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Metamath Proof Explorer

Theorem equsex

Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 . See equsexvw and equsexv for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsal . See equsexALT for an alternate proof. (Contributed by NM, 5-Aug-1993) (Revised by Mario Carneiro, 3-Oct-2016) (Proof shortened by Wolf Lammen, 6-Feb-2018) (New usage is discouraged.)

	Ref	Expression
Hypotheses	equsal.1	$⊢ Ⅎ x ψ$
	equsal.2	$⊢ x = y \to (φ \leftrightarrow ψ)$
Assertion	equsex	$⊢ \exists x (x = y \land φ) \leftrightarrow ψ$

Proof

Step	Hyp	Ref	Expression
1		equsal.1	$⊢ Ⅎ x ψ$
2		equsal.2	$⊢ x = y \to (φ \leftrightarrow ψ)$
3	2	biimpa	$⊢ (x = y \land φ) \to ψ$
4	1 3	exlimi	$⊢ \exists x (x = y \land φ) \to ψ$
5	1 2	equsal	$⊢ \forall x (x = y \to φ) \leftrightarrow ψ$
6		equs4	$⊢ \forall x (x = y \to φ) \to \exists x (x = y \land φ)$
7	5 6	sylbir	$⊢ ψ \to \exists x (x = y \land φ)$
8	4 7	impbii	$⊢ \exists x (x = y \land φ) \leftrightarrow ψ$