Metamath Proof Explorer


Theorem comfeqval

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

Ref Expression
Hypotheses comfeqval.b B=BaseC
comfeqval.h H=HomC
comfeqval.1 ·˙=compC
comfeqval.2 ˙=compD
comfeqval.3 φHom𝑓C=Hom𝑓D
comfeqval.4 φcomp𝑓C=comp𝑓D
comfeqval.x φXB
comfeqval.y φYB
comfeqval.z φZB
comfeqval.f φFXHY
comfeqval.g φGYHZ
Assertion comfeqval φGXY·˙ZF=GXY˙ZF

Proof

Step Hyp Ref Expression
1 comfeqval.b B=BaseC
2 comfeqval.h H=HomC
3 comfeqval.1 ·˙=compC
4 comfeqval.2 ˙=compD
5 comfeqval.3 φHom𝑓C=Hom𝑓D
6 comfeqval.4 φcomp𝑓C=comp𝑓D
7 comfeqval.x φXB
8 comfeqval.y φYB
9 comfeqval.z φZB
10 comfeqval.f φFXHY
11 comfeqval.g φGYHZ
12 6 oveqd φXYcomp𝑓CZ=XYcomp𝑓DZ
13 12 oveqd φGXYcomp𝑓CZF=GXYcomp𝑓DZF
14 eqid comp𝑓C=comp𝑓C
15 14 1 2 3 7 8 9 10 11 comfval φGXYcomp𝑓CZF=GXY·˙ZF
16 eqid comp𝑓D=comp𝑓D
17 eqid BaseD=BaseD
18 eqid HomD=HomD
19 5 homfeqbas φBaseC=BaseD
20 1 19 eqtrid φB=BaseD
21 7 20 eleqtrd φXBaseD
22 8 20 eleqtrd φYBaseD
23 9 20 eleqtrd φZBaseD
24 1 2 18 5 7 8 homfeqval φXHY=XHomDY
25 10 24 eleqtrd φFXHomDY
26 1 2 18 5 8 9 homfeqval φYHZ=YHomDZ
27 11 26 eleqtrd φGYHomDZ
28 16 17 18 4 21 22 23 25 27 comfval φGXYcomp𝑓DZF=GXY˙ZF
29 13 15 28 3eqtr3d φGXY·˙ZF=GXY˙ZF