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Theorem ereq2 7637
 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 2621 . . 3 (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵))
213anbi2d 1396 . 2 (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅)))
3 df-er 7629 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
4 df-er 7629 . 2 (𝑅 Er 𝐵 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
52, 3, 43bitr4g 302 1 (𝐴 = 𝐵 → (𝑅 Er 𝐴𝑅 Er 𝐵))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 195   ∧ w3a 1031   = wceq 1475   ∪ cun 3538   ⊆ wss 3540  ◡ccnv 5037  dom cdm 5038   ∘ ccom 5042  Rel wrel 5043   Er wer 7626 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-an 385  df-3an 1033  df-cleq 2603  df-er 7629 This theorem is referenced by:  iserd  7655  efgval  17953  frgp0  17996  frgpmhm  18001
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