Theorem ereq2 8737

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 2775	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝑅 = 𝐴 ↔ dom 𝑅 = 𝐵))
2	1	3anbi2d 1469	. 2 ⊢ (𝐴 = 𝐵 → ((Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅) ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅)))
3		df-er 8728	. 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
4		df-er 8728	. 2 ⊢ (𝑅 Er 𝐵 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐵 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
5	2, 3, 4	3bitr4g 317	1 ⊢ (𝐴 = 𝐵 → (𝑅 Er 𝐴 ↔ 𝑅 Er 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ w3a 1103   = wceq 1569   ⊆ wss 3899   ∪ cun 4151  ^◡ccnv 5653  dom cdm 5654   ∘ ccom 5658  Rel wrel 5659   Er wer 8725

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-ex 1812  df-cleq 2755  df-er 8728

This theorem is used by:  iserd  8755  efgval  19943  frgp0  19986  frgpmhm  19991