Metamath Proof Explorer


Theorem oppginv

Description: Inverses in a group are a symmetric notion. (Contributed by Stefan O'Rear, 26-Aug-2015)

Ref Expression
Hypotheses oppgbas.1 ⊢ O = opp 𝑔 ⁡ R
oppginv.2 ⊢ I = inv g ⁡ R
Assertion oppginv ⊢ R ∈ Grp → I = inv g ⁡ O

Proof

Step Hyp Ref Expression
1 oppgbas.1 ⊢ O = opp 𝑔 ⁡ R
2 oppginv.2 ⊢ I = inv g ⁡ R
3 eqid ⊢ Base R = Base R
4 3 2 grpinvf ⊢ R ∈ Grp → I : Base R ⟶ Base R
5 eqid ⊢ + R = + R
6 eqid ⊢ + O = + O
7 5 1 6 oppgplus ⊢ I ⁡ x + O x = x + R I ⁡ x
8 eqid ⊢ 0 R = 0 R
9 3 5 8 2 grprinv ⊢ R ∈ Grp ∧ x ∈ Base R → x + R I ⁡ x = 0 R
10 7 9 eqtrid ⊢ R ∈ Grp ∧ x ∈ Base R → I ⁡ x + O x = 0 R
11 10 ralrimiva ⊢ R ∈ Grp → ∀ x ∈ Base R I ⁡ x + O x = 0 R
12 1 oppggrp ⊢ R ∈ Grp → O ∈ Grp
13 1 3 oppgbas ⊢ Base R = Base O
14 1 8 oppgid ⊢ 0 R = 0 O
15 eqid ⊢ inv g ⁡ O = inv g ⁡ O
16 13 6 14 15 isgrpinv ⊢ O ∈ Grp → I : Base R ⟶ Base R ∧ ∀ x ∈ Base R I ⁡ x + O x = 0 R ↔ inv g ⁡ O = I
17 12 16 syl ⊢ R ∈ Grp → I : Base R ⟶ Base R ∧ ∀ x ∈ Base R I ⁡ x + O x = 0 R ↔ inv g ⁡ O = I
18 4 11 17 mpbi2and ⊢ R ∈ Grp → inv g ⁡ O = I
19 18 eqcomd ⊢ R ∈ Grp → I = inv g ⁡ O