Metamath Proof Explorer


Theorem diag11

Description: Value of the constant functor at an object. (Contributed by Mario Carneiro, 7-Jan-2017) (Revised by Mario Carneiro, 15-Jan-2017)

Ref Expression
Hypotheses diagval.l ⊢ L = C Δ func D
diagval.c ⊢ φ → C ∈ Cat
diagval.d ⊢ φ → D ∈ Cat
diag11.a ⊢ A = Base C
diag11.c ⊢ φ → X ∈ A
diag11.k ⊢ K = 1 st ⁡ L ⁡ X
diag11.b ⊢ B = Base D
diag11.y ⊢ φ → Y ∈ B
Assertion diag11 ⊢ φ → 1 st ⁡ K ⁡ Y = X

Proof

Step Hyp Ref Expression
1 diagval.l ⊢ L = C Δ func D
2 diagval.c ⊢ φ → C ∈ Cat
3 diagval.d ⊢ φ → D ∈ Cat
4 diag11.a ⊢ A = Base C
5 diag11.c ⊢ φ → X ∈ A
6 diag11.k ⊢ K = 1 st ⁡ L ⁡ X
7 diag11.b ⊢ B = Base D
8 diag11.y ⊢ φ → Y ∈ B
9 1 2 3 diagval ⊢ φ → L = C D curry F C 1 st F D
10 9 fveq2d ⊢ φ → 1 st ⁡ L = 1 st ⁡ C D curry F C 1 st F D
11 10 fveq1d ⊢ φ → 1 st ⁡ L ⁡ X = 1 st ⁡ C D curry F C 1 st F D ⁡ X
12 6 11 eqtrid ⊢ φ → K = 1 st ⁡ C D curry F C 1 st F D ⁡ X
13 12 fveq2d ⊢ φ → 1 st ⁡ K = 1 st ⁡ 1 st ⁡ C D curry F C 1 st F D ⁡ X
14 13 fveq1d ⊢ φ → 1 st ⁡ K ⁡ Y = 1 st ⁡ 1 st ⁡ C D curry F C 1 st F D ⁡ X ⁡ Y
15 eqid ⊢ C D curry F C 1 st F D = C D curry F C 1 st F D
16 eqid ⊢ C × c D = C × c D
17 eqid ⊢ C 1 st F D = C 1 st F D
18 16 2 3 17 1stfcl ⊢ φ → C 1 st F D ∈ C × c D Func C
19 eqid ⊢ 1 st ⁡ C D curry F C 1 st F D ⁡ X = 1 st ⁡ C D curry F C 1 st F D ⁡ X
20 15 4 2 3 18 7 5 19 8 curf11 ⊢ φ → 1 st ⁡ 1 st ⁡ C D curry F C 1 st F D ⁡ X ⁡ Y = X 1 st ⁡ C 1 st F D Y
21 df-ov ⊢ X 1 st ⁡ C 1 st F D Y = 1 st ⁡ C 1 st F D ⁡ X Y
22 16 4 7 xpcbas ⊢ A × B = Base C × c D
23 eqid ⊢ Hom ⁡ C × c D = Hom ⁡ C × c D
24 5 8 opelxpd ⊢ φ → X Y ∈ A × B
25 16 22 23 2 3 17 24 1stf1 ⊢ φ → 1 st ⁡ C 1 st F D ⁡ X Y = 1 st ⁡ X Y
26 21 25 eqtrid ⊢ φ → X 1 st ⁡ C 1 st F D Y = 1 st ⁡ X Y
27 op1stg ⊢ X ∈ A ∧ Y ∈ B → 1 st ⁡ X Y = X
28 5 8 27 syl2anc ⊢ φ → 1 st ⁡ X Y = X
29 26 28 eqtrd ⊢ φ → X 1 st ⁡ C 1 st F D Y = X
30 14 20 29 3eqtrd ⊢ φ → 1 st ⁡ K ⁡ Y = X