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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 → ( 𝜑 ↔ 𝜓 ) )
	Assertion	ceqsexgv	⊢ ( 𝐴 ∈ 𝑉 → ( ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ↔ 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		ceqsexgv.1	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2		id	⊢ ( 𝑥 = 𝐴 → 𝑥 = 𝐴 )
3	2 1	cgsexg	⊢ ( 𝐴 ∈ 𝑉 → ( ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ↔ 𝜓 ) )