Theorem ereq2 8732

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 2748	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵))
2	1	3anbi2d 1443	. 2 ⊢ (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅)))
3		df-er 8724	. 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
4		df-er 8724	. 2 ⊢ (𝑅 Er 𝐵 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
5	2, 3, 4	3bitr4g 314	1 ⊢ (𝐴 = 𝐵 → (𝑅 Er 𝐴 ↔ 𝑅 Er 𝐵))

Syntax hints:   → wi 4   ↔ wb 206   ∧ w3a 1086   = wceq 1540   ∪ cun 3929   ⊆ wss 3931  ^◡ccnv 5658  dom cdm 5659   ∘ ccom 5663  Rel wrel 5664   Er wer 8721

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-9 2119  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-ex 1780  df-cleq 2728  df-er 8724

This theorem is referenced by:  iserd  8750  efgval  19703  frgp0  19746  frgpmhm  19751