Metamath Proof Explorer


Theorem oppcbas

Description: Base set of an opposite category. (Contributed by Mario Carneiro, 2-Jan-2017)

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

Proof

Step Hyp Ref Expression
1 oppcbas.1 O = oppCat C
2 oppcbas.2 B = Base C
3 eqid Base C = Base C
4 eqid Hom C = Hom C
5 eqid comp C = comp C
6 3 4 5 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
7 6 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
8 baseid Base = Slot Base ndx
9 1re 1
10 1nn 1
11 4nn0 4 0
12 1nn0 1 0
13 1lt10 1 < 10
14 10 11 12 13 declti 1 < 14
15 9 14 ltneii 1 14
16 basendx Base ndx = 1
17 homndx Hom ndx = 14
18 16 17 neeq12i Base ndx Hom ndx 1 14
19 15 18 mpbir Base ndx Hom ndx
20 8 19 setsnid Base C = Base C sSet Hom ndx tpos Hom C
21 5nn 5
22 4lt5 4 < 5
23 12 11 21 22 declt 14 < 15
24 4nn 4
25 12 24 decnncl 14
26 25 nnrei 14
27 12 21 decnncl 15
28 27 nnrei 15
29 9 26 28 lttri 1 < 14 14 < 15 1 < 15
30 14 23 29 mp2an 1 < 15
31 9 30 ltneii 1 15
32 ccondx comp ndx = 15
33 16 32 neeq12i Base ndx comp ndx 1 15
34 31 33 mpbir Base ndx comp ndx
35 8 34 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
36 20 35 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
37 7 36 syl6reqr C V Base C = Base O
38 base0 = Base
39 fvprc ¬ C V Base C =
40 fvprc ¬ C V oppCat C =
41 1 40 syl5eq ¬ 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