Theorem fneq1d

Description: Equality deduction for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011)

	Ref	Expression
Hypothesis	fneq1d.1	⊢ ( 𝜑 → 𝐹 = 𝐺 )
Assertion	fneq1d	⊢ ( 𝜑 → ( 𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		fneq1d.1	⊢ ( 𝜑 → 𝐹 = 𝐺 )
2		fneq1	⊢ ( 𝐹 = 𝐺 → ( 𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴 ) )
3	1 2	syl	⊢ ( 𝜑 → ( 𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴 ) )