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 syl5eq 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 syl5eq 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