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 φA=B
Assertion fneq2d φFFnAFFnB

Proof

Step Hyp Ref Expression
1 fneq2d.1 φA=B
2 fneq2 A=BFFnAFFnB
3 1 2 syl φFFnAFFnB