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Theorem ereq2 7904
 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 2782 . . 3 (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵))
213anbi2d 1552 . 2 (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅)))
3 df-er 7896 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
4 df-er 7896 . 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 1071   = wceq 1631   ∪ cun 3721   ⊆ wss 3723  ◡ccnv 5248  dom cdm 5249   ∘ ccom 5253  Rel wrel 5254   Er wer 7893 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-9 2154  ax-ext 2751 This theorem depends on definitions:  df-bi 197  df-an 383  df-3an 1073  df-ex 1853  df-cleq 2764  df-er 7896 This theorem is referenced by:  iserd  7922  efgval  18337  frgp0  18380  frgpmhm  18385
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