Metamath Proof Explorer


Theorem opprabs

Description: The opposite ring of the opposite ring is the original ring. Note the conditions on this theorem, which makes it unpractical in case we only have e.g. R e. Ring as a premise. (Contributed by Thierry Arnoux, 9-Mar-2025)

Ref Expression
Hypotheses opprabs.o O = opp r R
opprabs.m · ˙ = R
opprabs.1 φ R V
opprabs.2 φ Fun R
opprabs.3 φ ndx dom R
opprabs.4 φ · ˙ Fn B × B
Assertion opprabs φ R = opp r O

Proof

Step Hyp Ref Expression
1 opprabs.o O = opp r R
2 opprabs.m · ˙ = R
3 opprabs.1 φ R V
4 opprabs.2 φ Fun R
5 opprabs.3 φ ndx dom R
6 opprabs.4 φ · ˙ Fn B × B
7 eqid Base R = Base R
8 eqid O = O
9 7 2 1 8 opprmulfval O = tpos · ˙
10 9 tposeqi tpos O = tpos tpos · ˙
11 fnrel · ˙ Fn B × B Rel · ˙
12 relxp Rel B × B
13 fndm · ˙ Fn B × B dom · ˙ = B × B
14 13 releqd · ˙ Fn B × B Rel dom · ˙ Rel B × B
15 12 14 mpbiri · ˙ Fn B × B Rel dom · ˙
16 tpostpos2 Rel · ˙ Rel dom · ˙ tpos tpos · ˙ = · ˙
17 11 15 16 syl2anc · ˙ Fn B × B tpos tpos · ˙ = · ˙
18 10 17 eqtrid · ˙ Fn B × B tpos O = · ˙
19 6 18 syl φ tpos O = · ˙
20 19 2 eqtrdi φ tpos O = R
21 20 opeq2d φ ndx tpos O = ndx R
22 21 oveq2d φ R sSet ndx tpos O = R sSet ndx R
23 1 7 opprbas Base R = Base O
24 eqid opp r O = opp r O
25 23 8 24 opprval opp r O = O sSet ndx tpos O
26 7 2 1 opprval O = R sSet ndx tpos · ˙
27 26 oveq1i O sSet ndx tpos O = R sSet ndx tpos · ˙ sSet ndx tpos O
28 25 27 eqtri opp r O = R sSet ndx tpos · ˙ sSet ndx tpos O
29 fvex O V
30 29 tposex tpos O V
31 setsabs R V tpos O V R sSet ndx tpos · ˙ sSet ndx tpos O = R sSet ndx tpos O
32 30 31 mpan2 R V R sSet ndx tpos · ˙ sSet ndx tpos O = R sSet ndx tpos O
33 28 32 eqtrid R V opp r O = R sSet ndx tpos O
34 3 33 syl φ opp r O = R sSet ndx tpos O
35 mulridx 𝑟 = Slot ndx
36 35 3 4 5 setsidvald φ R = R sSet ndx R
37 22 34 36 3eqtr4rd φ R = opp r O