Metamath Proof Explorer


Theorem hdmap14lem7

Description: Combine cases of F . TODO: Can this be done at once in hdmap14lem3 , in order to get rid of hdmap14lem6 ? Perhaps modify lspsneu to become E! k e. K instead of E! k e. ( K \ { .0. } ) ? (Contributed by NM, 1-Jun-2015)

Ref Expression
Hypotheses hdmap14lem7.h H=LHypK
hdmap14lem7.u U=DVecHKW
hdmap14lem7.v V=BaseU
hdmap14lem7.t ·˙=U
hdmap14lem7.o 0˙=0U
hdmap14lem7.r R=ScalarU
hdmap14lem7.b B=BaseR
hdmap14lem7.c C=LCDualKW
hdmap14lem7.e ˙=C
hdmap14lem7.p P=ScalarC
hdmap14lem7.a A=BaseP
hdmap14lem7.s S=HDMapKW
hdmap14lem7.k φKHLWH
hdmap14lem7.x φXV0˙
hdmap14lem7.f φFB
Assertion hdmap14lem7 φ∃!gASF·˙X=g˙SX

Proof

Step Hyp Ref Expression
1 hdmap14lem7.h H=LHypK
2 hdmap14lem7.u U=DVecHKW
3 hdmap14lem7.v V=BaseU
4 hdmap14lem7.t ·˙=U
5 hdmap14lem7.o 0˙=0U
6 hdmap14lem7.r R=ScalarU
7 hdmap14lem7.b B=BaseR
8 hdmap14lem7.c C=LCDualKW
9 hdmap14lem7.e ˙=C
10 hdmap14lem7.p P=ScalarC
11 hdmap14lem7.a A=BaseP
12 hdmap14lem7.s S=HDMapKW
13 hdmap14lem7.k φKHLWH
14 hdmap14lem7.x φXV0˙
15 hdmap14lem7.f φFB
16 eqid 0R=0R
17 eqid LSpanC=LSpanC
18 eqid 0P=0P
19 13 adantr φF=0RKHLWH
20 14 adantr φF=0RXV0˙
21 simpr φF=0RF=0R
22 1 2 3 4 5 6 7 16 8 9 17 10 11 18 12 19 20 21 hdmap14lem6 φF=0R∃!gASF·˙X=g˙SX
23 13 adantr φF0RKHLWH
24 14 adantr φF0RXV0˙
25 15 adantr φF0RFB
26 simpr φF0RF0R
27 eldifsn FB0RFBF0R
28 25 26 27 sylanbrc φF0RFB0R
29 1 2 3 4 5 6 7 16 8 9 17 10 11 18 12 23 24 28 hdmap14lem4 φF0R∃!gASF·˙X=g˙SX
30 22 29 pm2.61dane φ∃!gASF·˙X=g˙SX