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Metamath Proof Explorer

Theorem rexbidvaALT

Description: Alternate proof of rexbidva , shorter but requires more axioms. (Contributed by NM, 9-Mar-1997) (New usage is discouraged.) (Proof modification is discouraged.)

		Ref	Expression
	Hypothesis	rexbidvaALT.1	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
	Assertion	rexbidvaALT	⊢ ( 𝜑 → ( ∃ 𝑥 ∈ 𝐴 𝜓 ↔ ∃ 𝑥 ∈ 𝐴 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		rexbidvaALT.1	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
2		nfv	⊢ Ⅎ 𝑥 𝜑
3	2 1	rexbida	⊢ ( 𝜑 → ( ∃ 𝑥 ∈ 𝐴 𝜓 ↔ ∃ 𝑥 ∈ 𝐴 𝜒 ) )