Metamath Proof Explorer


Theorem cofu1st2nd

Description: Rewrite the functor composition with separated functor parts. (Contributed by Zhi Wang, 15-Nov-2025)

Ref Expression
Hypotheses cofu1st2nd.f ( 𝜑𝐹 ∈ ( 𝐶 Func 𝐷 ) )
cofu1st2nd.g ( 𝜑𝐺 ∈ ( 𝐷 Func 𝐸 ) )
Assertion cofu1st2nd ( 𝜑 → ( 𝐺func 𝐹 ) = ( ⟨ ( 1st𝐺 ) , ( 2nd𝐺 ) ⟩ ∘func ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ ) )

Proof

Step Hyp Ref Expression
1 cofu1st2nd.f ( 𝜑𝐹 ∈ ( 𝐶 Func 𝐷 ) )
2 cofu1st2nd.g ( 𝜑𝐺 ∈ ( 𝐷 Func 𝐸 ) )
3 relfunc Rel ( 𝐷 Func 𝐸 )
4 1st2nd ( ( Rel ( 𝐷 Func 𝐸 ) ∧ 𝐺 ∈ ( 𝐷 Func 𝐸 ) ) → 𝐺 = ⟨ ( 1st𝐺 ) , ( 2nd𝐺 ) ⟩ )
5 3 2 4 sylancr ( 𝜑𝐺 = ⟨ ( 1st𝐺 ) , ( 2nd𝐺 ) ⟩ )
6 relfunc Rel ( 𝐶 Func 𝐷 )
7 1st2nd ( ( Rel ( 𝐶 Func 𝐷 ) ∧ 𝐹 ∈ ( 𝐶 Func 𝐷 ) ) → 𝐹 = ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ )
8 6 1 7 sylancr ( 𝜑𝐹 = ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ )
9 5 8 oveq12d ( 𝜑 → ( 𝐺func 𝐹 ) = ( ⟨ ( 1st𝐺 ) , ( 2nd𝐺 ) ⟩ ∘func ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ ) )