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 ) )