Metamath Proof Explorer


Theorem xpccofval

Description: Value of composition in the binary product of categories. (Contributed by Mario Carneiro, 11-Jan-2017) (Proof shortened by AV, 2-Mar-2024)

Ref Expression
Hypotheses xpccofval.t T = C × c D
xpccofval.b B = Base T
xpccofval.k K = Hom T
xpccofval.o1 · ˙ = comp C
xpccofval.o2 ˙ = comp D
xpccofval.o O = comp T
Assertion xpccofval O = x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f

Proof

Step Hyp Ref Expression
1 xpccofval.t T = C × c D
2 xpccofval.b B = Base T
3 xpccofval.k K = Hom T
4 xpccofval.o1 · ˙ = comp C
5 xpccofval.o2 ˙ = comp D
6 xpccofval.o O = comp T
7 eqid Base C = Base C
8 eqid Base D = Base D
9 eqid Hom C = Hom C
10 eqid Hom D = Hom D
11 simpl C V D V C V
12 simpr C V D V D V
13 1 7 8 xpcbas Base C × Base D = Base T
14 2 13 eqtr4i B = Base C × Base D
15 14 a1i C V D V B = Base C × Base D
16 1 2 9 10 3 xpchomfval K = u B , v B 1 st u Hom C 1 st v × 2 nd u Hom D 2 nd v
17 16 a1i C V D V K = u B , v B 1 st u Hom C 1 st v × 2 nd u Hom D 2 nd v
18 eqidd C V D V x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f = x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f
19 1 7 8 9 10 4 5 11 12 15 17 18 xpcval C V D V T = Base ndx B Hom ndx K comp ndx x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f
20 catstr Base ndx B Hom ndx K comp ndx x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f Struct 1 15
21 ccoid comp = Slot comp ndx
22 snsstp3 comp ndx x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f Base ndx B Hom ndx K comp ndx x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f
23 2 fvexi B V
24 23 23 xpex B × B V
25 24 23 mpoex x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f V
26 25 a1i C V D V x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f V
27 19 20 21 22 26 6 strfv3 C V D V O = x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f
28 fnxpc × c Fn V × V
29 28 fndmi dom × c = V × V
30 29 ndmov ¬ C V D V C × c D =
31 1 30 eqtrid ¬ C V D V T =
32 31 fveq2d ¬ C V D V comp T = comp
33 21 str0 = comp
34 32 6 33 3eqtr4g ¬ C V D V O =
35 31 fveq2d ¬ C V D V Base T = Base
36 base0 = Base
37 35 2 36 3eqtr4g ¬ C V D V B =
38 37 olcd ¬ C V D V B × B = B =
39 0mpo0 B × B = B = x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f =
40 38 39 syl ¬ C V D V x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f =
41 34 40 eqtr4d ¬ C V D V O = x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f
42 27 41 pm2.61i O = x B × B , y B g 2 nd x K y , f K x 1 st g 1 st 1 st x 1 st 2 nd x · ˙ 1 st y 1 st f 2 nd g 2 nd 1 st x 2 nd 2 nd x ˙ 2 nd y 2 nd f