Metamath Proof Explorer


Theorem opprval

Description: Value of the opposite ring. (Contributed by Mario Carneiro, 1-Dec-2014)

Ref Expression
Hypotheses opprval.1 ⊢ B = Base R
opprval.2 ⊢ · ˙ = ⋅ R
opprval.3 ⊢ O = opp r ⁡ R
Assertion opprval ⊢ O = R sSet ⋅ ndx tpos · ˙

Proof

Step Hyp Ref Expression
1 opprval.1 ⊢ B = Base R
2 opprval.2 ⊢ · ˙ = ⋅ R
3 opprval.3 ⊢ O = opp r ⁡ R
4 id ⊢ x = R → x = R
5 fveq2 ⊢ x = R → ⋅ x = ⋅ R
6 5 2 eqtr4di ⊢ x = R → ⋅ x = · ˙
7 6 tposeqd ⊢ x = R → tpos ⋅ x = tpos · ˙
8 7 opeq2d ⊢ x = R → ⋅ ndx tpos ⋅ x = ⋅ ndx tpos · ˙
9 4 8 oveq12d ⊢ x = R → x sSet ⋅ ndx tpos ⋅ x = R sSet ⋅ ndx tpos · ˙
10 df-oppr ⊢ opp r = x ∈ V ⟼ x sSet ⋅ ndx tpos ⋅ x
11 ovex ⊢ R sSet ⋅ ndx tpos · ˙ ∈ V
12 9 10 11 fvmpt ⊢ R ∈ V → opp r ⁡ R = R sSet ⋅ ndx tpos · ˙
13 fvprc ⊢ ¬ R ∈ V → opp r ⁡ R = ∅
14 reldmsets ⊢ Rel ⁡ dom ⁡ sSet
15 14 ovprc1 ⊢ ¬ R ∈ V → R sSet ⋅ ndx tpos · ˙ = ∅
16 13 15 eqtr4d ⊢ ¬ R ∈ V → opp r ⁡ R = R sSet ⋅ ndx tpos · ˙
17 12 16 pm2.61i ⊢ opp r ⁡ R = R sSet ⋅ ndx tpos · ˙
18 3 17 eqtri ⊢ O = R sSet ⋅ ndx tpos · ˙