Theorem funeqd

Description: Equality deduction for the function predicate. (Contributed by NM, 23-Feb-2013)

	Ref	Expression
Hypothesis	funeqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	funeqd	⊢ ( 𝜑 → ( Fun 𝐴 ↔ Fun 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		funeqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		funeq	⊢ ( 𝐴 = 𝐵 → ( Fun 𝐴 ↔ Fun 𝐵 ) )
3	1 2	syl	⊢ ( 𝜑 → ( Fun 𝐴 ↔ Fun 𝐵 ) )