Metamath Proof Explorer


Theorem ringinveu

Description: If a ring unit element X admits both a left inverse Y and a right inverse Z , they are equal. (Contributed by Thierry Arnoux, 9-Mar-2025)

Ref Expression
Hypotheses isdrng4.b
|- B = ( Base ` R )
isdrng4.0
|- .0. = ( 0g ` R )
isdrng4.1
|- .1. = ( 1r ` R )
isdrng4.x
|- .x. = ( .r ` R )
isdrng4.u
|- U = ( Unit ` R )
isdrng4.r
|- ( ph -> R e. Ring )
ringinveu.1
|- ( ph -> X e. B )
ringinveu.2
|- ( ph -> Y e. B )
ringinveu.3
|- ( ph -> Z e. B )
ringinveu.4
|- ( ph -> ( Y .x. X ) = .1. )
ringinveu.5
|- ( ph -> ( X .x. Z ) = .1. )
Assertion ringinveu
|- ( ph -> Z = Y )

Proof

Step Hyp Ref Expression
1 isdrng4.b
 |-  B = ( Base ` R )
2 isdrng4.0
 |-  .0. = ( 0g ` R )
3 isdrng4.1
 |-  .1. = ( 1r ` R )
4 isdrng4.x
 |-  .x. = ( .r ` R )
5 isdrng4.u
 |-  U = ( Unit ` R )
6 isdrng4.r
 |-  ( ph -> R e. Ring )
7 ringinveu.1
 |-  ( ph -> X e. B )
8 ringinveu.2
 |-  ( ph -> Y e. B )
9 ringinveu.3
 |-  ( ph -> Z e. B )
10 ringinveu.4
 |-  ( ph -> ( Y .x. X ) = .1. )
11 ringinveu.5
 |-  ( ph -> ( X .x. Z ) = .1. )
12 11 oveq2d
 |-  ( ph -> ( Y .x. ( X .x. Z ) ) = ( Y .x. .1. ) )
13 10 oveq1d
 |-  ( ph -> ( ( Y .x. X ) .x. Z ) = ( .1. .x. Z ) )
14 1 4 6 8 7 9 ringassd
 |-  ( ph -> ( ( Y .x. X ) .x. Z ) = ( Y .x. ( X .x. Z ) ) )
15 1 4 3 6 9 ringlidmd
 |-  ( ph -> ( .1. .x. Z ) = Z )
16 13 14 15 3eqtr3d
 |-  ( ph -> ( Y .x. ( X .x. Z ) ) = Z )
17 1 4 3 6 8 ringridmd
 |-  ( ph -> ( Y .x. .1. ) = Y )
18 12 16 17 3eqtr3d
 |-  ( ph -> Z = Y )