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Theorem ereq2 8284
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 2813 . . 3 (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵))
213anbi2d 1438 . 2 (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅)))
3 df-er 8276 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
4 df-er 8276 . 2 (𝑅 Er 𝐵 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
52, 3, 43bitr4g 317 1 (𝐴 = 𝐵 → (𝑅 Er 𝐴𝑅 Er 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  w3a 1084   = wceq 1538  cun 3882  wss 3884  ccnv 5522  dom cdm 5523  ccom 5527  Rel wrel 5528   Er wer 8273
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 1911  ax-6 1970  ax-7 2015  ax-9 2122  ax-ext 2773
This theorem depends on definitions:  df-bi 210  df-an 400  df-3an 1086  df-ex 1782  df-cleq 2794  df-er 8276
This theorem is referenced by:  iserd  8302  efgval  18839  frgp0  18882  frgpmhm  18887
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