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Metamath Proof Explorer

Theorem ceqsexgv

Description: Elimination of an existential quantifier, using implicit substitution. (Contributed by NM, 29-Dec-1996) Drop ax-10 and ax-12 . (Revised by GG, 1-Dec-2023)

		Ref	Expression
	Hypothesis	ceqsexgv.1	$⊢ x = A \to (φ \leftrightarrow ψ)$
	Assertion	ceqsexgv	$⊢ A \in V \to (\exists x (x = A \land φ) \leftrightarrow ψ)$

Proof

Step	Hyp	Ref	Expression
1		ceqsexgv.1	$⊢ x = A \to (φ \leftrightarrow ψ)$
2		id	$⊢ x = A \to x = A$
3	2 1	cgsexg	$⊢ A \in V \to (\exists x (x = A \land φ) \leftrightarrow ψ)$