Metamath Proof Explorer


Theorem fneq2d

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

Ref Expression
Hypothesis fneq2d.1 φ A = B
Assertion fneq2d φ F Fn A F Fn B

Proof

Step Hyp Ref Expression
1 fneq2d.1 φ A = B
2 fneq2 A = B F Fn A F Fn B
3 1 2 syl φ F Fn A F Fn B