Metamath Proof Explorer


Theorem oppcbasOLD

Description: Obsolete version of oppcbas as of 18-Oct-2024. Base set of an opposite category. (Contributed by Mario Carneiro, 2-Jan-2017) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypotheses oppcbas.1 O = oppCat C
oppcbas.2 B = Base C
Assertion oppcbasOLD B = Base O

Proof

Step Hyp Ref Expression
1 oppcbas.1 O = oppCat C
2 oppcbas.2 B = Base C
3 baseid Base = Slot Base ndx
4 1re 1
5 1nn 1
6 4nn0 4 0
7 1nn0 1 0
8 1lt10 1 < 10
9 5 6 7 8 declti 1 < 14
10 4 9 ltneii 1 14
11 basendx Base ndx = 1
12 homndx Hom ndx = 14
13 11 12 neeq12i Base ndx Hom ndx 1 14
14 10 13 mpbir Base ndx Hom ndx
15 3 14 setsnid Base C = Base C sSet Hom ndx tpos Hom C
16 5nn 5
17 4lt5 4 < 5
18 7 6 16 17 declt 14 < 15
19 4nn 4
20 7 19 decnncl 14
21 20 nnrei 14
22 7 16 decnncl 15
23 22 nnrei 15
24 4 21 23 lttri 1 < 14 14 < 15 1 < 15
25 9 18 24 mp2an 1 < 15
26 4 25 ltneii 1 15
27 ccondx comp ndx = 15
28 11 27 neeq12i Base ndx comp ndx 1 15
29 26 28 mpbir Base ndx comp ndx
30 3 29 setsnid Base C sSet Hom ndx tpos Hom C = Base C sSet Hom ndx tpos Hom C sSet comp ndx u Base C × Base C , z Base C tpos z 2 nd u comp C 1 st u
31 15 30 eqtri Base C = Base C sSet Hom ndx tpos Hom C sSet comp ndx u Base C × Base C , z Base C tpos z 2 nd u comp C 1 st u
32 eqid Base C = Base C
33 eqid Hom C = Hom C
34 eqid comp C = comp C
35 32 33 34 1 oppcval C V O = C sSet Hom ndx tpos Hom C sSet comp ndx u Base C × Base C , z Base C tpos z 2 nd u comp C 1 st u
36 35 fveq2d C V Base O = Base C sSet Hom ndx tpos Hom C sSet comp ndx u Base C × Base C , z Base C tpos z 2 nd u comp C 1 st u
37 31 36 eqtr4id C V Base C = Base O
38 base0 = Base
39 fvprc ¬ C V Base C =
40 fvprc ¬ C V oppCat C =
41 1 40 eqtrid ¬ C V O =
42 41 fveq2d ¬ C V Base O = Base
43 38 39 42 3eqtr4a ¬ C V Base C = Base O
44 37 43 pm2.61i Base C = Base O
45 2 44 eqtri B = Base O