Metamath Proof Explorer


Theorem hdmaprnlem15N

Description: Lemma for hdmaprnN . Eliminate u . (Contributed by NM, 30-May-2015) (New usage is discouraged.)

Ref Expression
Hypotheses hdmaprnlem15.h H=LHypK
hdmaprnlem15.u U=DVecHKW
hdmaprnlem15.v V=BaseU
hdmaprnlem15.n N=LSpanU
hdmaprnlem15.c C=LCDualKW
hdmaprnlem15.d D=BaseC
hdmaprnlem15.q 0˙=0C
hdmaprnlem15.l L=LSpanC
hdmaprnlem15.m M=mapdKW
hdmaprnlem15.s S=HDMapKW
hdmaprnlem15.k φKHLWH
hdmaprnlem15.se φsD0˙
hdmaprnlem15.ve φvV
hdmaprnlem15.e φMNv=Ls
Assertion hdmaprnlem15N φsranS

Proof

Step Hyp Ref Expression
1 hdmaprnlem15.h H=LHypK
2 hdmaprnlem15.u U=DVecHKW
3 hdmaprnlem15.v V=BaseU
4 hdmaprnlem15.n N=LSpanU
5 hdmaprnlem15.c C=LCDualKW
6 hdmaprnlem15.d D=BaseC
7 hdmaprnlem15.q 0˙=0C
8 hdmaprnlem15.l L=LSpanC
9 hdmaprnlem15.m M=mapdKW
10 hdmaprnlem15.s S=HDMapKW
11 hdmaprnlem15.k φKHLWH
12 hdmaprnlem15.se φsD0˙
13 hdmaprnlem15.ve φvV
14 hdmaprnlem15.e φMNv=Ls
15 1 2 3 4 11 13 dvh2dim φtV¬tNv
16 11 3ad2ant1 φtV¬tNvKHLWH
17 12 3ad2ant1 φtV¬tNvsD0˙
18 13 3ad2ant1 φtV¬tNvvV
19 14 3ad2ant1 φtV¬tNvMNv=Ls
20 simp2 φtV¬tNvtV
21 simp3 φtV¬tNv¬tNv
22 eqid 0U=0U
23 eqid +C=+C
24 eqid +U=+U
25 1 2 3 4 5 8 9 10 16 17 18 19 20 21 6 7 22 23 24 hdmaprnlem11N φtV¬tNvsranS
26 25 rexlimdv3a φtV¬tNvsranS
27 15 26 mpd φsranS