Theorem releqd

Description: Equality deduction for the relation predicate. (Contributed by NM, 8-Mar-2014)

	Ref	Expression
Hypothesis	releqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	releqd	⊢ ( 𝜑 → ( Rel 𝐴 ↔ Rel 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		releqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		releq	⊢ ( 𝐴 = 𝐵 → ( Rel 𝐴 ↔ Rel 𝐵 ) )
3	1 2	syl	⊢ ( 𝜑 → ( Rel 𝐴 ↔ Rel 𝐵 ) )