Metamath Proof Explorer


Theorem funeqd

Description: Equality deduction for the function predicate. (Contributed by NM, 23-Feb-2013)

Ref Expression
Hypothesis funeqd.1 φA=B
Assertion funeqd φFunAFunB

Proof

Step Hyp Ref Expression
1 funeqd.1 φA=B
2 funeq A=BFunAFunB
3 1 2 syl φFunAFunB