Metamath Proof Explorer


Theorem cofu1

Description: Value of the object part of the functor composition. (Contributed by Mario Carneiro, 28-Jan-2017)

Ref Expression
Hypotheses cofuval.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
cofuval.f ⊢ ( 𝜑 → 𝐹 ∈ ( 𝐶 Func 𝐷 ) )
cofuval.g ⊢ ( 𝜑 → 𝐺 ∈ ( 𝐷 Func 𝐸 ) )
cofu2nd.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
Assertion cofu1 ( 𝜑 → ( ( 1st ‘ ( 𝐺 ∘func 𝐹 ) ) ‘ 𝑋 ) = ( ( 1st ‘ 𝐺 ) ‘ ( ( 1st ‘ 𝐹 ) ‘ 𝑋 ) ) )

Proof

Step Hyp Ref Expression
1 cofuval.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
2 cofuval.f ⊢ ( 𝜑 → 𝐹 ∈ ( 𝐶 Func 𝐷 ) )
3 cofuval.g ⊢ ( 𝜑 → 𝐺 ∈ ( 𝐷 Func 𝐸 ) )
4 cofu2nd.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
5 1 2 3 cofu1st ⊢ ( 𝜑 → ( 1st ‘ ( 𝐺 ∘func 𝐹 ) ) = ( ( 1st ‘ 𝐺 ) ∘ ( 1st ‘ 𝐹 ) ) )
6 5 fveq1d ⊢ ( 𝜑 → ( ( 1st ‘ ( 𝐺 ∘func 𝐹 ) ) ‘ 𝑋 ) = ( ( ( 1st ‘ 𝐺 ) ∘ ( 1st ‘ 𝐹 ) ) ‘ 𝑋 ) )
7 eqid ⊢ ( Base ‘ 𝐷 ) = ( Base ‘ 𝐷 )
8 relfunc ⊢ Rel ( 𝐶 Func 𝐷 )
9 1st2ndbr ⊢ ( ( Rel ( 𝐶 Func 𝐷 ) ∧ 𝐹 ∈ ( 𝐶 Func 𝐷 ) ) → ( 1st ‘ 𝐹 ) ( 𝐶 Func 𝐷 ) ( 2nd ‘ 𝐹 ) )
10 8 2 9 sylancr ⊢ ( 𝜑 → ( 1st ‘ 𝐹 ) ( 𝐶 Func 𝐷 ) ( 2nd ‘ 𝐹 ) )
11 1 7 10 funcf1 ⊢ ( 𝜑 → ( 1st ‘ 𝐹 ) : 𝐵 ⟶ ( Base ‘ 𝐷 ) )
12 fvco3 ⊢ ( ( ( 1st ‘ 𝐹 ) : 𝐵 ⟶ ( Base ‘ 𝐷 ) ∧ 𝑋 ∈ 𝐵 ) → ( ( ( 1st ‘ 𝐺 ) ∘ ( 1st ‘ 𝐹 ) ) ‘ 𝑋 ) = ( ( 1st ‘ 𝐺 ) ‘ ( ( 1st ‘ 𝐹 ) ‘ 𝑋 ) ) )
13 11 4 12 syl2anc ⊢ ( 𝜑 → ( ( ( 1st ‘ 𝐺 ) ∘ ( 1st ‘ 𝐹 ) ) ‘ 𝑋 ) = ( ( 1st ‘ 𝐺 ) ‘ ( ( 1st ‘ 𝐹 ) ‘ 𝑋 ) ) )
14 6 13 eqtrd ⊢ ( 𝜑 → ( ( 1st ‘ ( 𝐺 ∘func 𝐹 ) ) ‘ 𝑋 ) = ( ( 1st ‘ 𝐺 ) ‘ ( ( 1st ‘ 𝐹 ) ‘ 𝑋 ) ) )