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 = ( oppg𝑅 )
oppginv.2 𝐼 = ( invg𝑅 )
Assertion oppginv ( 𝑅 ∈ Grp → 𝐼 = ( invg𝑂 ) )

Proof

Step Hyp Ref Expression
1 oppgbas.1 𝑂 = ( oppg𝑅 )
2 oppginv.2 𝐼 = ( invg𝑅 )
3 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
4 3 2 grpinvf ( 𝑅 ∈ Grp → 𝐼 : ( Base ‘ 𝑅 ) ⟶ ( Base ‘ 𝑅 ) )
5 eqid ( +g𝑅 ) = ( +g𝑅 )
6 eqid ( +g𝑂 ) = ( +g𝑂 )
7 5 1 6 oppgplus ( ( 𝐼𝑥 ) ( +g𝑂 ) 𝑥 ) = ( 𝑥 ( +g𝑅 ) ( 𝐼𝑥 ) )
8 eqid ( 0g𝑅 ) = ( 0g𝑅 )
9 3 5 8 2 grprinv ( ( 𝑅 ∈ Grp ∧ 𝑥 ∈ ( Base ‘ 𝑅 ) ) → ( 𝑥 ( +g𝑅 ) ( 𝐼𝑥 ) ) = ( 0g𝑅 ) )
10 7 9 eqtrid ( ( 𝑅 ∈ Grp ∧ 𝑥 ∈ ( Base ‘ 𝑅 ) ) → ( ( 𝐼𝑥 ) ( +g𝑂 ) 𝑥 ) = ( 0g𝑅 ) )
11 10 ralrimiva ( 𝑅 ∈ Grp → ∀ 𝑥 ∈ ( Base ‘ 𝑅 ) ( ( 𝐼𝑥 ) ( +g𝑂 ) 𝑥 ) = ( 0g𝑅 ) )
12 1 oppggrp ( 𝑅 ∈ Grp → 𝑂 ∈ Grp )
13 1 3 oppgbas ( Base ‘ 𝑅 ) = ( Base ‘ 𝑂 )
14 1 8 oppgid ( 0g𝑅 ) = ( 0g𝑂 )
15 eqid ( invg𝑂 ) = ( invg𝑂 )
16 13 6 14 15 isgrpinv ( 𝑂 ∈ Grp → ( ( 𝐼 : ( Base ‘ 𝑅 ) ⟶ ( Base ‘ 𝑅 ) ∧ ∀ 𝑥 ∈ ( Base ‘ 𝑅 ) ( ( 𝐼𝑥 ) ( +g𝑂 ) 𝑥 ) = ( 0g𝑅 ) ) ↔ ( invg𝑂 ) = 𝐼 ) )
17 12 16 syl ( 𝑅 ∈ Grp → ( ( 𝐼 : ( Base ‘ 𝑅 ) ⟶ ( Base ‘ 𝑅 ) ∧ ∀ 𝑥 ∈ ( Base ‘ 𝑅 ) ( ( 𝐼𝑥 ) ( +g𝑂 ) 𝑥 ) = ( 0g𝑅 ) ) ↔ ( invg𝑂 ) = 𝐼 ) )
18 4 11 17 mpbi2and ( 𝑅 ∈ Grp → ( invg𝑂 ) = 𝐼 )
19 18 eqcomd ( 𝑅 ∈ Grp → 𝐼 = ( invg𝑂 ) )