Theorem rexbidvaALT 3283

Description: Alternate proof of rexbidva 3189, 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 1947	. 2 ⊢ Ⅎ𝑥𝜑
2		rexbidvaALT.1	. 2 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
3	1, 2	rexbida 3279	1 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∈ wcel 2146  ∃wrex 3091

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-12 2216

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-rex 3092

This theorem is used by: (None)