Theorem fneq1i

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

	Ref	Expression
Hypothesis	fneq1i.1	⊢ 𝐹 = 𝐺
Assertion	fneq1i	⊢ ( 𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		fneq1i.1	⊢ 𝐹 = 𝐺
2		fneq1	⊢ ( 𝐹 = 𝐺 → ( 𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴 ) )
3	1 2	ax-mp	⊢ ( 𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴 )