Theorem breqd

Description: Equality deduction for a binary relation. (Contributed by NM, 29-Oct-2011)

	Ref	Expression
Hypothesis	breq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	breqd	⊢ ( 𝜑 → ( 𝐶 𝐴 𝐷 ↔ 𝐶 𝐵 𝐷 ) )

Proof

Step	Hyp	Ref	Expression
1		breq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		breq	⊢ ( 𝐴 = 𝐵 → ( 𝐶 𝐴 𝐷 ↔ 𝐶 𝐵 𝐷 ) )
3	1 2	syl	⊢ ( 𝜑 → ( 𝐶 𝐴 𝐷 ↔ 𝐶 𝐵 𝐷 ) )