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 = 𝐶 ∧ 𝐵 = 𝐷 ) → ( 𝐹 : 𝐴 ⟶ 𝐵 ↔ 𝐹 : 𝐶 ⟶ 𝐷 ) )

Proof

Step Hyp Ref Expression
1 feq2 ⊢ ( 𝐴 = 𝐶 → ( 𝐹 : 𝐴 ⟶ 𝐵 ↔ 𝐹 : 𝐶 ⟶ 𝐵 ) )
2 feq3 ⊢ ( 𝐵 = 𝐷 → ( 𝐹 : 𝐶 ⟶ 𝐵 ↔ 𝐹 : 𝐶 ⟶ 𝐷 ) )
3 1 2 sylan9bb ⊢ ( ( 𝐴 = 𝐶 ∧ 𝐵 = 𝐷 ) → ( 𝐹 : 𝐴 ⟶ 𝐵 ↔ 𝐹 : 𝐶 ⟶ 𝐷 ) )