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 = LHyp K
hdmaprnlem15.u U = DVecH K W
hdmaprnlem15.v V = Base U
hdmaprnlem15.n N = LSpan U
hdmaprnlem15.c C = LCDual K W
hdmaprnlem15.d D = Base C
hdmaprnlem15.q 0 ˙ = 0 C
hdmaprnlem15.l L = LSpan C
hdmaprnlem15.m M = mapd K W
hdmaprnlem15.s S = HDMap K W
hdmaprnlem15.k φ K HL W H
hdmaprnlem15.se φ s D 0 ˙
hdmaprnlem15.ve φ v V
hdmaprnlem15.e φ M N v = L s
Assertion hdmaprnlem15N φ s ran S

Proof

Step Hyp Ref Expression
1 hdmaprnlem15.h H = LHyp K
2 hdmaprnlem15.u U = DVecH K W
3 hdmaprnlem15.v V = Base U
4 hdmaprnlem15.n N = LSpan U
5 hdmaprnlem15.c C = LCDual K W
6 hdmaprnlem15.d D = Base C
7 hdmaprnlem15.q 0 ˙ = 0 C
8 hdmaprnlem15.l L = LSpan C
9 hdmaprnlem15.m M = mapd K W
10 hdmaprnlem15.s S = HDMap K W
11 hdmaprnlem15.k φ K HL W H
12 hdmaprnlem15.se φ s D 0 ˙
13 hdmaprnlem15.ve φ v V
14 hdmaprnlem15.e φ M N v = L s
15 1 2 3 4 11 13 dvh2dim φ t V ¬ t N v
16 11 3ad2ant1 φ t V ¬ t N v K HL W H
17 12 3ad2ant1 φ t V ¬ t N v s D 0 ˙
18 13 3ad2ant1 φ t V ¬ t N v v V
19 14 3ad2ant1 φ t V ¬ t N v M N v = L s
20 simp2 φ t V ¬ t N v t V
21 simp3 φ t V ¬ t N v ¬ t N v
22 eqid 0 U = 0 U
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 φ t V ¬ t N v s ran S
26 25 rexlimdv3a φ t V ¬ t N v s ran S
27 15 26 mpd φ s ran S