Metamath Proof Explorer


Theorem feq3

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

Ref Expression
Assertion feq3 ( 𝐴 = 𝐵 → ( 𝐹 : 𝐶𝐴𝐹 : 𝐶𝐵 ) )

Proof

Step Hyp Ref Expression
1 sseq2 ( 𝐴 = 𝐵 → ( ran 𝐹𝐴 ↔ ran 𝐹𝐵 ) )
2 1 anbi2d ( 𝐴 = 𝐵 → ( ( 𝐹 Fn 𝐶 ∧ ran 𝐹𝐴 ) ↔ ( 𝐹 Fn 𝐶 ∧ ran 𝐹𝐵 ) ) )
3 df-f ( 𝐹 : 𝐶𝐴 ↔ ( 𝐹 Fn 𝐶 ∧ ran 𝐹𝐴 ) )
4 df-f ( 𝐹 : 𝐶𝐵 ↔ ( 𝐹 Fn 𝐶 ∧ ran 𝐹𝐵 ) )
5 2 3 4 3bitr4g ( 𝐴 = 𝐵 → ( 𝐹 : 𝐶𝐴𝐹 : 𝐶𝐵 ) )