Metamath Proof Explorer


Theorem lindsrng01

Description: Any subset of a module is always linearly independent if the underlying ring has at most one element. Since the underlying ring cannot be the empty set (see lmodsn0 ), this means that the underlying ring has only one element, so it is a zero ring. (Contributed by AV, 14-Apr-2019) (Revised by AV, 27-Apr-2019)

Ref Expression
Hypotheses lindsrng01.b B=BaseM
lindsrng01.r R=ScalarM
lindsrng01.e E=BaseR
Assertion lindsrng01 MLModE=0E=1S𝒫BSlinIndSM

Proof

Step Hyp Ref Expression
1 lindsrng01.b B=BaseM
2 lindsrng01.r R=ScalarM
3 lindsrng01.e E=BaseR
4 2 3 lmodsn0 MLModE
5 3 fvexi EV
6 hasheq0 EVE=0E=
7 5 6 ax-mp E=0E=
8 eqneqall E=ESlinIndSM
9 8 com12 EE=SlinIndSM
10 7 9 biimtrid EE=0SlinIndSM
11 4 10 syl MLModE=0SlinIndSM
12 11 adantr MLModS𝒫BE=0SlinIndSM
13 12 com12 E=0MLModS𝒫BSlinIndSM
14 2 lmodring MLModRRing
15 14 adantr MLModS𝒫BRRing
16 eqid 0R=0R
17 3 16 0ring RRingE=1E=0R
18 15 17 sylan MLModS𝒫BE=1E=0R
19 simpr MLModS𝒫BS𝒫B
20 19 adantr MLModS𝒫BE=1S𝒫B
21 20 adantl E=0RMLModS𝒫BE=1S𝒫B
22 snex 0RV
23 20 22 jctil MLModS𝒫BE=10RVS𝒫B
24 23 adantl E=0RMLModS𝒫BE=10RVS𝒫B
25 elmapg 0RVS𝒫Bf0RSf:S0R
26 24 25 syl E=0RMLModS𝒫BE=1f0RSf:S0R
27 fvex 0RV
28 27 fconst2 f:S0Rf=S×0R
29 fconstmpt S×0R=xS0R
30 29 eqeq2i f=S×0Rf=xS0R
31 28 30 bitri f:S0Rf=xS0R
32 eqidd E=0RMLModS𝒫BE=1vSxS0R=xS0R
33 eqidd E=0RMLModS𝒫BE=1vSx=v0R=0R
34 simpr E=0RMLModS𝒫BE=1vSvS
35 fvexd E=0RMLModS𝒫BE=1vS0RV
36 32 33 34 35 fvmptd E=0RMLModS𝒫BE=1vSxS0Rv=0R
37 36 ralrimiva E=0RMLModS𝒫BE=1vSxS0Rv=0R
38 37 a1d E=0RMLModS𝒫BE=1finSupp0RxS0RxS0RlinCMS=0MvSxS0Rv=0R
39 breq1 f=xS0RfinSupp0RffinSupp0RxS0R
40 oveq1 f=xS0RflinCMS=xS0RlinCMS
41 40 eqeq1d f=xS0RflinCMS=0MxS0RlinCMS=0M
42 39 41 anbi12d f=xS0RfinSupp0RfflinCMS=0MfinSupp0RxS0RxS0RlinCMS=0M
43 fveq1 f=xS0Rfv=xS0Rv
44 43 eqeq1d f=xS0Rfv=0RxS0Rv=0R
45 44 ralbidv f=xS0RvSfv=0RvSxS0Rv=0R
46 42 45 imbi12d f=xS0RfinSupp0RfflinCMS=0MvSfv=0RfinSupp0RxS0RxS0RlinCMS=0MvSxS0Rv=0R
47 38 46 syl5ibrcom E=0RMLModS𝒫BE=1f=xS0RfinSupp0RfflinCMS=0MvSfv=0R
48 31 47 biimtrid E=0RMLModS𝒫BE=1f:S0RfinSupp0RfflinCMS=0MvSfv=0R
49 26 48 sylbid E=0RMLModS𝒫BE=1f0RSfinSupp0RfflinCMS=0MvSfv=0R
50 49 ralrimiv E=0RMLModS𝒫BE=1f0RSfinSupp0RfflinCMS=0MvSfv=0R
51 oveq1 E=0RES=0RS
52 51 raleqdv E=0RfESfinSupp0RfflinCMS=0MvSfv=0Rf0RSfinSupp0RfflinCMS=0MvSfv=0R
53 52 adantr E=0RMLModS𝒫BE=1fESfinSupp0RfflinCMS=0MvSfv=0Rf0RSfinSupp0RfflinCMS=0MvSfv=0R
54 50 53 mpbird E=0RMLModS𝒫BE=1fESfinSupp0RfflinCMS=0MvSfv=0R
55 simpl MLModS𝒫BE=1MLModS𝒫B
56 55 ancomd MLModS𝒫BE=1S𝒫BMLMod
57 56 adantl E=0RMLModS𝒫BE=1S𝒫BMLMod
58 eqid 0M=0M
59 1 58 2 3 16 islininds S𝒫BMLModSlinIndSMS𝒫BfESfinSupp0RfflinCMS=0MvSfv=0R
60 57 59 syl E=0RMLModS𝒫BE=1SlinIndSMS𝒫BfESfinSupp0RfflinCMS=0MvSfv=0R
61 21 54 60 mpbir2and E=0RMLModS𝒫BE=1SlinIndSM
62 18 61 mpancom MLModS𝒫BE=1SlinIndSM
63 62 expcom E=1MLModS𝒫BSlinIndSM
64 13 63 jaoi E=0E=1MLModS𝒫BSlinIndSM
65 64 expd E=0E=1MLModS𝒫BSlinIndSM
66 65 com12 MLModE=0E=1S𝒫BSlinIndSM
67 66 3imp MLModE=0E=1S𝒫BSlinIndSM