Substitution Instance of Term is Term

Theorem

Let $\beta, \tau$ be terms of predicate logic.

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

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

Then $\map \beta {x \gets \tau}$ is a term.

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 either case, $\map \beta {x \gets \tau}$ is a term.

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

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

By the induction hypothesis, each $\map {\tau_i} {x \gets \tau}$ is a term.

Hence so is $\map \beta {x \gets \tau}$.

The result follows by the Principle of Structural Induction.

$\blacksquare$

Sources

• 2009: Kenneth Kunen: The Foundations of Mathematics ... (previous) ... (next): $\mathrm{II}.8$ Further Semantic Notions