Metamath Proof Explorer


Theorem mapdpglem2a

Description: Lemma for mapdpg . (Contributed by NM, 20-Mar-2015)

Ref Expression
Hypotheses mapdpglem.h H = LHyp K
mapdpglem.m M = mapd K W
mapdpglem.u U = DVecH K W
mapdpglem.v V = Base U
mapdpglem.s - ˙ = - U
mapdpglem.n N = LSpan U
mapdpglem.c C = LCDual K W
mapdpglem.k φ K HL W H
mapdpglem.x φ X V
mapdpglem.y φ Y V
mapdpglem1.p ˙ = LSSum C
mapdpglem2.j J = LSpan C
mapdpglem3.f F = Base C
mapdpglem3.te φ t M N X ˙ M N Y
Assertion mapdpglem2a φ t F

Proof

Step Hyp Ref Expression
1 mapdpglem.h H = LHyp K
2 mapdpglem.m M = mapd K W
3 mapdpglem.u U = DVecH K W
4 mapdpglem.v V = Base U
5 mapdpglem.s - ˙ = - U
6 mapdpglem.n N = LSpan U
7 mapdpglem.c C = LCDual K W
8 mapdpglem.k φ K HL W H
9 mapdpglem.x φ X V
10 mapdpglem.y φ Y V
11 mapdpglem1.p ˙ = LSSum C
12 mapdpglem2.j J = LSpan C
13 mapdpglem3.f F = Base C
14 mapdpglem3.te φ t M N X ˙ M N Y
15 1 7 8 lcdlmod φ C LMod
16 eqid LSubSp U = LSubSp U
17 eqid LSubSp C = LSubSp C
18 1 3 8 dvhlmod φ U LMod
19 4 16 6 lspsncl U LMod X V N X LSubSp U
20 18 9 19 syl2anc φ N X LSubSp U
21 1 2 3 16 7 17 8 20 mapdcl2 φ M N X LSubSp C
22 4 16 6 lspsncl U LMod Y V N Y LSubSp U
23 18 10 22 syl2anc φ N Y LSubSp U
24 1 2 3 16 7 17 8 23 mapdcl2 φ M N Y LSubSp C
25 17 11 lsmcl C LMod M N X LSubSp C M N Y LSubSp C M N X ˙ M N Y LSubSp C
26 15 21 24 25 syl3anc φ M N X ˙ M N Y LSubSp C
27 13 17 lssel M N X ˙ M N Y LSubSp C t M N X ˙ M N Y t F
28 26 14 27 syl2anc φ t F