Metamath Proof Explorer


Theorem mndractf1

Description: If an element X of a monoid E is right-invertible, with inverse Y , then its left-translation G is injective. See also grplactf1o . Remark in chapter I. of BourbakiAlg1 p. 17 . (Contributed by Thierry Arnoux, 3-Aug-2025)

Ref Expression
Hypotheses mndractfo.b B = Base E
mndractfo.z 0 ˙ = 0 E
mndractfo.p + ˙ = + E
mndractfo.f G = a B a + ˙ X
mndractfo.e φ E Mnd
mndractfo.x φ X B
mndractf1.1 φ Y B
mndractf1.2 φ X + ˙ Y = 0 ˙
Assertion mndractf1 φ G : B 1-1 B

Proof

Step Hyp Ref Expression
1 mndractfo.b B = Base E
2 mndractfo.z 0 ˙ = 0 E
3 mndractfo.p + ˙ = + E
4 mndractfo.f G = a B a + ˙ X
5 mndractfo.e φ E Mnd
6 mndractfo.x φ X B
7 mndractf1.1 φ Y B
8 mndractf1.2 φ X + ˙ Y = 0 ˙
9 5 adantr φ a B E Mnd
10 simpr φ a B a B
11 6 adantr φ a B X B
12 1 3 9 10 11 mndcld φ a B a + ˙ X B
13 12 4 fmptd φ G : B B
14 simpr φ i B j B G i = G j G i = G j
15 oveq1 a = i a + ˙ X = i + ˙ X
16 simpllr φ i B j B G i = G j i B
17 ovexd φ i B j B G i = G j i + ˙ X V
18 4 15 16 17 fvmptd3 φ i B j B G i = G j G i = i + ˙ X
19 oveq1 a = j a + ˙ X = j + ˙ X
20 simplr φ i B j B G i = G j j B
21 ovexd φ i B j B G i = G j j + ˙ X V
22 4 19 20 21 fvmptd3 φ i B j B G i = G j G j = j + ˙ X
23 14 18 22 3eqtr3d φ i B j B G i = G j i + ˙ X = j + ˙ X
24 23 oveq1d φ i B j B G i = G j i + ˙ X + ˙ Y = j + ˙ X + ˙ Y
25 5 ad3antrrr φ i B j B G i = G j E Mnd
26 6 ad3antrrr φ i B j B G i = G j X B
27 7 ad3antrrr φ i B j B G i = G j Y B
28 1 3 25 16 26 27 mndassd φ i B j B G i = G j i + ˙ X + ˙ Y = i + ˙ X + ˙ Y
29 1 3 25 20 26 27 mndassd φ i B j B G i = G j j + ˙ X + ˙ Y = j + ˙ X + ˙ Y
30 24 28 29 3eqtr3d φ i B j B G i = G j i + ˙ X + ˙ Y = j + ˙ X + ˙ Y
31 8 ad3antrrr φ i B j B G i = G j X + ˙ Y = 0 ˙
32 31 oveq2d φ i B j B G i = G j i + ˙ X + ˙ Y = i + ˙ 0 ˙
33 31 oveq2d φ i B j B G i = G j j + ˙ X + ˙ Y = j + ˙ 0 ˙
34 30 32 33 3eqtr3d φ i B j B G i = G j i + ˙ 0 ˙ = j + ˙ 0 ˙
35 1 3 2 mndrid E Mnd i B i + ˙ 0 ˙ = i
36 25 16 35 syl2anc φ i B j B G i = G j i + ˙ 0 ˙ = i
37 1 3 2 mndrid E Mnd j B j + ˙ 0 ˙ = j
38 25 20 37 syl2anc φ i B j B G i = G j j + ˙ 0 ˙ = j
39 34 36 38 3eqtr3d φ i B j B G i = G j i = j
40 39 ex φ i B j B G i = G j i = j
41 40 anasss φ i B j B G i = G j i = j
42 41 ralrimivva φ i B j B G i = G j i = j
43 dff13 G : B 1-1 B G : B B i B j B G i = G j i = j
44 13 42 43 sylanbrc φ G : B 1-1 B