Theorem ereq2 6566

Description: Equality theorem for equivalence predicate. (Contributed by Mario Carneiro, 12-Aug-2015.)

Assertion

Ref	Expression
ereq2	⊢ (𝐴 = 𝐵 → (𝑅 Er 𝐴 ↔ 𝑅 Er 𝐵))

Proof of Theorem ereq2

Step	Hyp	Ref	Expression
1		eqeq2 2199	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵))
2	1	3anbi2d 1328	. 2 ⊢ (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅)))
3		df-er 6558	. 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
4		df-er 6558	. 2 ⊢ (𝑅 Er 𝐵 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
5	2, 3, 4	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (𝑅 Er 𝐴 ↔ 𝑅 Er 𝐵))

Syntax hints:   → wi 4   ↔ wb 105   ∧ w3a 980   = wceq 1364   ∪ cun 3142   ⊆ wss 3144  ^◡ccnv 4643  dom cdm 4644   ∘ ccom 4648  Rel wrel 4649   Er wer 6555

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1458  ax-gen 1460  ax-4 1521  ax-17 1537  ax-ext 2171

This theorem depends on definitions:  df-bi 117  df-3an 982  df-cleq 2182  df-er 6558

This theorem is referenced by:  iserd  6584