Metamath Proof Explorer


Theorem fucval

Description: Value of the functor category. (Contributed by Mario Carneiro, 6-Jan-2017)

Ref Expression
Hypotheses fucval.q Q = C FuncCat D
fucval.b B = C Func D
fucval.n N = C Nat D
fucval.a A = Base C
fucval.o · ˙ = comp D
fucval.c φ C Cat
fucval.d φ D Cat
fucval.x φ ˙ = v B × B , h B 1 st v / f 2 nd v / g b g N h , a f N g x A b x 1 st f x 1 st g x · ˙ 1 st h x a x
Assertion fucval φ Q = Base ndx B Hom ndx N comp ndx ˙

Proof

Step Hyp Ref Expression
1 fucval.q Q = C FuncCat D
2 fucval.b B = C Func D
3 fucval.n N = C Nat D
4 fucval.a A = Base C
5 fucval.o · ˙ = comp D
6 fucval.c φ C Cat
7 fucval.d φ D Cat
8 fucval.x φ ˙ = v B × B , h B 1 st v / f 2 nd v / g b g N h , a f N g x A b x 1 st f x 1 st g x · ˙ 1 st h x a x
9 df-fuc FuncCat = t Cat , u Cat Base ndx t Func u Hom ndx t Nat u comp ndx v t Func u × t Func u , h t Func u 1 st v / f 2 nd v / g b g t Nat u h , a f t Nat u g x Base t b x 1 st f x 1 st g x comp u 1 st h x a x
10 9 a1i φ FuncCat = t Cat , u Cat Base ndx t Func u Hom ndx t Nat u comp ndx v t Func u × t Func u , h t Func u 1 st v / f 2 nd v / g b g t Nat u h , a f t Nat u g x Base t b x 1 st f x 1 st g x comp u 1 st h x a x
11 simprl φ t = C u = D t = C
12 simprr φ t = C u = D u = D
13 11 12 oveq12d φ t = C u = D t Func u = C Func D
14 13 2 eqtr4di φ t = C u = D t Func u = B
15 14 opeq2d φ t = C u = D Base ndx t Func u = Base ndx B
16 11 12 oveq12d φ t = C u = D t Nat u = C Nat D
17 16 3 eqtr4di φ t = C u = D t Nat u = N
18 17 opeq2d φ t = C u = D Hom ndx t Nat u = Hom ndx N
19 14 sqxpeqd φ t = C u = D t Func u × t Func u = B × B
20 17 oveqd φ t = C u = D g t Nat u h = g N h
21 17 oveqd φ t = C u = D f t Nat u g = f N g
22 11 fveq2d φ t = C u = D Base t = Base C
23 22 4 eqtr4di φ t = C u = D Base t = A
24 12 fveq2d φ t = C u = D comp u = comp D
25 24 5 eqtr4di φ t = C u = D comp u = · ˙
26 25 oveqd φ t = C u = D 1 st f x 1 st g x comp u 1 st h x = 1 st f x 1 st g x · ˙ 1 st h x
27 26 oveqd φ t = C u = D b x 1 st f x 1 st g x comp u 1 st h x a x = b x 1 st f x 1 st g x · ˙ 1 st h x a x
28 23 27 mpteq12dv φ t = C u = D x Base t b x 1 st f x 1 st g x comp u 1 st h x a x = x A b x 1 st f x 1 st g x · ˙ 1 st h x a x
29 20 21 28 mpoeq123dv φ t = C u = D b g t Nat u h , a f t Nat u g x Base t b x 1 st f x 1 st g x comp u 1 st h x a x = b g N h , a f N g x A b x 1 st f x 1 st g x · ˙ 1 st h x a x
30 29 csbeq2dv φ t = C u = D 2 nd v / g b g t Nat u h , a f t Nat u g x Base t b x 1 st f x 1 st g x comp u 1 st h x a x = 2 nd v / g b g N h , a f N g x A b x 1 st f x 1 st g x · ˙ 1 st h x a x
31 30 csbeq2dv φ t = C u = D 1 st v / f 2 nd v / g b g t Nat u h , a f t Nat u g x Base t b x 1 st f x 1 st g x comp u 1 st h x a x = 1 st v / f 2 nd v / g b g N h , a f N g x A b x 1 st f x 1 st g x · ˙ 1 st h x a x
32 19 14 31 mpoeq123dv φ t = C u = D v t Func u × t Func u , h t Func u 1 st v / f 2 nd v / g b g t Nat u h , a f t Nat u g x Base t b x 1 st f x 1 st g x comp u 1 st h x a x = v B × B , h B 1 st v / f 2 nd v / g b g N h , a f N g x A b x 1 st f x 1 st g x · ˙ 1 st h x a x
33 8 adantr φ t = C u = D ˙ = v B × B , h B 1 st v / f 2 nd v / g b g N h , a f N g x A b x 1 st f x 1 st g x · ˙ 1 st h x a x
34 32 33 eqtr4d φ t = C u = D v t Func u × t Func u , h t Func u 1 st v / f 2 nd v / g b g t Nat u h , a f t Nat u g x Base t b x 1 st f x 1 st g x comp u 1 st h x a x = ˙
35 34 opeq2d φ t = C u = D comp ndx v t Func u × t Func u , h t Func u 1 st v / f 2 nd v / g b g t Nat u h , a f t Nat u g x Base t b x 1 st f x 1 st g x comp u 1 st h x a x = comp ndx ˙
36 15 18 35 tpeq123d φ t = C u = D Base ndx t Func u Hom ndx t Nat u comp ndx v t Func u × t Func u , h t Func u 1 st v / f 2 nd v / g b g t Nat u h , a f t Nat u g x Base t b x 1 st f x 1 st g x comp u 1 st h x a x = Base ndx B Hom ndx N comp ndx ˙
37 tpex Base ndx B Hom ndx N comp ndx ˙ V
38 37 a1i φ Base ndx B Hom ndx N comp ndx ˙ V
39 10 36 6 7 38 ovmpod φ C FuncCat D = Base ndx B Hom ndx N comp ndx ˙
40 1 39 syl5eq φ Q = Base ndx B Hom ndx N comp ndx ˙