# 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

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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 | |||||||||||

\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \exists y: \, \) | \(\ds \paren {y = x \land \map P y}\) | \(\implies\) | \(\ds \map P x\) | 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 |

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$\Box$

Similarly:

\(\ds \) | \(\) | \(\ds \map P x\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds y = x\) | \(\implies\) | \(\ds \map P y\) | 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.

$\blacksquare$

## Sources

- 1963: Willard Van Orman Quine:
*Set Theory and Its Logic*: $\S 6.1$ - 1963: Willard Van Orman Quine:
*Set Theory and Its Logic*: $\S 6.2$