Metamath Proof Explorer


Theorem feq2

Description: Equality theorem for functions. (Contributed by NM, 1-Aug-1994)

Ref Expression
Assertion feq2 ( 𝐴 = 𝐵 → ( 𝐹 : 𝐴𝐶𝐹 : 𝐵𝐶 ) )

Proof

Step Hyp Ref Expression
1 fneq2 ( 𝐴 = 𝐵 → ( 𝐹 Fn 𝐴𝐹 Fn 𝐵 ) )
2 1 anbi1d ( 𝐴 = 𝐵 → ( ( 𝐹 Fn 𝐴 ∧ ran 𝐹𝐶 ) ↔ ( 𝐹 Fn 𝐵 ∧ ran 𝐹𝐶 ) ) )
3 df-f ( 𝐹 : 𝐴𝐶 ↔ ( 𝐹 Fn 𝐴 ∧ ran 𝐹𝐶 ) )
4 df-f ( 𝐹 : 𝐵𝐶 ↔ ( 𝐹 Fn 𝐵 ∧ ran 𝐹𝐶 ) )
5 2 3 4 3bitr4g ( 𝐴 = 𝐵 → ( 𝐹 : 𝐴𝐶𝐹 : 𝐵𝐶 ) )