Theorem rexeqbidva 3330

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 3187	. 2 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))
3		raleqbidva.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
4	3	rexeqdv 3324	. 2 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜒 ↔ ∃𝑥 ∈ 𝐵 𝜒))
5	2, 4	bitrd 282	1 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐵 𝜒))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1570   ∈ wcel 2143  ∃wrex 3089

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1810  df-cleq 2755  df-rex 3090

This theorem is referenced by:  rexeqbidvv  3332  catpropd  17760  addsval  28155  istrkgcb  28725  isperp  28992  perpcom  28993  eengtrkg  29336  eengtrkge  29337  opprqusdrng  33775  fldextrspunlsplem  34063  afsval  35061  matunitlindflem2  38268  rrxlines  49513