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Theorem ereq2 6050
 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 2046 . . 3 (A = B → (dom 𝑅 = A ↔ dom 𝑅 = B))
213anbi2d 1211 . 2 (A = B → ((Rel 𝑅 dom 𝑅 = A (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 dom 𝑅 = B (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅)))
3 df-er 6042 . 2 (𝑅 Er A ↔ (Rel 𝑅 dom 𝑅 = A (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
4 df-er 6042 . 2 (𝑅 Er B ↔ (Rel 𝑅 dom 𝑅 = B (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
52, 3, 43bitr4g 212 1 (A = B → (𝑅 Er A𝑅 Er B))
 Colors of variables: wff set class Syntax hints:   → wi 4   ↔ wb 98   ∧ w3a 884   = wceq 1242   ∪ cun 2909   ⊆ wss 2911  ◡ccnv 4287  dom cdm 4288   ∘ ccom 4292  Rel wrel 4293   Er wer 6039 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-4 1397  ax-17 1416  ax-ext 2019 This theorem depends on definitions:  df-bi 110  df-3an 886  df-cleq 2030  df-er 6042 This theorem is referenced by:  iserd  6068
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