Substitution Theorem for Terms

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Theorem

Let $\beta, \tau$ be terms.

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

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

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

Let $\sigma$ be an assignment for $\beta$ 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:

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

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


Proof

Proceed by the Principle of Structural Induction on the definition of term, applied to $\beta$.


If $\beta = y$ for some variable $y$, then:

$\map \beta {x \gets \tau} = \begin {cases} \tau & : \text{if $y = x$} \\ y & : \text {otherwise} \end {cases}$

In the first case:

\(\ds \map {\operatorname{val}_\AA} {\map \beta {x \gets \tau} } \sqbrk \sigma\) \(=\) \(\ds \map {\operatorname{val}_\AA} \tau \sqbrk \sigma\)
\(\ds \) \(=\) \(\ds a\) Definition of $a$
\(\ds \) \(=\) \(\ds \map {\paren {\sigma + \paren {x / a} } } x\) Definition of Extension of Assignment
\(\ds \) \(=\) \(\ds \map {\operatorname{val}_\AA} \beta \sqbrk {\sigma + \paren {x / a} }\) Definition of Value under $\sigma + \paren {x / a}$

In the second case:

\(\ds \map {\operatorname{val}_\AA} {\map \beta {x \gets \tau} } \sqbrk \sigma\) \(=\) \(\ds \map {\operatorname{val}_\AA} y \sqbrk \sigma\)
\(\ds \) \(=\) \(\ds \map \sigma y\) Definition of Value under $\sigma$
\(\ds \) \(=\) \(\ds \map {\paren {\sigma + \paren {x / a} } } y\) Definition of Extension of Assignment
\(\ds \) \(=\) \(\ds \map {\operatorname{val}_\AA} \beta \sqbrk {\sigma + \paren {x / a} }\) Definition of Value under $\sigma + \paren {x / a}$

as desired.


If $\beta = \map f {\tau_1, \ldots, \tau_n}$ and the induction hypothesis holds for $\tau_1, \ldots, \tau_n$, then:

$\map \beta {x \gets \tau} = \map f {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} }$

Now:

\(\ds \map {\operatorname{val}_\AA} {\map \beta {x \gets \tau} } \sqbrk \sigma\) \(=\) \(\ds \map {\operatorname{val}_\AA} {\map f {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} } } \sqbrk \sigma\)
\(\ds \) \(=\) \(\ds \map {f_\AA} {\map {\operatorname{val}_\AA} {\map {\tau_1} {x \gets \tau} } \sqbrk \sigma, \ldots, \map {\operatorname{val}_\AA} {\map {\tau_n} {x \gets \tau} } \sqbrk \sigma}\) Definition of Value under $\sigma$
\(\ds \) \(=\) \(\ds \map {f_\AA} {\map {\operatorname{val}_\AA} {\tau_1} \sqbrk {\sigma + \paren {x / a} }, \ldots, \map {\operatorname{val}_\AA} {\tau_n} \sqbrk {\sigma + \paren {x / a} } }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \map {\operatorname{val}_\AA} \beta \sqbrk {\sigma + \paren {x / a} }\) Definition of Value under $\sigma + \paren {x / a}$

as desired.


The result follows by the Principle of Structural Induction.

$\blacksquare$


Sources

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