Metamath Proof Explorer


Theorem remulinvcom

Description: A left multiplicative inverse is a right multiplicative inverse. Proven without ax-mulcom . (Contributed by SN, 5-Feb-2024)

Ref Expression
Hypotheses remulinvcom.1 φA
remulinvcom.2 φB
remulinvcom.3 φAB=1
Assertion remulinvcom φBA=1

Proof

Step Hyp Ref Expression
1 remulinvcom.1 φA
2 remulinvcom.2 φB
3 remulinvcom.3 φAB=1
4 ax-1ne0 10
5 4 a1i φ10
6 3 5 eqnetrd φAB0
7 simpr φB=0B=0
8 7 oveq2d φB=0AB=A0
9 1 adantr φB=0A
10 remul01 AA0=0
11 9 10 syl φB=0A0=0
12 8 11 eqtrd φB=0AB=0
13 6 12 mteqand φB0
14 ax-rrecex BB0xBx=1
15 2 13 14 syl2anc φxBx=1
16 simprl φxBx=1x
17 simprr φxBx=1Bx=1
18 4 a1i φxBx=110
19 17 18 eqnetrd φxBx=1Bx0
20 simpr φxBx=1x=0x=0
21 20 oveq2d φxBx=1x=0Bx=B0
22 2 ad2antrr φxBx=1x=0B
23 remul01 BB0=0
24 22 23 syl φxBx=1x=0B0=0
25 21 24 eqtrd φxBx=1x=0Bx=0
26 19 25 mteqand φxBx=1x0
27 ax-rrecex xx0yxy=1
28 16 26 27 syl2anc φxBx=1yxy=1
29 simplrr φxBx=1yxy=1Bx=1
30 29 oveq2d φxBx=1yxy=1ABx=A1
31 30 oveq1d φxBx=1yxy=1ABxy=A1y
32 1 ad2antrr φxBx=1yxy=1A
33 2 ad2antrr φxBx=1yxy=1B
34 32 33 remulcld φxBx=1yxy=1AB
35 34 recnd φxBx=1yxy=1AB
36 simplrl φxBx=1yxy=1x
37 36 recnd φxBx=1yxy=1x
38 simprl φxBx=1yxy=1y
39 38 recnd φxBx=1yxy=1y
40 35 37 39 mulassd φxBx=1yxy=1ABxy=ABxy
41 32 recnd φxBx=1yxy=1A
42 33 recnd φxBx=1yxy=1B
43 41 42 37 mulassd φxBx=1yxy=1ABx=ABx
44 43 oveq1d φxBx=1yxy=1ABxy=ABxy
45 3 ad2antrr φxBx=1yxy=1AB=1
46 simprr φxBx=1yxy=1xy=1
47 45 46 oveq12d φxBx=1yxy=1ABxy=11
48 1t1e1ALT 11=1
49 47 48 eqtrdi φxBx=1yxy=1ABxy=1
50 40 44 49 3eqtr3d φxBx=1yxy=1ABxy=1
51 ax-1rid AA1=A
52 32 51 syl φxBx=1yxy=1A1=A
53 52 oveq1d φxBx=1yxy=1A1y=Ay
54 31 50 53 3eqtr3rd φxBx=1yxy=1Ay=1
55 54 46 eqtr4d φxBx=1yxy=1Ay=xy
56 4 a1i φxBx=1yxy=110
57 46 56 eqnetrd φxBx=1yxy=1xy0
58 simpr φxBx=1yxy=1y=0y=0
59 58 oveq2d φxBx=1yxy=1y=0xy=x0
60 36 adantr φxBx=1yxy=1y=0x
61 remul01 xx0=0
62 60 61 syl φxBx=1yxy=1y=0x0=0
63 59 62 eqtrd φxBx=1yxy=1y=0xy=0
64 57 63 mteqand φxBx=1yxy=1y0
65 32 36 38 64 remulcan2d φxBx=1yxy=1Ay=xyA=x
66 55 65 mpbid φxBx=1yxy=1A=x
67 simpr φxBx=1yxy=1A=xA=x
68 67 oveq2d φxBx=1yxy=1A=xBA=Bx
69 17 ad2antrr φxBx=1yxy=1A=xBx=1
70 68 69 eqtrd φxBx=1yxy=1A=xBA=1
71 66 70 mpdan φxBx=1yxy=1BA=1
72 28 71 rexlimddv φxBx=1BA=1
73 15 72 rexlimddv φBA=1