Theorem rexeqbidva 3332

Description: Equality deduction for restricted universal quantifier. (Contributed by Mario Carneiro, 5-Jan-2017.)

Hypotheses

Ref	Expression
raleqbidva.1	⊢ (𝜑 → 𝐴 = 𝐵)
raleqbidva.2	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))

Assertion

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

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜑,𝑥

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

Proof of Theorem rexeqbidva

Step	Hyp	Ref	Expression
1		raleqbidva.2	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
2	1	rexbidva 3189	. 2 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))
3		raleqbidva.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
4	3	rexeqdv 3326	. 2 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜒 ↔ ∃𝑥 ∈ 𝐵 𝜒))
5	2, 4	bitrd 282	1 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐵 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ 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-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2757  df-rex 3092

This theorem is used by:  rexeqbidvv  3334  catpropd  17787  addsval  28206  istrkgcb  28776  isperp  29043  perpcom  29044  eengtrkg  29391  eengtrkge  29392  opprqusdrng  33839  fldextrspunlsplem  34127  afsval  35126  matunitlindflem2  38325  rrxlines  49570