \lozenge q)$ \hfill $\mathbf{PL}: 9, \mathbf{DUAL}, (\mathbf{US: DUAL})$ 76 | \end{enumerate} 77 | 78 | \section*{7} 79 | 易验证$\mathbf{S4}$对于所有自反且传递的框架均是可靠的（新增加的$p \rightarrow \lozenge p$ 要求框架具有自反性）。为证明$p \rightarrow \square \lozenge p$在$\mathbf{S4}$中不可证，构造模型$\mathfrak{M} = (\mathfrak{F}, L)$，其中$\mathfrak{F} = (\{a, b\}, \{(a,a), (a,b), (b,b)\})$, $L = \{(a, \{p\}), (b, \varnothing)\}$。易验证框架$\mathfrak{F}$自反且传递，但$p \rightarrow \square \lozenge p$在该模型的状态$a$上不为真，从而该公式在$\mathbf{S4}$中不可证。 80 | 81 | 记$\mathbf{S5}$系统中新增加的公理为： 82 | \begin{quote} 83 | \begin{description} 84 | \item[$\mathbf{T'}$] $p \rightarrow \lozenge p$ 85 | \item[$\mathbf{4'}$] $\lozenge \lozenge p \rightarrow \lozenge p$ 86 | \item[$\mathbf{B}$] $p \rightarrow \square \lozenge p$ 87 | \end{description} 88 | \end{quote} 89 | 90 | 则$\lozenge \square p \rightarrow \square p$在该系统中的证明为： 91 | \begin{enumerate} 92 | \item $p \rightarrow \square \lozenge p$ \hfill $\mathbf{B}$ 93 | \item $\lozenge \square p \rightarrow \square \lozenge \lozenge \square p$ \hfill $\mathbf{US}: 1$ 94 | \item $\lozenge \lozenge p \rightarrow \lozenge p$ \hfill $\mathbf{4'}$ 95 | \item $\lozenge \lozenge \square p \rightarrow \lozenge \square p$ \hfill $\mathbf{US}: 3$ 96 | \item $\square(\lozenge \lozenge \square p \rightarrow \lozenge \square p)$ \hfill $\mathbf{N}: 4$ 97 | 98 | \item $\square(\lozenge \lozenge \square p \rightarrow \lozenge \square p) \rightarrow (\square \lozenge \lozenge \square p \rightarrow \square \lozenge \square p)$ \hfill $\mathbf{US: K}$ 99 | \item $\square \lozenge \lozenge \square p \rightarrow \square \lozenge \square p$ \hfill $\mathbf{MP}: 5, 6$ 100 | \item $\neg p \rightarrow \square \lozenge \neg p$ \hfill $\mathbf{US}: 1$ 101 | \item $\square \neg \neg p \rightarrow \neg \lozenge \neg p$ \hfill $\mathbf{PL: (US: DUAL)}$ 102 | \item $\square (p \rightarrow \neg \neg p)$ \hfill $\mathbf{N: TAUT}$ 103 | 104 | \item $\square (p \rightarrow \neg \neg p) \rightarrow (\square p \rightarrow \square \neg \neg p)$ \hfill $\mathbf{US: K}$ 105 | \item $\square p \rightarrow \square \neg \neg p$ \hfill $\mathbf{MP}: 10, 11$ 106 | \item $\square p \rightarrow \neg \lozenge \neg p$ \hfill $\mathbf{PL}: 12, 9$ 107 | \item $\lozenge \neg p \rightarrow \neg \square p$ \hfill $\mathbf{PL}: 13$ 108 | \item $\square \lozenge \neg p \rightarrow \square \neg \square p$ \hfill $\mathbf{MP:(N}: 14 \mathbf{, US:K)}$ 109 | 110 | \item $\neg \square \neg \square p \rightarrow p$ \hfill $\mathbf{PL}: 8, 15$ 111 | \item $\neg \square \neg \square p \leftrightarrow \lozenge \square p$ \hfill $\mathbf{US: DUAL}$ 112 | \item $\lozenge \square p \rightarrow p$ \hfill $\mathbf{PL}: 17, 16$ 113 | \item $\square \lozenge \square p \rightarrow \square p$ \hfill $\mathbf{MP:(N}: 18 \mathbf{, US:K)}$ 114 | \item $\lozenge \square p \rightarrow \square p$ \hfill $\mathbf{PL}: 2, 7, 19$ 115 | \end{enumerate} 116 | \end{document} 117 | -------------------------------------------------------------------------------- /Solutions/Chap3/Lecture3.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/sleepycoke/Mathematical_Logic_NJUCS/88e4550da4597d5343fe5a595f94a544ed475d70/Solutions/Chap3/Lecture3.pdf -------------------------------------------------------------------------------- /Solutions/Chap3/Lecture3.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \usepackage{amsmath} 3 | \usepackage{amsthm} 4 | \usepackage{hyperref} 5 | \usepackage{amssymb} 6 | \author{DING Chao} 7 | 8 | \title{Selected Answers to Exercise 3\footnote{ver 2.2. This document is maintained on \url{https://github.com/sleepycoke/Mathematical_Logic_NJUCS}}} 9 | \begin{document} 10 | \maketitle 11 | \section{P47-13(3)} 12 | \begin{align*} 13 | &((x_1*x_2)^{-1})_{\frak{M}[\sigma]}\\ 14 | =&I(^{-1})((x_1*x_2)_{\frak{M}[\sigma]})\\ 15 | =&(n - (x_1*x_2)_{\frak{M}[\sigma]})) \mbox{ mod } n\\ 16 | =&(n - I(*)(x_{1_{\frak{M}[\sigma]}}, x_{2_{\frak{M}[\sigma]}})\text{ mod } n\\ 17 | =&(n - (x_{1_{\frak{M}[\sigma]}} +_n x_{2_{\frak{M}[\sigma]}})) \mbox{ mod } n\\ 18 | =&(n - (\sigma(x_1) +_n \sigma(x_2)) \mbox{ mod } n\\ 19 | =&(n - (1 \text{ mod } n) +_n (2 \text{ mod } n) ) \mbox{ mod } n\\ 20 | =&-3 \mbox{ mod } n 21 | \end{align*} 22 | 23 | \bigbreak 24 | 25 | \begin{align*} 26 | &((x_1)^{-1}*(x_2)^{-1})_{\frak{M}[\sigma]}\\ 27 | =&I(*)[(x_1^{-1})_{\frak{M}[\sigma]}, (x_2^{-1})_{\frak{M}[\sigma]}]\\ 28 | =&I(*)[I(^{-1})(\sigma(x_1)), I(^{-1})(\sigma(x_2))]\\ 29 | =&((n - 1) \mbox{ mod } n) +_n ((n - 2) \mbox{ mod } n)\\ 30 | =&-3 \mbox{ mod } n 31 | \end{align*} 32 | 33 | Since $((x_1*x_2)^{-1})_{\frak{M}[\sigma]} = ((x_1)^{-1}*(x_2)^{-1})_{\frak{M}[\sigma]}$, by Definition 3.12(2), p36, $A_{\frak{M}[\sigma]} = T$. 34 | 35 | \section{P47-14(2)} 36 | Lemma 1: $\models A \Rightarrow \models \forall x. A$. ($\Leftrightarrow$ indeed). 37 | 38 | \begin{proof} 39 | By $\models A$, we have for any $(M, \sigma)$, $A_{\frak{M}[\sigma]} = T$. Therefore for any $a \in M, A_{\frak{M}[\sigma[x:=a]]} = T$(note that $\sigma[x:=a]$ is one specific assignment of all the $\sigma$), which implies $\forall x.A _{\frak{M}[\sigma]} = T$, aka $\frak{M} \models _\sigma \forall x.A$ . As a result, since $(M, \sigma)$ is arbitrary, $\models \forall x. A$. 40 | \end{proof} 41 | 42 | Let $A \triangleq (x \doteq y \rightarrow y \doteq x)$, 43 | \begin{align*} 44 | &A_{\frak{M}[\sigma]}\\ 45 | =&B_\rightarrow[(x \doteq y)_{\frak{M}[\sigma]}, (y \doteq x) _{\frak{M}[\sigma]}]\\ 46 | =&\begin{cases} 47 | F \quad \mbox{ if } (x \doteq y)_{\frak{M}[\sigma]} = T \mbox{ and } (y \doteq x)_{\frak{M}[\sigma]} = F\\ 48 | T \quad \mbox{otherwise}%\quad \mbox{ if } (x \doteq y)_{\frak{M}[\sigma]} = F \mbox{ or } (x \doteq y)_{\frak{M}[\sigma]} = (y \doteq x) _{\frak{M}[\sigma]} = T\\ 49 | \end{cases}\\ 50 | \end{align*} 51 | 52 | However, $(x\doteq y)_{\frak{M}[\sigma]} = T$ implies $\sigma(x) = \sigma(y)$, which implies $\sigma(y) = \sigma(x)$ and $(y \doteq x)_{\frak{M}[\sigma]} = T$. Thus $A _{\frak{M}[\sigma]} = T$. Applying Lemma 1 twice we get $$\models \forall x \forall y. A.$$ 53 | 54 | \section{P49-24} 55 | \subsection*{$\Rightarrow$} 56 | \begin{align*} 57 | & \frak{M} \models _\sigma \forall x. A\\ 58 | \Rightarrow & \text{For all }a \in M, A_{\frak{M}[\sigma[x:=a]]} = T\\ 59 | \Rightarrow & \text{For all }a \in M, A_{\frak{M} [\rho[x:=a]]} = T, \text{ since $z$ is fresh in $A$, } \text{ where $\rho \triangleq \sigma[z:=a]$. } \\ 60 | \Rightarrow & \text{For all }a \in M, A_{\frak{M} [\rho[x := z_{\frak{M} [\rho]}]]} = T\\ 61 | \Rightarrow & \text{For all }a \in M, A[\frac{z}{x}]_{\frak{M} [\rho]} = T \text{, by Lemma 3.24, p40. } \\ 62 | \Rightarrow &\frak{M} \models _\rho\forall z. A[\frac{z}{x}] \\ 63 | \Rightarrow &\frak{M} \models _\sigma\forall z. A[\frac{z}{x}] \text{, since $z \notin FV(\forall z. A[\frac{z}{x}])$. } 64 | \end{align*} 65 | \subsection*{$\Leftarrow$} 66 | \begin{align*} 67 | &\frak{M} \models _\sigma\forall z. A[\frac{z}{x}]\\ 68 | \Rightarrow &\frak{M} \models _\rho\forall z. A[\frac{z}{x}] \text{, since $z \notin FV(\forall z. A[\frac{z}{x}])$, }\text{ where $\rho \triangleq \sigma[z:=a]$. } \\ 69 | \Rightarrow & \text{For all }a \in M, A[\frac{z}{x}]_{\frak{M} [\rho]} = T \\ 70 | \Rightarrow & \text{For all }a \in M, A_{\frak{M} [\rho[x := z_{\frak{M} [\rho]}]]} = T, \text{ by Lemma 3.24, p40. } \\ 71 | \Rightarrow & \text{For all }a \in M, A_{\frak{M} [\rho[x:=a]]} = T \\ 72 | \Rightarrow & \text{For all }a \in M, A_{\frak{M}[\sigma[x:=a]]} = T, \text{ since $z$ is fresh in $A$. } \\ 73 | \Rightarrow & \frak{M} \models _\sigma \forall x. A 74 | \end{align*} 75 | 76 | 77 | \end{document} -------------------------------------------------------------------------------- /Solutions/Chap6/Lecture6.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/sleepycoke/Mathematical_Logic_NJUCS/88e4550da4597d5343fe5a595f94a544ed475d70/Solutions/Chap6/Lecture6.pdf -------------------------------------------------------------------------------- /Solutions/Chap6/Lecture6.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \usepackage{CJKutf8} 3 | \usepackage{amsmath} 4 | \usepackage{amssymb} 5 | \usepackage{mathrsfs} 6 | \usepackage{enumerate} 7 | \usepackage{amsthm} 8 | \usepackage{bussproofs} 9 | \usepackage{hyperref} 10 | 11 | \renewcommand{\today}{\number\year 年 \number\month 月 \number\day 日} 12 | 13 | \setlength\parindent{0pt} 14 | \vspace{-8ex} 15 | %\date{} 16 | \begin{document} 17 | 18 | \begin{CJK*}{UTF8}{gbsn} 19 | \title{数理逻辑第六讲习题参考答案\footnote{Ver. 1.2. 原答案由宋方敏教授给出手稿, 乔羽同学录入, 最后由丁超同学修订补充. 此文档来源为\mbox{\url{https://github.com/sleepycoke/Mathematical_Logic_NJUCS}}, 20 | 欢迎各位同学提出意见共同维护. 21 | }} 22 | \maketitle 23 | \section{} 24 | \subsection{} 25 | \begin{proof}[\emph{证明: }] 26 | 设 $\Gamma := \Phi \cap \Psi $. 27 | 28 | 反设 $Incon(\Gamma$), 则存在非空有穷公式集 $\Delta \subseteq \Gamma$ 使 $\Delta \vdash$ 可证. 29 | 30 | 因为 $\Delta \subseteq \Phi $ 所以 $Incon(\Phi)$, 与 $Con (\Phi)$ 矛盾. 31 | \end{proof} 32 | 33 | 34 | \subsection{} 35 | 由于等词的存在, 我们可以确保公式集非空. 36 | 37 | 设$A$为任一公式, 38 | $\Phi := \{ A\}$, $\Psi := \{ \neg A \}$. 39 | 40 | 易见 $Con ({\Phi})$且 $Con ({\Psi})$, 但 $Incon (\Phi \cup \Psi)$. 41 | 42 | \section{} 43 | \subsection{} 44 | \begin{proof}[\emph{证明: }] 45 | $A , A \rightarrow B \vdash B$ 可证; 又 $A,A\rightarrow B \in \Phi$, 46 | 由命题6.7知 $B \in \Phi$. 47 | \end{proof} 48 | 49 | \subsection{} 50 | \begin{proof}[\emph{证明: }] 51 | 52 | 因为 $\forall x. A \vdash A [\frac{t}{x}]$ 可证. 53 | 54 | 所以, 由命题6.7可知 $A [\frac{t}{x}]\in \Phi$. 55 | \end{proof} 56 | 57 | \begin{samepage} 58 | \section{} 59 | \begin{proof}[\emph{证明: }] 60 | 61 | 首先注意根据本书的定义, 一阶语言都是可数的. 那么 62 | 设 $\Phi$为$\mathscr{L}$的公式集且$Con(\Phi)$, 63 | 令$\mathscr{L}$的全体公式集为 $\{\varphi_n | n \in N\}$, 64 | 65 | $$\Gamma_0 := \Phi$$ 66 | 67 | \begin{equation*} 68 | \Gamma_{n+1} := \left\{ {\begin{array}{*{20}{c}} 69 | {\Gamma_n \cup \{\varphi_n\}}&{Con (\Gamma\cup \{\varphi_n\})}\\ 70 | \Gamma_n&{Incon ( \Gamma\cup \{\varphi_n\})} 71 | \end{array}} \right. 72 | \end{equation*} 73 | 74 | $$\Gamma := \cup_{n\in N} \Gamma_n,$$ 75 | 76 | 以下证明 77 | \begin{enumerate}[(1)] 78 | 79 | \item $\Phi \subseteq \Gamma$, 即$\Gamma$为$\Phi$的扩张, 易见. 80 | 81 | \item $Con(\Gamma)$. 对$n$作归纳易见$Con(\Gamma_n)$. 82 | 83 | 若$Incon(\Gamma)$那么存在有穷集$\Delta \subseteq \Gamma$使得$\Delta \vdash$可证. 注意到对于任意$\psi \in \Delta$, 存在$k\in \mathbb{N}$使得$\psi \in \Gamma_k$, 也就是有函数$f : \Delta \rightarrow \mathbb{N}$, $f(\psi) = k$. 取$m := \max\{f(\psi)|\psi \in \Delta\}$, 从而$\Delta \subseteq \Gamma_{m}$. 于是$Incon(\Gamma_{m})$, 矛盾. 84 | 85 | 86 | 因此 $Con(\Gamma)$. 87 | 88 | \item $\Gamma$为极大协调. 即若$Con(\Gamma \cup \{\varphi_n\})$，则$\varphi_n \in \Gamma$. 89 | 90 | 设$Con(\Gamma \cup \{\varphi_n\})$. 91 | 92 | Case1. $Con(\Gamma_n \cup \{\varphi_n\})$, 从而$\varphi \in \Gamma_n \cup \{\varphi_n\} =\Gamma_{n+1} $. 又$\Gamma_{n+1} \subseteq \Gamma$, 有$\varphi \in \Gamma$ 93 | 94 | Case2. $Incon(\Gamma_n \cup \{\varphi_n\})$, 从而有穷集$\Delta \subseteq \Gamma_n$使得$\Delta, \varphi \vdash$可证.又$\Delta \subseteq \Gamma$, 有$Incon(\Gamma \cup \{\varphi_n\})$, 矛盾. 95 | \end{enumerate} 96 | 97 | 因此$\Gamma$为$\Phi$的扩张且$\Gamma$是极大协调的. 98 | \end{proof} 99 | \end{samepage} 100 | 101 | \section{} 102 | 103 | \subsection{} 104 | \begin{proof}[\emph{证明: }] 105 | 106 | 令$P(x)$为$x \doteq s$. 107 | 108 | 从而$s \doteq t , P(s) \vdash P(t)$可证, 事实上为公理(定义6.10(3)). 109 | 110 | 即$s \doteq t , s\doteq s \vdash t\doteq s$可证. 111 | 112 | 又$\vdash s\doteq s$可证. 113 | 114 | 故$s\doteq t \vdash t\doteq s$可证(由前两条Cut得). 115 | 116 | 因此 $\vdash (s\doteq t )\rightarrow (t \doteq s)$可证(对上条使用$\rightarrow R$). 117 | \end{proof} 118 | 119 | \subsection{} 120 | \begin{proof}[\emph{证明: }] 121 | 122 | 令 $P(x)$为$c \doteq u$, 123 | 124 | 从而 125 | $t\doteq s , P(t) \vdash P(s)$可证, 126 | 127 | 即$t\doteq s , t \doteq u \vdash s\doteq u$, 可证. 128 | 129 | 又 $s \doteq t \vdash t \doteq s$ 可证, 130 | 131 | 故 $s\doteq t , t \doteq u \vdash s\doteq u$ 可证(由前两条Cut得), 132 | 133 | 从而 134 | $\vdash (s \doteq t)\rightarrow ((t \doteq u)\rightarrow(s\doteq u))$可证(对上条使用两次$\rightarrow R$). 135 | \end{proof} 136 | \section{} 137 | \begin{proof}[\emph{证明: }] 138 | 139 | 因为$||\mathscr{L}|| = \aleph_0$(注：事实上按书上定义的一阶语言都是可数的) 140 | 141 | 所以$||\mathcal{T}erm|| = \aleph_0$, $||\mathcal{F}ormula|| = \aleph_0$. 142 | 令$\mathscr{L'} := \mathscr{L} \cup \{c_n | n \in \mathbb{N}\}$, 143 | 144 | $\Phi$\text{有模型} 145 | $\Rightarrow Con(\Phi)$ 146 | $\Rightarrow \text{存在}\Phi\text{在}\mathscr{L'}\text{中的Henkin集扩张}\Psi. $\text{(定理6.17)}. 147 | 148 | 又$\Psi$是Hintikka集(定理6.18), 可根据定义3.11和引理3.33找到$\Psi$的一个$\mathscr{L'}$中的可数模型$\mathbb{H}$. 由于$\Phi \subseteq \Psi$, $\mathbb{H}$也同时满足了$\Phi$中的所有公式. 那么我们最后只要定义$\mathbb{M}$为$\mathbb{H}$去掉$\mathscr{L'} \backslash \mathscr{L}$中额外加入的常元的解释后得到的结构, $\mathbb{M}$就是$\Phi$的一个(在$\mathscr{L}$中的)可数模型. 149 | \end{proof} 150 | 151 | 此问题有更一般的结果，参见 L\"owenheim-Skolem Theorem. 152 | \section{} 153 | \begin{proof}[\emph{证明: }] 154 | 155 | $\Gamma , A[\frac{c}{x}] \models B$ 156 | 157 | $\Rightarrow \Gamma , A[\frac{c}{x}] \vdash B$可证(完全性). 158 | 159 | $\Rightarrow \Gamma , A[\frac{c}{x}] \vdash B$ 有证明树, 设为$T(c)$. 160 | 161 | $\Rightarrow \Gamma , A[\frac{y}{x}] \vdash B$ 有证明树$T(y)$. 162 | 163 | 这里$T(y)$为在$T(c)$中由$y$替换$c$而得, 且$y$为新变元. 理由同命题10.10. 可由结构归纳得. 164 | 165 | $\Rightarrow\Gamma , \exists xA\vdash B$可证(规则$\exists L$). 166 | 167 | $\Rightarrow \Gamma , \exists xA\models B$(有效性). 168 | \end{proof} 169 | \section{} 170 | 本题借用了LK中的$WL,WR$规则, 其定义参见定义10.6, 在G中的正确性由命题4.12保证. 171 | \begin{enumerate}[(1)] 172 | \item 反例: $M:=\{0,1\}$, $P_{\frak{M}}:=\{0\}$. 173 | \item 174 | 175 | \begin{proof}[\emph{证明: }] ~ 176 | \begin{prooftree} 177 | \AxiomC{$Q(z), P(z)\vdash Q(z)$} 178 | \RightLabel{$\rightarrow R$} 179 | 180 | \UnaryInfC{$ Q(z)\vdash P(z)\rightarrow Q(z) $} 181 | \RightLabel{$WL$} 182 | \UnaryInfC{$ \forall y Q(y) , Q(z) \vdash P(z) \rightarrow Q(z)$} 183 | \RightLabel{$\forall L, \forall R$} 184 | \UnaryInfC{$\forall y Q(y) \vdash \forall x (P(x) \rightarrow Q(x))$} 185 | 186 | \AxiomC{$P(z)\vdash Q(z), P(z)$} 187 | \RightLabel{$\rightarrow R$} 188 | \UnaryInfC{$\vdash P(z) \rightarrow Q(z), P(z)$} 189 | \RightLabel{$WR$} 190 | \UnaryInfC{$\vdash P(z) \rightarrow Q(z), P(z),\exists x P(x)$} 191 | \RightLabel{$\exists R, \forall R$} 192 | \UnaryInfC{$\vdash \forall x(P(x) \rightarrow Q(x)), \exists x P(x)$} 193 | 194 | \RightLabel{$\rightarrow L$} 195 | \BinaryInfC{$\exists x P(x) \rightarrow \forall yQ(y) \vdash \forall x(P(x) \rightarrow Q(x))$} 196 | \RightLabel{$\rightarrow R$} 197 | \UnaryInfC{$\vdash (\exists x P(x) \rightarrow \forall yQ(y)) \rightarrow (\forall x(P(x) \rightarrow Q(x))$} 198 | \end{prooftree} 199 | %这里$z$是新变元. 不必说明 200 | \end{proof} 201 | 202 | 203 | 204 | 205 | \item 反例: $M:=\{0,1\}$, $P_{\frak{M}}:= Q_{\frak{M}} := \{0\}$. 206 | 207 | \item 208 | \begin{proof}[\emph{证明: }] 209 | 设$A$为任一公式, 我们有: 210 | \begin{prooftree} 211 | \AxiomC{$\neg A\vdash \neg A$} 212 | \AxiomC{$A\vdash A$} 213 | \RightLabel{$\neg R$} 214 | \UnaryInfC{$\vdash A, \neg A$} 215 | \RightLabel{$\rightarrow L$} 216 | \BinaryInfC{$A \rightarrow \neg A \vdash \neg A$} 217 | 218 | \AxiomC{$A\vdash A$} 219 | \AxiomC{$A\vdash A$} 220 | \RightLabel{$\neg R$} 221 | \UnaryInfC{$\vdash A, \neg A$} 222 | \RightLabel{$\rightarrow L$} 223 | \BinaryInfC{$\neg A \rightarrow A \vdash A$} 224 | \RightLabel{$\neg L$} 225 | \UnaryInfC{$\neg A \rightarrow A, \neg A \vdash $} 226 | 227 | 228 | \RightLabel{$Cut, \wedge L$} 229 | \BinaryInfC{$A \leftrightarrow \neg A \vdash$} 230 | \end{prooftree} 231 | 232 | 从而有$R(z, z)\leftrightarrow \neg R(z, z)\vdash$可证. 再注意到 233 | $$R(z, z) = R(x, z)[\frac{z}{x}], \quad \neg R(z, z) = \neg R(x, x) [\frac{z}{x}]$$ 234 | 继而有 235 | $$R(z, z)\leftrightarrow \neg R(z, z) = (R(x, z)\leftrightarrow \neg R(x, x))[\frac{z}{x}]. $$ 236 | 那么我们有证明树 237 | \begin{prooftree} 238 | \AxiomC{$R(z, z)\leftrightarrow \neg R(z, z)\vdash$} 239 | \RightLabel{$WL$} 240 | \UnaryInfC{$\forall x(R(x, z)\leftrightarrow \neg R(x, x)), R(z, z)\leftrightarrow \neg R(z, z)\vdash$} 241 | \RightLabel{$\forall L$} 242 | \UnaryInfC{$\forall x(R(x, z)\leftrightarrow \neg R(x, x))\vdash$}\RightLabel{$\exists L$} 243 | \UnaryInfC{$\exists y\forall x(R(x, y)\leftrightarrow \neg R(x, x))\vdash$} 244 | \RightLabel{$\neg R$} 245 | \UnaryInfC{$\vdash \neg \exists y\forall x(R(x, y)\leftrightarrow \neg R(x, x))$} 246 | 247 | \end{prooftree} 248 | \end{proof} 249 | \end{enumerate} 250 | 251 | 252 | \end{CJK*} 253 | 254 | 255 | \end{document} -------------------------------------------------------------------------------- /Solutions/Chap7/Lecture7.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/sleepycoke/Mathematical_Logic_NJUCS/88e4550da4597d5343fe5a595f94a544ed475d70/Solutions/Chap7/Lecture7.pdf -------------------------------------------------------------------------------- /Solutions/Chap7/Lecture7.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \usepackage{ctex} 3 | \usepackage{fontspec, xunicode, xltxtra} 4 | \usepackage{amsmath} 5 | \usepackage{amssymb} 6 | \usepackage{mathrsfs} 7 | \usepackage{enumerate} 8 | \usepackage{amsthm} 9 | \usepackage{bussproofs} 10 | \usepackage{hyperref} 11 | %\usepackage{mathabx} 12 | \usepackage[a4paper,top=30mm,bottom=30mm,left=30mm,right=30mm]{geometry} 13 | 14 | \newcommand{\vv}[0]{\vdash\dashv} 15 | \renewcommand{\today}{\number\year 年 \number\month 月 \number\day 日} 16 | 17 | \setlength\parindent{0pt} 18 | \vspace{-8ex} 19 | %\date{} 20 | \begin{document} 21 | 22 | \title{数理逻辑第七讲习题参考答案\footnote{Ver. 1.4. 原答案由宋方敏教授给出手稿, 乔羽同学录入, 最后由丁超同学修订补充. 此文档来源为{\url{https://github.com/sleepycoke/Mathematical_Logic_NJUCS}}, 23 | 欢迎各位同学提出意见共同维护. 24 | }} 25 | \maketitle 26 | 经黄毅飞提醒, 默认$P,Q,R$为谓词符, $f$为函数符, $c$为常元符, $A$是公式, $x, y, z, u$是变元. 27 | \section{} 28 | 此题答案由陈劭源给出. 29 | \begin{align*} 30 | &(\forall x \exists y \forall z \exists u P(x, y, z, u))^s \\ 31 | =& \forall x (\exists y \forall z \exists u P(x, y, z, u))^s \\ 32 | =& \forall x (\forall z \exists u P(x, f(x), z, u))^s\quad \text{这里$f$为一元新函数}\\ 33 | =& \forall x \forall z (\exists u P(x, f(x), z, u))^s \\ 34 | =& \forall x \forall z (P(x, f(x), z, g(x, z)))^s \quad \text{这里$g$为二元新函数}\\ 35 | =& \forall x \forall z P(x, f(x), z, g(x, z)) 36 | \end{align*} 37 | 38 | \section{} 39 | 此题答案由杨嘉文板书补充得. 40 | \begin{proof} 41 | 引入记号``$\vv$'': 42 | $$A \vv B \Leftrightarrow A\vdash B \text{且} B \vdash A. $$ 43 | 那么显然有 44 | \[\vdash A \leftrightarrow B \Leftrightarrow A\vv B \Rightarrow A \rightarrow C \vv B \rightarrow C \label{lemma} \tag{*}\] 45 | 且它是一个等价关系. 46 | \begin{align*} 47 | &(\forall x P(x) \wedge \forall y Q(y)) \rightarrow \exists z P(z) \\ 48 | \vv & (\forall x \forall y (P(x) \wedge Q(y))) \rightarrow \exists z P(z) \quad &\text{命题7.4(3), (\ref{lemma}})\\ 49 | \vv & \exists x \exists y (P(x) \wedge Q(y) \rightarrow \exists z P(z)) \quad &\text{命题7.4(5), 命题7.3(1)}\\ 50 | \vv & \exists x \exists y \exists z (P(x) \wedge Q(y) \rightarrow P(z))\quad &\text{命题7.4(8), 命题7.3(1)}\\ 51 | \end{align*} 52 | 于是 53 | $$\vdash [(\forall x P(x) \wedge \forall y Q(y)) \rightarrow \exists z P(z)]\leftrightarrow [\exists x \exists y \exists z (P(x) \wedge Q(y) \rightarrow P(z))].$$ 54 | \end{proof} 55 | 56 | 57 | \section{} 58 | 丁超在简写孙旭东的板书时引入了一个明显且致命的错误, 感谢黄毅飞指出它并给出另一个证明: 我们证明一个更强的结论，$FV(A^s) \subseteq FV(A)$ 59 | \begin{proof} 60 | 设$A$的量词数为$n$, 我们对$n$进行归纳证明. 61 | \begin{itemize} 62 | \item[Basis:] $n = 0$. 此时$A^s = A$, 从而$FV(A^s) \subseteq FV(A)$. 63 | \item[I.H.:] $n \le k$时$FV(A^s) \subseteq FV(A)$. 64 | \item[I.S.:] 设 $n = k + 1$, $A$呈形$Q x B$, 从而$B$的量词个数为$k$, 以及$FV(A) = FV(B) \backslash \{x\}$. 65 | 66 | 若$Q$为$\forall$, 那么$A^s = \forall x B^s$，则$FV(A^s) = FV(B^s) \backslash \{x\}$. 又根据归纳假设有$FV(B^s) \subseteq FV(B)$, 有$FV(A^s) = FV(B^s) \backslash \{x\} \subseteq FV(B) \backslash \{x\} = FV(A)$. 67 | 68 | 若$Q$为$\exists$, 那么不妨设$FV(A) = \{x_1,x_2,...,x_i\}$($FV(A)=\emptyset$情况类似, 不赘述), 从而$A^s = (B[\frac{f(x_1,x_2,...,x_i)}{x}])^s$. 注意到$B[\frac{f(x_1,x_2,...,x_i)}{x}]$的量词只有$k$个, 因此根据归纳假设我们有$FV((B[\frac{f(x_1,x_2,...,x_i)}{x}])^s) \subseteq FV(B[\frac{f(x_1,x_2,...,x_i)}{x}])$. 另有$FV(B[\frac{f(x_1,x_2,...,x_i)}{x}]) \subseteq FV(A)$, 整体有 69 | $$FV(A^s)=FV((B[\frac{f(x_1,x_2,...,x_i)}{x}])^s) \subseteq FV(A).$$ 70 | \end{itemize} 71 | 综上, $FV(A^s) \subseteq FV(A)$，从而有$FV(A) = \emptyset \Rightarrow FV(A^s) = \emptyset$ 72 | \end{proof} 73 | \section{} 74 | 先给出一个语义证明: 75 | 76 | \begin{proof} 77 | 78 | 设$\mathfrak{M}$为任意结构. 79 | 80 | 设 $\mathfrak{M} \vDash \exists x \forall y P(x,y)$,以下证明$\mathfrak{M} \vDash \forall y \exists x P(x,y)$. 81 | 82 | $\mathfrak{M} \vDash \exists x \forall y P(x,y)$ 83 | 84 | $\Rightarrow$存在$a \in M$ 对任何$b\in M$, 有$(a,b)\in P_{\frak{M}}$ 85 | 86 | $ \Rightarrow$对任何$b\in M$, 有$a\in M$使得$(a,b)\in P_{\frak{M}}$ 87 | 88 | $\Rightarrow \mathfrak{M}\vDash \forall y \exists x P(x,y)$ 89 | \end{proof} 90 | 91 | 再给出陈劭源在$G$中的语法证明: 92 | \begin{proof} 93 | ~ 94 | \begin{prooftree} 95 | \AxiomC{Axiom} 96 | \noLine 97 | \UnaryInfC{$P(u, v) \vdash P(u, v)$} \RightLabel{$\forall L, \exists R$} 98 | \doubleLine 99 | \UnaryInfC{$\forall y P(u, y) \vdash \exists x P(x, v)$} \RightLabel{$\forall R, \exists L$} 100 | \doubleLine 101 | \UnaryInfC{$\exists x \forall y P(x, y) \vdash \forall y \exists x P(x, y)$} \RightLabel{$\rightarrow R$} 102 | \UnaryInfC{$\vdash \exists x \forall y P(x, y) \rightarrow \forall y \exists x P(x, y)$} 103 | \end{prooftree} 104 | \end{proof} 105 | 106 | \section{} 107 | 直接引用陈劭源的答案: 108 | \begin{proof} 109 | 构造模型$\mathfrak{M} = (M, \sigma)$，其中$M = \{1, 2\}$，$P_{\frak{M}} = \{(1,1), (2,2)\}$，则易验证\\${\mathfrak{M} \nvDash_\sigma \forall x \exists y P(x, y) \rightarrow \exists y \forall x P(x, y)}$，从而$\nvDash\forall x \exists y P(x, y) \rightarrow \exists y \forall x P(x, y)$. 110 | \end{proof} 111 | 112 | 113 | \section{} 114 | 先给出一个语义证明: 115 | \begin{proof} 116 | 设$\mathfrak{M}$为任意结构 117 | 118 | $\mathfrak{M} \vDash \forall x P(x,f(x)) $ 119 | 120 | $\Rightarrow$对任何$a\in M, (a, f_\frak{M}(a))\in P_{\frak{M}}$ 121 | 122 | $\Rightarrow$对任何$a\in M $有$b \in M$使$(a, b)\in P_{\frak{M}}$ 123 | 124 | $\Rightarrow \mathfrak{M} \vDash \forall x \exists y P(x, y)$ 125 | 126 | 所以$\mathfrak{M} \vDash \forall x P(x, f(x)) \Rightarrow \mathfrak{M} \vDash \forall x \exists y P(x, y)$ 127 | 128 | 即$\vDash \forall x P(x, f(x)) \rightarrow \forall x \exists y P(x, y)$ 129 | \end{proof} 130 | 131 | 再引用陈劭源在$G$中的语法证明: 132 | \begin{proof} 133 | ~ 134 | \begin{prooftree} 135 | \AxiomC{Axiom} 136 | \noLine 137 | \UnaryInfC{$P(z, f(z)) \vdash P(z, f(z))$} \RightLabel{$\exists R$} 138 | \UnaryInfC{$P(z, f(z)) \vdash \exists y P(z, y)$} \RightLabel{$\forall L$} 139 | \UnaryInfC{$\forall x P(x, f(x)) \vdash \exists y P(z, y)$} \RightLabel{$\forall R$} 140 | \UnaryInfC{$\forall x P(x, f(x)) \vdash \forall x \exists y P(x, y)$} \RightLabel{$\rightarrow R$} 141 | \UnaryInfC{$\vdash \forall x P(x, f(x)) \rightarrow \forall x \exists y P(x, y)$} 142 | \end{prooftree} 143 | \end{proof} 144 | 145 | \section{} 146 | 此题实际上是定理7.7的一个直接结果. 147 | 148 | 首先给出宋方敏教授的证明: 149 | \begin{proof} 150 | $ \forall x \exists y P(x, y) $可满足 151 | 152 | $\Rightarrow$存在结构$\mathfrak{M} \models \forall x \exists y P(x, y)$. 设其论域为$M$. 153 | 154 | $\Rightarrow$对任何$a \in M$有$b \in M$使得$(a, b) \in P_{\frak{M}}$. 155 | 156 | $\Rightarrow$对任何$a\in M$, $M_a := \{b|(a, b)\in P_{\frak{M}}\} \neq \emptyset$. 157 | 158 | 由选择公理知,存在$\tau : \mathcal{P} (M) \backslash \{\emptyset\} \rightarrow M$. 159 | 160 | 161 | 162 | $\Rightarrow$对任何$a\in M$, $\tau(M_a) \in M_a$ 163 | 164 | $\Rightarrow$对任何$a\in M$, $(a,\tau(M_a)) \in P_{\frak{M}}$, 165 | 166 | $\Rightarrow$令$f_{\frak{M'}}=\{(a,\tau(M_a))|a \in M\}$, $\mathfrak{M'} = \mathfrak{M} +f_{\frak{M'}}$. 167 | 168 | 那么有$\mathfrak{M'} \vDash \forall x P(x, f(x))$, 169 | 即$\forall x P(x, f(x))$可满足. 170 | 171 | \end{proof} 172 | 173 | 下面给出陈劭源的一个更具体的证明: 174 | \begin{proof} 175 | 构造模型$\mathfrak{M} = (M, \sigma)$, 其中$M = \{0\}$，$P_{\frak{M}} = \{(0, 0)\}$，$f_{\frak{M'}}(0) = 0$, $\mathfrak{M'} = \mathfrak{M} + f_{\frak{M'}}$. 则易验证$\mathfrak{M} \models \forall x \exists y P(x, y)$和$\mathfrak{M'} \models \forall x P(x, f(x))$. 从而$\forall x \exists y P(x, y)$可满足$\Rightarrow$ $\forall x P(x, f(x))$可满足. 176 | \end{proof} 177 | 178 | 读到这里想必你已经发现，由于此题的两个公式给得太具体了，我们只要证出以下命题之一就足够 179 | 180 | \begin{itemize} 181 | \item $\forall x \exists y P(x, y)$矛盾; 182 | \item $\forall x P(x, f(x))$可满足. 183 | \end{itemize} 184 | 185 | 而事实上后者很容易就可以构造出模型. 186 | 187 | \section{} 188 | 189 | 设$A$为$P(f(c))$, 190 | 191 | $H_0=\{c\}$, 192 | 193 | $H_1= H_0 \cup \{f(t)|t\in H_0\}= \{c\}\cup \{f(c)\} =\{c, f(c)\}$, 194 | 195 | $H_2=\{c,f(c),f^2(c)\}$, 196 | 197 | ... 198 | 199 | $H_n=\{c,f(c),...,f^n(c)\}$, 200 | 201 | $H_A=\{f^n(c)|n \in N\}$. 202 | 203 | 204 | \section{} 205 | 丁超认为此题结论要修正为$|H_n| < \aleph_0$而且$|H_A| \le \aleph_0$. 206 | \begin{proof} 207 | 若$A$中无函数符，那么对任何$n$, $H_A = H_n = H_0$, 结论成立. 208 | 209 | 若$A$中有函数符, 对$n$归纳证明$H_n$有穷. 210 | \begin{itemize} 211 | \item[Basis:]$n=0, H_0=\{c_0\}$或$\{c|c$为$\alpha$中常元$\}$, 易见$H_n$有穷. 212 | 213 | \item[I.H.:] $H_k$有穷 214 | 215 | \item[I.S.:] $H_{k+1} = H_k \cup \{ f(t_1,...,t_m)|t_i \in H_k$, $f$为$A$中$m$元函数符.\} 216 | 217 | 注意到对于特定的函数符$f_0$，$|\{f_0(t_1,...,t_m)|t_i \in H_k\}| = |H_k^m|$. 218 | 由归纳假设$H_k$有穷，从而上式$< \aleph_0$. 又由于$A$的函数符集是有穷的，$\{ f(t_1,...,t_m)|t_i \in H_k$, $f$为$A$中$m$元函数符.\}有穷. 219 | 220 | 从而$H_{k+1}$有穷. 221 | 222 | 223 | \end{itemize} 224 | 于是$H_n$有穷. 225 | 226 | 最后$H_A$是可数个有穷集的并，一定可数. 而由第8题知$H_A$又是无穷的, 那么就一定有$|H_A| = \aleph_0$. 227 | 228 | \end{proof} 229 | 230 | 231 | \end{document} -------------------------------------------------------------------------------- /Solutions/README.MD: -------------------------------------------------------------------------------- 1 | Since cheaters cheat anyway, why not make it fair? 2 | --------------------------------------------------------------------------------