Metamath Proof Explorer


Theorem hdmaprnlem4tN

Description: Lemma for hdmaprnN . TODO: This lemma doesn't quite pay for itself even though used six times. Maybe prove this directly instead. (Contributed by NM, 27-May-2015) (New usage is discouraged.)

Ref Expression
Hypotheses hdmaprnlem1.h H = LHyp K
hdmaprnlem1.u U = DVecH K W
hdmaprnlem1.v V = Base U
hdmaprnlem1.n N = LSpan U
hdmaprnlem1.c C = LCDual K W
hdmaprnlem1.l L = LSpan C
hdmaprnlem1.m M = mapd K W
hdmaprnlem1.s S = HDMap K W
hdmaprnlem1.k φ K HL W H
hdmaprnlem1.se φ s D Q
hdmaprnlem1.ve φ v V
hdmaprnlem1.e φ M N v = L s
hdmaprnlem1.ue φ u V
hdmaprnlem1.un φ ¬ u N v
hdmaprnlem1.d D = Base C
hdmaprnlem1.q Q = 0 C
hdmaprnlem1.o 0 ˙ = 0 U
hdmaprnlem1.a ˙ = + C
hdmaprnlem1.t2 φ t N v 0 ˙
Assertion hdmaprnlem4tN φ t V

Proof

Step Hyp Ref Expression
1 hdmaprnlem1.h H = LHyp K
2 hdmaprnlem1.u U = DVecH K W
3 hdmaprnlem1.v V = Base U
4 hdmaprnlem1.n N = LSpan U
5 hdmaprnlem1.c C = LCDual K W
6 hdmaprnlem1.l L = LSpan C
7 hdmaprnlem1.m M = mapd K W
8 hdmaprnlem1.s S = HDMap K W
9 hdmaprnlem1.k φ K HL W H
10 hdmaprnlem1.se φ s D Q
11 hdmaprnlem1.ve φ v V
12 hdmaprnlem1.e φ M N v = L s
13 hdmaprnlem1.ue φ u V
14 hdmaprnlem1.un φ ¬ u N v
15 hdmaprnlem1.d D = Base C
16 hdmaprnlem1.q Q = 0 C
17 hdmaprnlem1.o 0 ˙ = 0 U
18 hdmaprnlem1.a ˙ = + C
19 hdmaprnlem1.t2 φ t N v 0 ˙
20 1 2 9 dvhlmod φ U LMod
21 11 snssd φ v V
22 3 4 lspssv U LMod v V N v V
23 20 21 22 syl2anc φ N v V
24 23 ssdifssd φ N v 0 ˙ V
25 24 19 sseldd φ t V