Theorem ereq2 6509

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 2175	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵))
2	1	3anbi2d 1307	. 2 ⊢ (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅)))
3		df-er 6501	. 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
4		df-er 6501	. 2 ⊢ (𝑅 Er 𝐵 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
5	2, 3, 4	3bitr4g 222	1 ⊢ (𝐴 = 𝐵 → (𝑅 Er 𝐴 ↔ 𝑅 Er 𝐵))

Syntax hints:   → wi 4   ↔ wb 104   ∧ w3a 968   = wceq 1343   ∪ cun 3114   ⊆ wss 3116  ^◡ccnv 4603  dom cdm 4604   ∘ ccom 4608  Rel wrel 4609   Er wer 6498

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-gen 1437  ax-4 1498  ax-17 1514  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-3an 970  df-cleq 2158  df-er 6501

This theorem is referenced by:  iserd  6527