Metamath Proof Explorer


Theorem lcfrlem42

Description: Lemma for lcfr . Eliminate nonzero condition. (Contributed by NM, 11-Mar-2015)

Ref Expression
Hypotheses lcfrlem38.h H = LHyp K
lcfrlem38.o ˙ = ocH K W
lcfrlem38.u U = DVecH K W
lcfrlem38.p + ˙ = + U
lcfrlem38.f F = LFnl U
lcfrlem38.l L = LKer U
lcfrlem38.d D = LDual U
lcfrlem38.q Q = LSubSp D
lcfrlem38.c C = f LFnl U | ˙ ˙ L f = L f
lcfrlem38.e E = g G ˙ L g
lcfrlem38.k φ K HL W H
lcfrlem38.g φ G Q
lcfrlem38.gs φ G C
lcfrlem38.xe φ X E
lcfrlem38.ye φ Y E
Assertion lcfrlem42 φ X + ˙ Y E

Proof

Step Hyp Ref Expression
1 lcfrlem38.h H = LHyp K
2 lcfrlem38.o ˙ = ocH K W
3 lcfrlem38.u U = DVecH K W
4 lcfrlem38.p + ˙ = + U
5 lcfrlem38.f F = LFnl U
6 lcfrlem38.l L = LKer U
7 lcfrlem38.d D = LDual U
8 lcfrlem38.q Q = LSubSp D
9 lcfrlem38.c C = f LFnl U | ˙ ˙ L f = L f
10 lcfrlem38.e E = g G ˙ L g
11 lcfrlem38.k φ K HL W H
12 lcfrlem38.g φ G Q
13 lcfrlem38.gs φ G C
14 lcfrlem38.xe φ X E
15 lcfrlem38.ye φ Y E
16 1 3 11 dvhlmod φ U LMod
17 eqid Base U = Base U
18 1 2 3 17 6 7 8 10 11 12 14 lcfrlem4 φ X Base U
19 1 2 3 17 6 7 8 10 11 12 15 lcfrlem4 φ Y Base U
20 17 4 lmodcom U LMod X Base U Y Base U X + ˙ Y = Y + ˙ X
21 16 18 19 20 syl3anc φ X + ˙ Y = Y + ˙ X
22 21 adantr φ X = 0 U X + ˙ Y = Y + ˙ X
23 11 adantr φ X = 0 U K HL W H
24 12 adantr φ X = 0 U G Q
25 15 adantr φ X = 0 U Y E
26 eqid 0 U = 0 U
27 simpr φ X = 0 U X = 0 U
28 1 2 3 4 6 7 8 23 24 10 25 26 27 lcfrlem7 φ X = 0 U Y + ˙ X E
29 22 28 eqeltrd φ X = 0 U X + ˙ Y E
30 11 adantr φ Y = 0 U K HL W H
31 12 adantr φ Y = 0 U G Q
32 14 adantr φ Y = 0 U X E
33 simpr φ Y = 0 U Y = 0 U
34 1 2 3 4 6 7 8 30 31 10 32 26 33 lcfrlem7 φ Y = 0 U X + ˙ Y E
35 11 adantr φ X 0 U Y 0 U K HL W H
36 12 adantr φ X 0 U Y 0 U G Q
37 13 adantr φ X 0 U Y 0 U G C
38 14 adantr φ X 0 U Y 0 U X E
39 15 adantr φ X 0 U Y 0 U Y E
40 simprl φ X 0 U Y 0 U X 0 U
41 simprr φ X 0 U Y 0 U Y 0 U
42 1 2 3 4 5 6 7 8 9 10 35 36 37 38 39 26 40 41 lcfrlem41 φ X 0 U Y 0 U X + ˙ Y E
43 29 34 42 pm2.61da2ne φ X + ˙ Y E