Theorem rexbidvaALT 3278

Description: Alternate proof of rexbidva 3184, shorter but requires more axioms. (Contributed by NM, 9-Mar-1997.) (New usage is discouraged.) (Proof modification is discouraged.)

Hypothesis

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

Assertion

Ref	Expression
rexbidvaALT	⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))

Distinct variable group:   𝜑,𝑥

Allowed substitution hints:   𝜓(𝑥)   𝜒(𝑥)   𝐴(𝑥)

Proof of Theorem rexbidvaALT

Step	Hyp	Ref	Expression
1		nfv 1946	. 2 ⊢ Ⅎ𝑥𝜑
2		rexbidvaALT.1	. 2 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
3	1, 2	rexbida 3274	1 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∈ wcel 2144  ∃wrex 3086

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-12 2212

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816  df-rex 3087

This theorem is used by: (None)