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Theorem ereq2 6430
 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 2147 . . 3 (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵))
213anbi2d 1295 . 2 (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅)))
3 df-er 6422 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
4 df-er 6422 . 2 (𝑅 Er 𝐵 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
52, 3, 43bitr4g 222 1 (𝐴 = 𝐵 → (𝑅 Er 𝐴𝑅 Er 𝐵))
 Colors of variables: wff set class Syntax hints:   → wi 4   ↔ wb 104   ∧ w3a 962   = wceq 1331   ∪ cun 3064   ⊆ wss 3066  ◡ccnv 4533  dom cdm 4534   ∘ ccom 4538  Rel wrel 4539   Er wer 6419 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-gen 1425  ax-4 1487  ax-17 1506  ax-ext 2119 This theorem depends on definitions:  df-bi 116  df-3an 964  df-cleq 2130  df-er 6422 This theorem is referenced by:  iserd  6448
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