Metamath Proof Explorer


Theorem fucoco2

Description: Composition in the source category is mapped to composition in the target. See also fucoco . (Contributed by Zhi Wang, 3-Oct-2025)

Ref Expression
Hypotheses fucoco2.t T = D FuncCat E × c C FuncCat D
fucoco2.q Q = C FuncCat E
fucoco2.o No typesetting found for |- ( ph -> ( <. C , D >. o.F E ) = <. O , P >. ) with typecode |-
fucoco2.1 · ˙ = comp T
fucoco2.2 ˙ = comp Q
fucoco2.w φ W = D Func E × C Func D
fucoco2.x φ X W
fucoco2.y φ Y W
fucoco2.z φ Z W
fucoco2.j J = Hom T
fucoco2.a φ A X J Y
fucoco2.b φ B Y J Z
Assertion fucoco2 φ X P Z B X Y · ˙ Z A = Y P Z B O X O Y ˙ O Z X P Y A

Proof

Step Hyp Ref Expression
1 fucoco2.t T = D FuncCat E × c C FuncCat D
2 fucoco2.q Q = C FuncCat E
3 fucoco2.o Could not format ( ph -> ( <. C , D >. o.F E ) = <. O , P >. ) : No typesetting found for |- ( ph -> ( <. C , D >. o.F E ) = <. O , P >. ) with typecode |-
4 fucoco2.1 · ˙ = comp T
5 fucoco2.2 ˙ = comp Q
6 fucoco2.w φ W = D Func E × C Func D
7 fucoco2.x φ X W
8 fucoco2.y φ Y W
9 fucoco2.z φ Z W
10 fucoco2.j J = Hom T
11 fucoco2.a φ A X J Y
12 fucoco2.b φ B Y J Z
13 1 xpcfucbas D Func E × C Func D = Base T
14 7 6 eleqtrd φ X D Func E × C Func D
15 8 6 eleqtrd φ Y D Func E × C Func D
16 1 13 10 14 15 xpcfuchom φ X J Y = 1 st X D Nat E 1 st Y × 2 nd X C Nat D 2 nd Y
17 11 16 eleqtrd φ A 1 st X D Nat E 1 st Y × 2 nd X C Nat D 2 nd Y
18 xp1st A 1 st X D Nat E 1 st Y × 2 nd X C Nat D 2 nd Y 1 st A 1 st X D Nat E 1 st Y
19 17 18 syl φ 1 st A 1 st X D Nat E 1 st Y
20 xp2nd A 1 st X D Nat E 1 st Y × 2 nd X C Nat D 2 nd Y 2 nd A 2 nd X C Nat D 2 nd Y
21 17 20 syl φ 2 nd A 2 nd X C Nat D 2 nd Y
22 9 6 eleqtrd φ Z D Func E × C Func D
23 1 13 10 15 22 xpcfuchom φ Y J Z = 1 st Y D Nat E 1 st Z × 2 nd Y C Nat D 2 nd Z
24 12 23 eleqtrd φ B 1 st Y D Nat E 1 st Z × 2 nd Y C Nat D 2 nd Z
25 xp1st B 1 st Y D Nat E 1 st Z × 2 nd Y C Nat D 2 nd Z 1 st B 1 st Y D Nat E 1 st Z
26 24 25 syl φ 1 st B 1 st Y D Nat E 1 st Z
27 xp2nd B 1 st Y D Nat E 1 st Z × 2 nd Y C Nat D 2 nd Z 2 nd B 2 nd Y C Nat D 2 nd Z
28 24 27 syl φ 2 nd B 2 nd Y C Nat D 2 nd Z
29 1st2nd2 X D Func E × C Func D X = 1 st X 2 nd X
30 14 29 syl φ X = 1 st X 2 nd X
31 1st2nd2 Y D Func E × C Func D Y = 1 st Y 2 nd Y
32 15 31 syl φ Y = 1 st Y 2 nd Y
33 1st2nd2 Z D Func E × C Func D Z = 1 st Z 2 nd Z
34 22 33 syl φ Z = 1 st Z 2 nd Z
35 1st2nd2 A 1 st X D Nat E 1 st Y × 2 nd X C Nat D 2 nd Y A = 1 st A 2 nd A
36 17 35 syl φ A = 1 st A 2 nd A
37 1st2nd2 B 1 st Y D Nat E 1 st Z × 2 nd Y C Nat D 2 nd Z B = 1 st B 2 nd B
38 24 37 syl φ B = 1 st B 2 nd B
39 19 21 26 28 3 30 32 34 36 38 2 5 1 4 fucoco φ X P Z B X Y · ˙ Z A = Y P Z B O X O Y ˙ O Z X P Y A