Theorem breq 2064

Description: Equality theorem for a binary relation.

Assertion

Ref	Expression
breq	⊢ (R = S → (ARB ↔ ASB))

Proof of Theorem breq

Step	Hyp	Ref	Expression
1		eleq2 1150	. 2 ⊢ (R = S → (⟨A, B⟩ ∈ R ↔ ⟨A, B⟩ ∈ S))
2		df-br 2063	. 2 ⊢ (ARB ↔ ⟨A, B⟩ ∈ R)
3		df-br 2063	. 2 ⊢ (ASB ↔ ⟨A, B⟩ ∈ S)
4	1, 2, 3	3bitr4g 428	1 ⊢ (R = S → (ARB ↔ ASB))

Syntax hints:   → wi 2   ↔ wb 127   = wceq 1091   ∈ wcel 1092  ⟨cop 1810   class class class wbr 2054

This theorem is referenced by:  poeq1 2130  soeq1 2141  freq1 2174  coeq1 2502  coeq2 2503  isoeq2 2926  isoeq3 2927

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-gen 677  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-cleq 1097  df-clel 1099  df-br 2063

metamath.org