Metamath Proof Explorer


Theorem comfeqval

Description: Equality of two compositions. (Contributed by Mario Carneiro, 4-Jan-2017)

Ref Expression
Hypotheses comfeqval.b B = Base C
comfeqval.h H = Hom C
comfeqval.1 · ˙ = comp C
comfeqval.2 ˙ = comp D
comfeqval.3 φ Hom 𝑓 C = Hom 𝑓 D
comfeqval.4 φ comp 𝑓 C = comp 𝑓 D
comfeqval.x φ X B
comfeqval.y φ Y B
comfeqval.z φ Z B
comfeqval.f φ F X H Y
comfeqval.g φ G Y H Z
Assertion comfeqval φ G X Y · ˙ Z F = G X Y ˙ Z F

Proof

Step Hyp Ref Expression
1 comfeqval.b B = Base C
2 comfeqval.h H = Hom C
3 comfeqval.1 · ˙ = comp C
4 comfeqval.2 ˙ = comp D
5 comfeqval.3 φ Hom 𝑓 C = Hom 𝑓 D
6 comfeqval.4 φ comp 𝑓 C = comp 𝑓 D
7 comfeqval.x φ X B
8 comfeqval.y φ Y B
9 comfeqval.z φ Z B
10 comfeqval.f φ F X H Y
11 comfeqval.g φ G Y H Z
12 6 oveqd φ X Y comp 𝑓 C Z = X Y comp 𝑓 D Z
13 12 oveqd φ G X Y comp 𝑓 C Z F = G X Y comp 𝑓 D Z F
14 eqid comp 𝑓 C = comp 𝑓 C
15 14 1 2 3 7 8 9 10 11 comfval φ G X Y comp 𝑓 C Z F = G X Y · ˙ Z F
16 eqid comp 𝑓 D = comp 𝑓 D
17 eqid Base D = Base D
18 eqid Hom D = Hom D
19 5 homfeqbas φ Base C = Base D
20 1 19 eqtrid φ B = Base D
21 7 20 eleqtrd φ X Base D
22 8 20 eleqtrd φ Y Base D
23 9 20 eleqtrd φ Z Base D
24 1 2 18 5 7 8 homfeqval φ X H Y = X Hom D Y
25 10 24 eleqtrd φ F X Hom D Y
26 1 2 18 5 8 9 homfeqval φ Y H Z = Y Hom D Z
27 11 26 eleqtrd φ G Y Hom D Z
28 16 17 18 4 21 22 23 25 27 comfval φ G X Y comp 𝑓 D Z F = G X Y ˙ Z F
29 13 15 28 3eqtr3d φ G X Y · ˙ Z F = G X Y ˙ Z F