Metamath Proof Explorer


Theorem oppchomfval

Description: Hom-sets of the opposite category. (Contributed by Mario Carneiro, 2-Jan-2017)

Ref Expression
Hypotheses oppchom.h H = Hom C
oppchom.o O = oppCat C
Assertion oppchomfval tpos H = Hom O

Proof

Step Hyp Ref Expression
1 oppchom.h H = Hom C
2 oppchom.o O = oppCat C
3 homid Hom = Slot Hom ndx
4 1nn0 1 0
5 4nn 4
6 4 5 decnncl 14
7 6 nnrei 14
8 4nn0 4 0
9 5nn 5
10 4lt5 4 < 5
11 4 8 9 10 declt 14 < 15
12 7 11 ltneii 14 15
13 homndx Hom ndx = 14
14 ccondx comp ndx = 15
15 13 14 neeq12i Hom ndx comp ndx 14 15
16 12 15 mpbir Hom ndx comp ndx
17 3 16 setsnid Hom C sSet Hom ndx tpos H = Hom C sSet Hom ndx tpos H sSet comp ndx u Base C × Base C , z Base C tpos z 2 nd u comp C 1 st u
18 1 fvexi H V
19 18 tposex tpos H V
20 3 setsid C V tpos H V tpos H = Hom C sSet Hom ndx tpos H
21 19 20 mpan2 C V tpos H = Hom C sSet Hom ndx tpos H
22 eqid Base C = Base C
23 eqid comp C = comp C
24 22 1 23 2 oppcval C V O = C sSet Hom ndx tpos H sSet comp ndx u Base C × Base C , z Base C tpos z 2 nd u comp C 1 st u
25 24 fveq2d C V Hom O = Hom C sSet Hom ndx tpos H sSet comp ndx u Base C × Base C , z Base C tpos z 2 nd u comp C 1 st u
26 17 21 25 3eqtr4a C V tpos H = Hom O
27 tpos0 tpos =
28 fvprc ¬ C V Hom C =
29 1 28 syl5eq ¬ C V H =
30 29 tposeqd ¬ C V tpos H = tpos
31 fvprc ¬ C V oppCat C =
32 2 31 syl5eq ¬ C V O =
33 32 fveq2d ¬ C V Hom O = Hom
34 df-hom Hom = Slot 14
35 34 str0 = Hom
36 33 35 eqtr4di ¬ C V Hom O =
37 27 30 36 3eqtr4a ¬ C V tpos H = Hom O
38 26 37 pm2.61i tpos H = Hom O