Metamath Proof Explorer


Theorem feq23d

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

Ref Expression
Hypotheses feq23d.1 φA=C
feq23d.2 φB=D
Assertion feq23d φF:ABF:CD

Proof

Step Hyp Ref Expression
1 feq23d.1 φA=C
2 feq23d.2 φB=D
3 eqidd φF=F
4 3 1 2 feq123d φF:ABF:CD