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 → 1 st ⁡ G ∘ func F ⁡ X = 1 st ⁡ G ⁡ 1 st ⁡ 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 → 1 st ⁡ G ∘ func F = 1 st ⁡ G ∘ 1 st ⁡ F
6 5 fveq1d ⊢ φ → 1 st ⁡ G ∘ func F ⁡ X = 1 st ⁡ G ∘ 1 st ⁡ F ⁡ X
7 eqid ⊢ Base D = Base D
8 relfunc ⊢ Rel ⁡ C Func D
9 1st2ndbr ⊢ Rel ⁡ C Func D ∧ F ∈ C Func D → 1 st ⁡ F C Func D 2 nd ⁡ F
10 8 2 9 sylancr ⊢ φ → 1 st ⁡ F C Func D 2 nd ⁡ F
11 1 7 10 funcf1 ⊢ φ → 1 st ⁡ F : B ⟶ Base D
12 fvco3 ⊢ 1 st ⁡ F : B ⟶ Base D ∧ X ∈ B → 1 st ⁡ G ∘ 1 st ⁡ F ⁡ X = 1 st ⁡ G ⁡ 1 st ⁡ F ⁡ X
13 11 4 12 syl2anc ⊢ φ → 1 st ⁡ G ∘ 1 st ⁡ F ⁡ X = 1 st ⁡ G ⁡ 1 st ⁡ F ⁡ X
14 6 13 eqtrd ⊢ φ → 1 st ⁡ G ∘ func F ⁡ X = 1 st ⁡ G ⁡ 1 st ⁡ F ⁡ X