Metamath Proof Explorer


Theorem fuco11

Description: The object part of the functor composition bifunctor maps two functors to their composition. (Contributed by Zhi Wang, 30-Sep-2025)

Ref Expression
Hypotheses fuco11.o No typesetting found for |- ( ph -> ( <. C , D >. o.F E ) = <. O , P >. ) with typecode |-
fuco11.f ⊢ φ → F C Func D G
fuco11.k ⊢ φ → K D Func E L
fuco11.u ⊢ φ → U = K L F G
Assertion fuco11 ⊢ φ → O ⁡ U = K L ∘ func F G

Proof

Step Hyp Ref Expression
1 fuco11.o Could not format ( ph -> ( <. C , D >. o.F E ) = <. O , P >. ) : No typesetting found for |- ( ph -> ( <. C , D >. o.F E ) = <. O , P >. ) with typecode |-
2 fuco11.f ⊢ φ → F C Func D G
3 fuco11.k ⊢ φ → K D Func E L
4 fuco11.u ⊢ φ → U = K L F G
5 2 funcrcl2 ⊢ φ → C ∈ Cat
6 3 funcrcl2 ⊢ φ → D ∈ Cat
7 3 funcrcl3 ⊢ φ → E ∈ Cat
8 eqidd ⊢ φ → D Func E × C Func D = D Func E × C Func D
9 5 6 7 1 8 fuco1 ⊢ φ → O = ∘ func ↾ D Func E × C Func D
10 9 fveq1d ⊢ φ → O ⁡ U = ∘ func ↾ D Func E × C Func D ⁡ U
11 8 4 3 2 fuco2eld ⊢ φ → U ∈ D Func E × C Func D
12 11 fvresd ⊢ φ → ∘ func ↾ D Func E × C Func D ⁡ U = ∘ func ⁡ U
13 4 fveq2d ⊢ φ → ∘ func ⁡ U = ∘ func ⁡ K L F G
14 df-ov ⊢ K L ∘ func F G = ∘ func ⁡ K L F G
15 13 14 eqtr4di ⊢ φ → ∘ func ⁡ U = K L ∘ func F G
16 10 12 15 3eqtrd ⊢ φ → O ⁡ U = K L ∘ func F G