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	$$⊢FFnA\leftrightarrow GFnA$$

Proof

Step	Hyp	Ref	Expression
1		fneq1i.1	$$⊢F=G$$
2		fneq1	$$⊢F=G\to \left(FFnA\leftrightarrow GFnA\right)$$
3	1 2	ax-mp	$$⊢FFnA\leftrightarrow GFnA$$