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 ˙ = 0 R
isdrng4.1 1 ˙ = 1 R
isdrng4.x · ˙ = R
isdrng4.u U = Unit R
isdrng4.r φ R Ring
ringinveu.1 φ X B
ringinveu.2 φ Y B
ringinveu.3 φ Z B
ringinveu.4 φ Y · ˙ X = 1 ˙
ringinveu.5 φ X · ˙ Z = 1 ˙
Assertion ringinveu φ Z = Y

Proof

Step Hyp Ref Expression
1 isdrng4.b B = Base R
2 isdrng4.0 0 ˙ = 0 R
3 isdrng4.1 1 ˙ = 1 R
4 isdrng4.x · ˙ = R
5 isdrng4.u U = Unit R
6 isdrng4.r φ R Ring
7 ringinveu.1 φ X B
8 ringinveu.2 φ Y B
9 ringinveu.3 φ Z B
10 ringinveu.4 φ Y · ˙ X = 1 ˙
11 ringinveu.5 φ X · ˙ Z = 1 ˙
12 11 oveq2d φ Y · ˙ X · ˙ Z = Y · ˙ 1 ˙
13 10 oveq1d φ Y · ˙ X · ˙ Z = 1 ˙ · ˙ Z
14 1 4 6 8 7 9 ringassd φ Y · ˙ X · ˙ Z = Y · ˙ X · ˙ Z
15 1 4 3 6 9 ringlidmd φ 1 ˙ · ˙ Z = Z
16 13 14 15 3eqtr3d φ Y · ˙ X · ˙ Z = Z
17 1 4 3 6 8 ringridmd φ Y · ˙ 1 ˙ = Y
18 12 16 17 3eqtr3d φ Z = Y