Metamath Proof Explorer


Theorem feq23

Description: Equality theorem for functions. (Contributed by FL, 14-Jul-2007) (Proof shortened by Andrew Salmon, 17-Sep-2011)

Ref Expression
Assertion feq23 ( ( 𝐴 = 𝐶𝐵 = 𝐷 ) → ( 𝐹 : 𝐴𝐵𝐹 : 𝐶𝐷 ) )

Proof

Step Hyp Ref Expression
1 feq2 ( 𝐴 = 𝐶 → ( 𝐹 : 𝐴𝐵𝐹 : 𝐶𝐵 ) )
2 feq3 ( 𝐵 = 𝐷 → ( 𝐹 : 𝐶𝐵𝐹 : 𝐶𝐷 ) )
3 1 2 sylan9bb ( ( 𝐴 = 𝐶𝐵 = 𝐷 ) → ( 𝐹 : 𝐴𝐵𝐹 : 𝐶𝐷 ) )