Theorem rexeqbidva 3305

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 3160	. 2 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))
3		raleqbidva.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
4	3	rexeqdv 3299	. 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 1542   ∈ wcel 2114  ∃wrex 3062

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1782  df-cleq 2729  df-rex 3063

This theorem is referenced by:  rexeqbidvv  3308  catpropd  17644  addsval  27970  istrkgcb  28540  isperp  28796  perpcom  28797  eengtrkg  29071  eengtrkge  29072  opprqusdrng  33586  fldextrspunlsplem  33851  afsval  34849  matunitlindflem2  37868  rrxlines  49093