Metamath Proof Explorer


Theorem feq23d

Description: Equality deduction for functions. (Contributed by NM, 8-Jun-2013)

Ref Expression
Hypotheses feq23d.1 φ A = C
feq23d.2 φ B = D
Assertion feq23d φ F : A B F : C D

Proof

Step Hyp Ref Expression
1 feq23d.1 φ A = C
2 feq23d.2 φ B = D
3 eqidd φ F = F
4 3 1 2 feq123d φ F : A B F : C D