Metamath Proof Explorer


Theorem 2ndfval

Description: Value of the first projection functor. (Contributed by Mario Carneiro, 11-Jan-2017)

Ref Expression
Hypotheses 1stfval.t T = C × c D
1stfval.b B = Base T
1stfval.h H = Hom T
1stfval.c φ C Cat
1stfval.d φ D Cat
2ndfval.p Q = C 2 nd F D
Assertion 2ndfval φ Q = 2 nd B x B , y B 2 nd x H y

Proof

Step Hyp Ref Expression
1 1stfval.t T = C × c D
2 1stfval.b B = Base T
3 1stfval.h H = Hom T
4 1stfval.c φ C Cat
5 1stfval.d φ D Cat
6 2ndfval.p Q = C 2 nd F D
7 fvex Base c V
8 fvex Base d V
9 7 8 xpex Base c × Base d V
10 9 a1i c = C d = D Base c × Base d V
11 simpl c = C d = D c = C
12 11 fveq2d c = C d = D Base c = Base C
13 simpr c = C d = D d = D
14 13 fveq2d c = C d = D Base d = Base D
15 12 14 xpeq12d c = C d = D Base c × Base d = Base C × Base D
16 eqid Base C = Base C
17 eqid Base D = Base D
18 1 16 17 xpcbas Base C × Base D = Base T
19 18 2 eqtr4i Base C × Base D = B
20 15 19 eqtrdi c = C d = D Base c × Base d = B
21 simpr c = C d = D b = B b = B
22 21 reseq2d c = C d = D b = B 2 nd b = 2 nd B
23 simpll c = C d = D b = B c = C
24 simplr c = C d = D b = B d = D
25 23 24 oveq12d c = C d = D b = B c × c d = C × c D
26 25 1 eqtr4di c = C d = D b = B c × c d = T
27 26 fveq2d c = C d = D b = B Hom c × c d = Hom T
28 27 3 eqtr4di c = C d = D b = B Hom c × c d = H
29 28 oveqd c = C d = D b = B x Hom c × c d y = x H y
30 29 reseq2d c = C d = D b = B 2 nd x Hom c × c d y = 2 nd x H y
31 21 21 30 mpoeq123dv c = C d = D b = B x b , y b 2 nd x Hom c × c d y = x B , y B 2 nd x H y
32 22 31 opeq12d c = C d = D b = B 2 nd b x b , y b 2 nd x Hom c × c d y = 2 nd B x B , y B 2 nd x H y
33 10 20 32 csbied2 c = C d = D Base c × Base d / b 2 nd b x b , y b 2 nd x Hom c × c d y = 2 nd B x B , y B 2 nd x H y
34 df-2ndf 2 nd F = c Cat , d Cat Base c × Base d / b 2 nd b x b , y b 2 nd x Hom c × c d y
35 opex 2 nd B x B , y B 2 nd x H y V
36 33 34 35 ovmpoa C Cat D Cat C 2 nd F D = 2 nd B x B , y B 2 nd x H y
37 4 5 36 syl2anc φ C 2 nd F D = 2 nd B x B , y B 2 nd x H y
38 6 37 eqtrid φ Q = 2 nd B x B , y B 2 nd x H y