Metamath Proof Explorer


Theorem fneq2

Description: Equality theorem for function predicate with domain. (Contributed by NM, 1-Aug-1994)

Ref Expression
Assertion fneq2 ( 𝐴 = 𝐵 → ( 𝐹 Fn 𝐴𝐹 Fn 𝐵 ) )

Proof

Step Hyp Ref Expression
1 eqeq2 ( 𝐴 = 𝐵 → ( dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵 ) )
2 1 anbi2d ( 𝐴 = 𝐵 → ( ( Fun 𝐹 ∧ dom 𝐹 = 𝐴 ) ↔ ( Fun 𝐹 ∧ dom 𝐹 = 𝐵 ) ) )
3 df-fn ( 𝐹 Fn 𝐴 ↔ ( Fun 𝐹 ∧ dom 𝐹 = 𝐴 ) )
4 df-fn ( 𝐹 Fn 𝐵 ↔ ( Fun 𝐹 ∧ dom 𝐹 = 𝐵 ) )
5 2 3 4 3bitr4g ( 𝐴 = 𝐵 → ( 𝐹 Fn 𝐴𝐹 Fn 𝐵 ) )