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Theorem ereq2 7915
Description: Equality theorem for equivalence predicate. (Contributed by Mario Carneiro, 12-Aug-2015.)
Assertion
Ref Expression
ereq2 (𝐴 = 𝐵 → (𝑅 Er 𝐴𝑅 Er 𝐵))

Proof of Theorem ereq2
StepHypRef Expression
1 eqeq2 2767 . . 3 (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵))
213anbi2d 1549 . 2 (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅)))
3 df-er 7907 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
4 df-er 7907 . 2 (𝑅 Er 𝐵 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
52, 3, 43bitr4g 303 1 (𝐴 = 𝐵 → (𝑅 Er 𝐴𝑅 Er 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  w3a 1072   = wceq 1628  cun 3709  wss 3711  ccnv 5261  dom cdm 5262  ccom 5266  Rel wrel 5267   Er wer 7904
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1867  ax-4 1882  ax-5 1984  ax-6 2050  ax-7 2086  ax-9 2144  ax-ext 2736
This theorem depends on definitions:  df-bi 197  df-an 385  df-3an 1074  df-ex 1850  df-cleq 2749  df-er 7907
This theorem is referenced by:  iserd  7933  efgval  18326  frgp0  18369  frgpmhm  18374
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