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 ) )