Metamath Proof Explorer


Theorem feq2d

Description: Equality deduction for functions. (Contributed by Paul Chapman, 22-Jun-2011)

Ref Expression
Hypothesis feq2d.1 ( 𝜑𝐴 = 𝐵 )
Assertion feq2d ( 𝜑 → ( 𝐹 : 𝐴𝐶𝐹 : 𝐵𝐶 ) )

Proof

Step Hyp Ref Expression
1 feq2d.1 ( 𝜑𝐴 = 𝐵 )
2 feq2 ( 𝐴 = 𝐵 → ( 𝐹 : 𝐴𝐶𝐹 : 𝐵𝐶 ) )
3 1 2 syl ( 𝜑 → ( 𝐹 : 𝐴𝐶𝐹 : 𝐵𝐶 ) )