Metamath Proof Explorer


Theorem ringinvnzdiv

Description: In a unital ring, a left invertible element is not a zero divisor. (Contributed by FL, 18-Apr-2010) (Revised by Jeff Madsen, 18-Apr-2010) (Revised by AV, 24-Aug-2021)

Ref Expression
Hypotheses ringinvnzdiv.b B=BaseR
ringinvnzdiv.t ·˙=R
ringinvnzdiv.u 1˙=1R
ringinvnzdiv.z 0˙=0R
ringinvnzdiv.r φRRing
ringinvnzdiv.x φXB
ringinvnzdiv.a φaBa·˙X=1˙
ringinvnzdiv.y φYB
Assertion ringinvnzdiv φX·˙Y=0˙Y=0˙

Proof

Step Hyp Ref Expression
1 ringinvnzdiv.b B=BaseR
2 ringinvnzdiv.t ·˙=R
3 ringinvnzdiv.u 1˙=1R
4 ringinvnzdiv.z 0˙=0R
5 ringinvnzdiv.r φRRing
6 ringinvnzdiv.x φXB
7 ringinvnzdiv.a φaBa·˙X=1˙
8 ringinvnzdiv.y φYB
9 1 2 3 ringlidm RRingYB1˙·˙Y=Y
10 5 8 9 syl2anc φ1˙·˙Y=Y
11 10 eqcomd φY=1˙·˙Y
12 11 ad3antrrr φaBa·˙X=1˙X·˙Y=0˙Y=1˙·˙Y
13 oveq1 1˙=a·˙X1˙·˙Y=a·˙X·˙Y
14 13 eqcoms a·˙X=1˙1˙·˙Y=a·˙X·˙Y
15 14 adantl φaBa·˙X=1˙1˙·˙Y=a·˙X·˙Y
16 5 adantr φaBRRing
17 simpr φaBaB
18 6 adantr φaBXB
19 8 adantr φaBYB
20 17 18 19 3jca φaBaBXBYB
21 16 20 jca φaBRRingaBXBYB
22 21 adantr φaBa·˙X=1˙RRingaBXBYB
23 1 2 ringass RRingaBXBYBa·˙X·˙Y=a·˙X·˙Y
24 22 23 syl φaBa·˙X=1˙a·˙X·˙Y=a·˙X·˙Y
25 15 24 eqtrd φaBa·˙X=1˙1˙·˙Y=a·˙X·˙Y
26 25 adantr φaBa·˙X=1˙X·˙Y=0˙1˙·˙Y=a·˙X·˙Y
27 oveq2 X·˙Y=0˙a·˙X·˙Y=a·˙0˙
28 1 2 4 ringrz RRingaBa·˙0˙=0˙
29 5 28 sylan φaBa·˙0˙=0˙
30 29 adantr φaBa·˙X=1˙a·˙0˙=0˙
31 27 30 sylan9eqr φaBa·˙X=1˙X·˙Y=0˙a·˙X·˙Y=0˙
32 12 26 31 3eqtrd φaBa·˙X=1˙X·˙Y=0˙Y=0˙
33 32 exp31 φaBa·˙X=1˙X·˙Y=0˙Y=0˙
34 33 rexlimdva φaBa·˙X=1˙X·˙Y=0˙Y=0˙
35 7 34 mpd φX·˙Y=0˙Y=0˙
36 oveq2 Y=0˙X·˙Y=X·˙0˙
37 1 2 4 ringrz RRingXBX·˙0˙=0˙
38 5 6 37 syl2anc φX·˙0˙=0˙
39 36 38 sylan9eqr φY=0˙X·˙Y=0˙
40 39 ex φY=0˙X·˙Y=0˙
41 35 40 impbid φX·˙Y=0˙Y=0˙