Theorem rexeqbidva 3301

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 3156	. 2 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))
3		raleqbidva.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
4	3	rexeqdv 3295	. 2 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜒 ↔ ∃𝑥 ∈ 𝐵 𝜒))
5	2, 4	bitrd 279	1 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐵 𝜒))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1541   ∈ wcel 2113  ∃wrex 3058

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-9 2123  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1781  df-cleq 2726  df-rex 3059

This theorem is referenced by:  rexeqbidvv  3304  catpropd  17630  addsval  27932  istrkgcb  28477  isperp  28733  perpcom  28734  eengtrkg  29008  eengtrkge  29009  opprqusdrng  33523  fldextrspunlsplem  33779  afsval  34777  matunitlindflem2  37757  rrxlines  48921