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69 | 69 | ponens, then there are !!{formula}s $!B$ and $!B \lif !A$ in the |
70 | 70 | !!{derivation}, and the induction hypothesis applies to the part of |
71 | 71 | the !!{derivation} ending in those !!{formula}s (since they contain at |
72 | | -least one fewer steps justified by an inference). So, by induction |
| 72 | +least one fewer step justified by an inference). So, by induction |
73 | 73 | hypothesis, $\Gamma \Entails !B$ and $\Gamma \Entails !B \lif |
74 | 74 | !A$. Then $\Gamma \Entails !A$ by |
75 | 75 | \iftag{FOL} |
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80 | 80 | \QR. Then that step has the form $!C \lif \lforall[x][B(x)]$ and |
81 | 81 | there is a preceding step $!C \lif !B(c)$ with $c$ not in $\Gamma$, |
82 | 82 | $!C$, or $\lforall[x][B(x)]$. By induction hypothesis, $\Gamma |
83 | | -\Entails !C \lif \lforall[x][B(x)]$. By |
| 83 | +\Entails !C \lif !B(c)$. By |
84 | 84 | \olref[syn][sem]{thm:sem-deduction}, $\Gamma \cup \{!C\} \Entails |
85 | 85 | !B(c)$. |
86 | 86 |
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95 | 95 | not occur in~$\Gamma$ or~$!C$, $\Sat{M'}{\Gamma \cup \{!C\}}$ by |
96 | 96 | \olref[syn][ext]{cor:extensionality-sent}. Since $\Gamma \cup \{!C\} |
97 | 97 | \Entails !B(c)$, $\Sat{M'}{B(c)}$. Since $!B(c)$ is !!a{sentence}, |
98 | | -$\Sat{M}{!B(c)}[s]$ by |
| 98 | +$\Sat{M'}{!B(c)}[s]$ by |
99 | 99 | \olref[syn][ass]{prop:sentence-sat-true}. $\Sat{M'}{!B(x)}[s]$ iff |
100 | | -$\Sat{M'}{!B(c)}$ by \olref[syn][ext]{prop:ext-formulas} (recall that |
| 100 | +$\Sat{M'}{!B(c)}[s]$ by \olref[syn][ext]{prop:ext-formulas} (recall that |
101 | 101 | $!B(c)$ is just $\Subst{!B(x)}{c}{x}$). So, |
102 | 102 | $\Sat{M'}{!B(x)}[s]$. Since $c$ does not occur in~$!B(x)$, by |
103 | 103 | \olref[syn][ext]{prop:extensionality}, $\Sat{M}{!B(x)}[s]$. But $s$ |
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