Substitution Theorem for Well-Formed Formulas

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Theorem

Let $\mathbf A$ be a WFF of predicate logic.

Let $x \in \mathrm{VAR}$ be a variable.

Let $\tau$ be a term of predicate logic which is freely substitutable for $x$ in $\mathbf A$.

Let $\map {\mathbf A} {x \gets \tau}$ be the substitution instance of $\mathbf A$ substituting $\tau$ for $x$.

Let $\AA$ be a structure for predicate logic.

Let $\sigma$ be an assignment for $\mathbf A$ and $\tau$.

Suppose that:

$\map {\operatorname{val}_\AA} \tau \sqbrk \sigma = a$

where $\map {\operatorname{val}_\AA} \tau \sqbrk \sigma$ is the value of $\tau$ under $\sigma$.


Then:

$\map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }$

where $\sigma + \paren {x / a}$ is the extension of $\sigma$ by mapping $x$ to $a$.


Corollary

Let $\map {\mathbf A} {x_1, \ldots, x_n}$ be a WFF whose free variables are among $x_1, \ldots, x_n$.

Let $\tau_1, \ldots, \tau_n$ be closed terms, and let each $\tau_i$ be freely substitutable for $x_i$.

Let $a_i = \map {\operatorname{val}_\AA} {\tau_i}$, where $\operatorname{val}_\AA$ denotes the value of $\tau$ in $\AA$.


Then:

$\AA \models_{\mathrm{PL}} \map {\mathbf A} {\tau_1, \ldots, \tau_n}$ if and only if $\AA \models_{\mathrm{PL_A}} \mathbf A \sqbrk {a_1, \ldots, a_n}$

where $\models_{\mathrm{PL}}$ denotes model of sentence, and $\models_{\mathrm{PL_A}}$ denotes model of formula.


Proof

Proceed by the Principle of Structural Induction on the bottom-up specification of predicate logic, applied to $\mathbf A$.


If $\mathbf A = \map p {\tau_1, \ldots, \tau_n}$, then:

$\map {\mathbf A} {x \gets \tau} = \map p {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} }$

where $\map {\tau_i} {x \gets \tau}$ is the substitution instance of $\tau_i$.

Now:

\(\ds \map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma\) \(=\) \(\ds \map {\operatorname{val}_\AA} {\map p {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} } } \sqbrk \sigma\)
\(\ds \) \(=\) \(\ds \map {p_\AA} {\map {\operatorname{val}_\AA} {\map {\tau_1} {x \gets \tau} } \sqbrk \sigma, \ldots, \map {\operatorname{val}_\AA} {\map {\tau_n} {x \gets \tau} } \sqbrk \sigma}\) Definition of Value under $\sigma$
\(\ds \) \(=\) \(\ds \map {p_\AA} {\map {\operatorname{val}_\AA} {\tau_1} \sqbrk {\sigma + \paren {x / a} }, \ldots, \map {\operatorname{val}_\AA} {\tau_n} \sqbrk {\sigma + \paren {x / a} } }\) Substitution Theorem for Terms
\(\ds \) \(=\) \(\ds \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }\) Definition of Value under $\sigma + \paren {x / a}$


Suppose $\mathbf A = \neg \mathbf B$ and the induction hypothesis applies to $\mathbf B$.

Then since $\tau$ is also free for $x$ in $\mathbf B$:

\(\ds \map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma\) \(=\) \(\ds \map {f^\neg} {\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma}\) Definition of Value under $\sigma$
\(\ds \) \(=\) \(\ds \map {f^\neg} {\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {x / a} } }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }\) Definition of Value under $\sigma + \paren {x / a}$


Suppose $\mathbf A = \mathbf B \circ \mathbf B'$ for $\circ$ one of $\land, \lor, \implies, \iff$ and the induction hypothesis applies to $\mathbf B, \mathbf B'$.

Then since $\tau$ is also free for $x$ in $\mathbf B$ and $\mathbf B'$:

\(\ds \map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma\) \(=\) \(\ds \map {f^\circ} {\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma, \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma}\) Definition of Value under $\sigma$
\(\ds \) \(=\) \(\ds \map {f^\circ} {\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {x / a} }, \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {x / a} } }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }\) Definition of Value under $\sigma + \paren {x / a}$


Suppose $\mathbf A = \exists y: \mathbf B$ or $\mathbf A = \forall y: \mathbf B$, and the induction hypothesis applies to $\mathbf B$.

Because $\tau$ is free for $x$ in $\mathbf A$, it must be that either $x$ does not occur freely in $\mathbf A$, or $y$ does not occur in $\tau$.

In the first case:

$\map {\mathbf A} {x \gets \tau} = \mathbf A$

and by Value of Formula under Assignment Determined by Free Variables:

$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }$


Now consider the case where $y$ does not occur in $\tau$.

From the definition of value under $\sigma$, $\map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma$ is determined by:

$\map {\operatorname{val}_\AA} {\map {\mathbf B} {x \gets \tau} } \sqbrk {\sigma + \paren {y / a'} }$

where $a'$ ranges over $\AA$.

Now from Value of Term under Assignment Determined by Variables, since $y$ does not occur in $\tau$:

$\map {\operatorname{val}_\AA} \tau \sqbrk {\sigma + \paren {y / a'} } = \map {\operatorname{val}_\AA} \tau \sqbrk \sigma = a$

for all $a'$.

Hence the induction hypothesis also applies to the assignment $\sigma + \paren {y / a'}$.

Thus, for all $a'$:

\(\ds \map {\operatorname{val}_\AA} {\map {\mathbf B} {x \gets \tau} } \sqbrk {\sigma + \paren {y / a'} }\) \(=\) \(\ds \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {y / a'} + \paren {x / a} }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {x / a} + \paren {y / a'} }\) Definition of Extension of Assignment

from which we infer:

$\map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }$

as desired.


The result follows by the Principle of Structural Induction.

$\blacksquare$


Sources

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