Metamath Proof Explorer


Theorem feq1d

Description: Equality deduction for functions. (Contributed by NM, 19-Feb-2008)

Ref Expression
Hypothesis feq1d.1 φ F = G
Assertion feq1d φ F : A B G : A B

Proof

Step Hyp Ref Expression
1 feq1d.1 φ F = G
2 feq1 F = G F : A B G : A B
3 1 2 syl φ F : A B G : A B