Equality implies Substitution

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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.


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

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


\(\ds \) \(\) \(\ds x = x\) 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


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



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

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

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