Theorem eqbrb

Description: Substitution of equal classes in a binary relation. (Contributed by Peter Mazsa, 14-Jun-2024)

	Ref	Expression
Assertion	eqbrb	|- ( ( A = B /\ A R C ) <-> ( A = B /\ B R C ) )

Proof

Step	Hyp	Ref	Expression
1		simpl	|- ( ( B = A /\ A R C ) -> B = A )
2		eqbrtr	|- ( ( B = A /\ A R C ) -> B R C )
3	1 2	jca	|- ( ( B = A /\ A R C ) -> ( B = A /\ B R C ) )
4		eqcom	|- ( B = A <-> A = B )
5	4	anbi1i	|- ( ( B = A /\ A R C ) <-> ( A = B /\ A R C ) )
6	4	anbi1i	|- ( ( B = A /\ B R C ) <-> ( A = B /\ B R C ) )
7	3 5 6	3imtr3i	|- ( ( A = B /\ A R C ) -> ( A = B /\ B R C ) )
8		simpl	|- ( ( A = B /\ B R C ) -> A = B )
9		eqbrtr	|- ( ( A = B /\ B R C ) -> A R C )
10	8 9	jca	|- ( ( A = B /\ B R C ) -> ( A = B /\ A R C ) )
11	7 10	impbii	|- ( ( A = B /\ A R C ) <-> ( A = B /\ B R C ) )