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 ) )