Equality implies Substitution

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Theorem

Let $\map P x$ denote a Well-Formed Formula which contains $x$ as a free variable.

Then the following are tautologies:

$\forall x: \paren {\map P x \iff \exists y: \paren {y = x \land \map P y} }$
$\forall x: \paren {\map P x \iff \forall y: \paren {y = x \implies \map P y} }$

Note that when $y$ is substituted for $x$ in either formula, it is false in general; compare Confusion of Bound Variables.


Proof

By Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous:

$\paren {\exists y: y = x \land \forall y: \paren {y = x \implies \map P x} } \implies \exists y: \paren {y = x \land \map P x}$

Then:

\(\ds \) \(\) \(\ds x = x\) Equality is Reflexive
\(\ds \) \(\leadsto\) \(\ds \exists y: y = x\) Existential Generalisation
\(\ds \) \(\leadsto\) \(\ds \paren {\forall y: \paren {y = x \implies \map P x} \implies \exists y: \paren {y = x \land \map P x} }\) Modus Ponendo Ponens

$\Box$


\(\ds \paren { y = x \land \map P y}\) \(\implies\) \(\ds \map P x\) Substitutivity of Equality
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \exists y: \paren {y = x \land \map P y}\) \(\implies\) \(\ds \map P x\) Universal Generalisation
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \forall y: \paren {y = x \implies \map P y}\) \(\implies\) \(\ds \map P x\) Hypothetical Syllogism with first lemma

$\Box$


Similarly:

\(\ds \map P x\) \(\leadsto\) \(\ds \paren {y = x \implies \map P y}\) Substitutivity of Equality
\(\text {(3)}: \quad\) \(\ds \) \(\leadsto\) \(\ds \forall y: \paren {y = x \implies \map P y}\) Universal Generalisation
\(\text {(4)}: \quad\) \(\ds \) \(\leadsto\) \(\ds \exists y: \paren {y = x \land \map P y}\) First lemma

The above two statements comprise the other direction of the biconditional assertions.

Together, $(1)$, $(2)$, $(3)$, and $(4)$ prove the two assertions.

$\blacksquare$


Sources