ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  ereq2 GIF version

Theorem ereq2 6314
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 2098 . . 3 (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵))
213anbi2d 1254 . 2 (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅)))
3 df-er 6306 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
4 df-er 6306 . 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 925   = wceq 1290  cun 2998  wss 3000  ccnv 4451  dom cdm 4452  ccom 4456  Rel wrel 4457   Er wer 6303
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1382  ax-gen 1384  ax-4 1446  ax-17 1465  ax-ext 2071
This theorem depends on definitions:  df-bi 116  df-3an 927  df-cleq 2082  df-er 6306
This theorem is referenced by:  iserd  6332
  Copyright terms: Public domain W3C validator