Metamath Proof Explorer


Theorem dia2dimlem13

Description: Lemma for dia2dim . Eliminate U =/= V condition. (Contributed by NM, 8-Sep-2014)

Ref Expression
Hypotheses dia2dimlem12.l ˙=K
dia2dimlem12.j ˙=joinK
dia2dimlem12.m ˙=meetK
dia2dimlem12.a A=AtomsK
dia2dimlem12.h H=LHypK
dia2dimlem12.t T=LTrnKW
dia2dimlem12.r R=trLKW
dia2dimlem12.y Y=DVecAKW
dia2dimlem12.s S=LSubSpY
dia2dimlem12.pl ˙=LSSumY
dia2dimlem12.n N=LSpanY
dia2dimlem12.i I=DIsoAKW
dia2dimlem12.k φKHLWH
dia2dimlem12.u φUAU˙W
dia2dimlem12.v φVAV˙W
Assertion dia2dimlem13 φIU˙VIU˙IV

Proof

Step Hyp Ref Expression
1 dia2dimlem12.l ˙=K
2 dia2dimlem12.j ˙=joinK
3 dia2dimlem12.m ˙=meetK
4 dia2dimlem12.a A=AtomsK
5 dia2dimlem12.h H=LHypK
6 dia2dimlem12.t T=LTrnKW
7 dia2dimlem12.r R=trLKW
8 dia2dimlem12.y Y=DVecAKW
9 dia2dimlem12.s S=LSubSpY
10 dia2dimlem12.pl ˙=LSSumY
11 dia2dimlem12.n N=LSpanY
12 dia2dimlem12.i I=DIsoAKW
13 dia2dimlem12.k φKHLWH
14 dia2dimlem12.u φUAU˙W
15 dia2dimlem12.v φVAV˙W
16 oveq2 U=VU˙U=U˙V
17 16 adantl φU=VU˙U=U˙V
18 13 simpld φKHL
19 14 simpld φUA
20 2 4 hlatjidm KHLUAU˙U=U
21 18 19 20 syl2anc φU˙U=U
22 21 adantr φU=VU˙U=U
23 17 22 eqtr3d φU=VU˙V=U
24 23 fveq2d φU=VIU˙V=IU
25 ssid IUIU
26 24 25 eqsstrdi φU=VIU˙VIU
27 5 8 dvalvec KHLWHYLVec
28 lveclmod YLVecYLMod
29 13 27 28 3syl φYLMod
30 eqid BaseK=BaseK
31 30 4 atbase UAUBaseK
32 19 31 syl φUBaseK
33 14 simprd φU˙W
34 30 1 5 8 12 9 dialss KHLWHUBaseKU˙WIUS
35 13 32 33 34 syl12anc φIUS
36 9 lsssubg YLModIUSIUSubGrpY
37 29 35 36 syl2anc φIUSubGrpY
38 10 lsmidm IUSubGrpYIU˙IU=IU
39 37 38 syl φIU˙IU=IU
40 fveq2 U=VIU=IV
41 40 oveq2d U=VIU˙IU=IU˙IV
42 39 41 sylan9req φU=VIU=IU˙IV
43 26 42 sseqtrd φU=VIU˙VIU˙IV
44 13 adantr φUVKHLWH
45 14 adantr φUVUAU˙W
46 15 adantr φUVVAV˙W
47 simpr φUVUV
48 1 2 3 4 5 6 7 8 9 10 11 12 44 45 46 47 dia2dimlem12 φUVIU˙VIU˙IV
49 43 48 pm2.61dane φIU˙VIU˙IV