Metamath Proof Explorer


Theorem oppccofval

Description: Composition in the opposite category. (Contributed by Mario Carneiro, 2-Jan-2017)

Ref Expression
Hypotheses oppcco.b B=BaseC
oppcco.c ·˙=compC
oppcco.o O=oppCatC
oppcco.x φXB
oppcco.y φYB
oppcco.z φZB
Assertion oppccofval φXYcompOZ=tposZY·˙X

Proof

Step Hyp Ref Expression
1 oppcco.b B=BaseC
2 oppcco.c ·˙=compC
3 oppcco.o O=oppCatC
4 oppcco.x φXB
5 oppcco.y φYB
6 oppcco.z φZB
7 elfvex XBaseCCV
8 7 1 eleq2s XBCV
9 eqid HomC=HomC
10 1 9 2 3 oppcval CVO=CsSetHomndxtposHomCsSetcompndxuB×B,zBtposz2ndu·˙1stu
11 4 8 10 3syl φO=CsSetHomndxtposHomCsSetcompndxuB×B,zBtposz2ndu·˙1stu
12 11 fveq2d φcompO=compCsSetHomndxtposHomCsSetcompndxuB×B,zBtposz2ndu·˙1stu
13 ovex CsSetHomndxtposHomCV
14 1 fvexi BV
15 14 14 xpex B×BV
16 15 14 mpoex uB×B,zBtposz2ndu·˙1stuV
17 ccoid comp=Slotcompndx
18 17 setsid CsSetHomndxtposHomCVuB×B,zBtposz2ndu·˙1stuVuB×B,zBtposz2ndu·˙1stu=compCsSetHomndxtposHomCsSetcompndxuB×B,zBtposz2ndu·˙1stu
19 13 16 18 mp2an uB×B,zBtposz2ndu·˙1stu=compCsSetHomndxtposHomCsSetcompndxuB×B,zBtposz2ndu·˙1stu
20 12 19 eqtr4di φcompO=uB×B,zBtposz2ndu·˙1stu
21 simprr φu=XYz=Zz=Z
22 simprl φu=XYz=Zu=XY
23 22 fveq2d φu=XYz=Z2ndu=2ndXY
24 5 adantr φu=XYz=ZYB
25 op2ndg XBYB2ndXY=Y
26 4 24 25 syl2an2r φu=XYz=Z2ndXY=Y
27 23 26 eqtrd φu=XYz=Z2ndu=Y
28 21 27 opeq12d φu=XYz=Zz2ndu=ZY
29 22 fveq2d φu=XYz=Z1stu=1stXY
30 op1stg XBYB1stXY=X
31 4 24 30 syl2an2r φu=XYz=Z1stXY=X
32 29 31 eqtrd φu=XYz=Z1stu=X
33 28 32 oveq12d φu=XYz=Zz2ndu·˙1stu=ZY·˙X
34 33 tposeqd φu=XYz=Ztposz2ndu·˙1stu=tposZY·˙X
35 4 5 opelxpd φXYB×B
36 ovex ZY·˙XV
37 36 tposex tposZY·˙XV
38 37 a1i φtposZY·˙XV
39 20 34 35 6 38 ovmpod φXYcompOZ=tposZY·˙X