Metamath Proof Explorer


Theorem oppcbas

Description: Base set of an opposite category. (Contributed by Mario Carneiro, 2-Jan-2017) (Proof shortened by AV, 18-Oct-2024)

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 baseid Base = Slot Base ndx
4 slotsbhcdif Base ndx Hom ndx Base ndx comp ndx Hom ndx comp ndx
5 4 simp1i Base ndx Hom ndx
6 3 5 setsnid Base C = Base C sSet Hom ndx tpos Hom C
7 4 simp2i Base ndx comp ndx
8 3 7 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
9 6 8 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
10 eqid Base C = Base C
11 eqid Hom C = Hom C
12 eqid comp C = comp C
13 10 11 12 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
14 13 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
15 9 14 eqtr4id C V Base C = Base O
16 base0 = Base
17 16 eqcomi Base =
18 17 1 fveqprc ¬ C V Base C = Base O
19 15 18 pm2.61i Base C = Base O
20 2 19 eqtri B = Base O