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Metamath Proof Explorer

Theorem rexbidvALT

Description: Alternate proof of rexbidv , shorter but requires more axioms. (Contributed by NM, 20-Nov-1994) (New usage is discouraged.) (Proof modification is discouraged.)

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

Proof

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