Existential Generalisation/Model

Theorem
Let $\mathbf A \left({x}\right)$ be a WFF of predicate logic.

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

Then $\mathbf A \left({\tau}\right) \implies \exists x: \mathbf A \left({x}\right)$ is a tautology.

Proof
Let $\mathcal A$ be a structure on a set $A$, and let $\sigma$ be an assignment for $\forall x: \mathbf A \left({x}\right) \implies \mathbf A \left({\tau}\right)$.

Define:


 * $a_\tau := \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]$

the value of $\tau$ under $\sigma$.

From the definition of value under $\sigma$:


 * $\mathop{ \operatorname{val}_{\mathcal A} \left({\forall x: \mathbf A \left({x}\right) \implies \mathbf A \left({\tau}\right) }\right) } \left[{\sigma}\right] = f^\to \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\forall x: \mathbf A \left({x}\right)}\right) } \left[{\sigma}\right], \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({\tau}\right) }\right) } \left[{\sigma}\right] }\right)$

where $f^\to$ is the truth function of $\implies$.

We thus need to ascertain that if:


 * $\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({\tau}\right)}\right) } \left[{\sigma}\right] = T$

then also:


 * $\mathop{ \operatorname{val}_{\mathcal A} \left({\exists x: \mathbf A \left({x}\right) }\right) } \left[{\sigma}\right] = T$

By the Substitution Theorem for Well-Formed Formulas, the former amounts to:


 * $\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({x}\right)}\right) } \left[{\sigma + \left({x / a_\tau}\right)}\right] = T$

By the definition of value under $\sigma$, the latter reduces to:


 * $\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({x}\right)}\right) } \left[{\sigma + \left({x / a}\right)}\right] = T$

for some $a \in A$.

Since $a_\tau \in A$, the conclusion follows, and $\mathbf A \left({\tau}\right) \implies \exists x: \mathbf A \left({x}\right)$ is a tautology.