Metamath Proof Explorer


Theorem hdmap14lem4a

Description: Simplify ( A \ { Q } ) in hdmap14lem3 to provide a slightly simpler definition later. (Contributed by NM, 31-May-2015)

Ref Expression
Hypotheses hdmap14lem1.h H=LHypK
hdmap14lem1.u U=DVecHKW
hdmap14lem1.v V=BaseU
hdmap14lem1.t ·˙=U
hdmap14lem3.o 0˙=0U
hdmap14lem1.r R=ScalarU
hdmap14lem1.b B=BaseR
hdmap14lem1.z Z=0R
hdmap14lem1.c C=LCDualKW
hdmap14lem2.e ˙=C
hdmap14lem1.l L=LSpanC
hdmap14lem2.p P=ScalarC
hdmap14lem2.a A=BaseP
hdmap14lem2.q Q=0P
hdmap14lem1.s S=HDMapKW
hdmap14lem1.k φKHLWH
hdmap14lem3.x φXV0˙
hdmap14lem1.f φFBZ
Assertion hdmap14lem4a φ∃!gAQSF·˙X=g˙SX∃!gASF·˙X=g˙SX

Proof

Step Hyp Ref Expression
1 hdmap14lem1.h H=LHypK
2 hdmap14lem1.u U=DVecHKW
3 hdmap14lem1.v V=BaseU
4 hdmap14lem1.t ·˙=U
5 hdmap14lem3.o 0˙=0U
6 hdmap14lem1.r R=ScalarU
7 hdmap14lem1.b B=BaseR
8 hdmap14lem1.z Z=0R
9 hdmap14lem1.c C=LCDualKW
10 hdmap14lem2.e ˙=C
11 hdmap14lem1.l L=LSpanC
12 hdmap14lem2.p P=ScalarC
13 hdmap14lem2.a A=BaseP
14 hdmap14lem2.q Q=0P
15 hdmap14lem1.s S=HDMapKW
16 hdmap14lem1.k φKHLWH
17 hdmap14lem3.x φXV0˙
18 hdmap14lem1.f φFBZ
19 eqid 0C=0C
20 eqid BaseC=BaseC
21 1 2 16 dvhlmod φULMod
22 18 eldifad φFB
23 17 eldifad φXV
24 3 6 4 7 lmodvscl ULModFBXVF·˙XV
25 21 22 23 24 syl3anc φF·˙XV
26 eldifsni FBZFZ
27 18 26 syl φFZ
28 eldifsni XV0˙X0˙
29 17 28 syl φX0˙
30 1 2 16 dvhlvec φULVec
31 3 4 6 7 8 5 30 22 23 lvecvsn0 φF·˙X0˙FZX0˙
32 27 29 31 mpbir2and φF·˙X0˙
33 eldifsn F·˙XV0˙F·˙XVF·˙X0˙
34 25 32 33 sylanbrc φF·˙XV0˙
35 1 2 3 5 9 19 20 15 16 34 hdmapnzcl φSF·˙XBaseC0C
36 eldifsni SF·˙XBaseC0CSF·˙X0C
37 35 36 syl φSF·˙X0C
38 37 adantr φgQSF·˙X0C
39 elsni gQg=Q
40 39 oveq1d gQg˙SX=Q˙SX
41 1 9 16 lcdlmod φCLMod
42 1 2 3 9 20 15 16 23 hdmapcl φSXBaseC
43 20 12 10 14 19 lmod0vs CLModSXBaseCQ˙SX=0C
44 41 42 43 syl2anc φQ˙SX=0C
45 40 44 sylan9eqr φgQg˙SX=0C
46 38 45 neeqtrrd φgQSF·˙Xg˙SX
47 46 neneqd φgQ¬SF·˙X=g˙SX
48 47 nrexdv φ¬gQSF·˙X=g˙SX
49 reuun2 ¬gQSF·˙X=g˙SX∃!gAQQSF·˙X=g˙SX∃!gAQSF·˙X=g˙SX
50 48 49 syl φ∃!gAQQSF·˙X=g˙SX∃!gAQSF·˙X=g˙SX
51 12 13 14 lmod0cl CLModQA
52 difsnid QAAQQ=A
53 reueq1 AQQ=A∃!gAQQSF·˙X=g˙SX∃!gASF·˙X=g˙SX
54 41 51 52 53 4syl φ∃!gAQQSF·˙X=g˙SX∃!gASF·˙X=g˙SX
55 50 54 bitr3d φ∃!gAQSF·˙X=g˙SX∃!gASF·˙X=g˙SX