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Theorem rexeqbidva 3327
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
StepHypRef Expression
1 raleqbidva.2 . . 3 ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
21rexbidva 3185 . 2 (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))
3 raleqbidva.1 . . 3 (𝜑 → 𝐴 = 𝐵)
43rexeqdv 3321 . 2 (𝜑 → (∃𝑥 ∈ 𝐴 𝜒 ↔ ∃𝑥 ∈ 𝐵 𝜒))
52, 4bitrd 282 1 (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐵 𝜒))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  ∃wrex 3087
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 2155  ax-ext 2733
This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2753  df-rex 3088
This theorem is used by:  rexeqbidvv  3329  mpt3eqdv  7684  catpropd  17876  matunitlindflem2  22988  addsval  28341  istrkgcb  28911  isperp  29180  perpcom  29181  eengtrkg  29557  eengtrkge  29558  opprqusdrng  34010  fldextrspunlsplem  34298  afsval  35296  rrxlines  49814
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