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