Metamath Proof Explorer


Theorem lsslsp

Description: Spans in submodules correspond to spans in the containing module. (Contributed by Stefan O'Rear, 12-Dec-2014) TODO: Shouldn't we swap MG and NG since we are computing a property of NG ? (Like we say sin 0 = 0 and not 0 = sin 0.) - NM 15-Mar-2015.

Ref Expression
Hypotheses lsslsp.x X = W 𝑠 U
lsslsp.m M = LSpan W
lsslsp.n N = LSpan X
lsslsp.l L = LSubSp W
Assertion lsslsp W LMod U L G U M G = N G

Proof

Step Hyp Ref Expression
1 lsslsp.x X = W 𝑠 U
2 lsslsp.m M = LSpan W
3 lsslsp.n N = LSpan X
4 lsslsp.l L = LSubSp W
5 simp1 W LMod U L G U W LMod
6 1 4 lsslmod W LMod U L X LMod
7 6 3adant3 W LMod U L G U X LMod
8 simp3 W LMod U L G U G U
9 eqid Base W = Base W
10 9 4 lssss U L U Base W
11 10 3ad2ant2 W LMod U L G U U Base W
12 1 9 ressbas2 U Base W U = Base X
13 11 12 syl W LMod U L G U U = Base X
14 8 13 sseqtrd W LMod U L G U G Base X
15 eqid Base X = Base X
16 eqid LSubSp X = LSubSp X
17 15 16 3 lspcl X LMod G Base X N G LSubSp X
18 7 14 17 syl2anc W LMod U L G U N G LSubSp X
19 1 4 16 lsslss W LMod U L N G LSubSp X N G L N G U
20 19 3adant3 W LMod U L G U N G LSubSp X N G L N G U
21 18 20 mpbid W LMod U L G U N G L N G U
22 21 simpld W LMod U L G U N G L
23 15 3 lspssid X LMod G Base X G N G
24 7 14 23 syl2anc W LMod U L G U G N G
25 4 2 lspssp W LMod N G L G N G M G N G
26 5 22 24 25 syl3anc W LMod U L G U M G N G
27 8 11 sstrd W LMod U L G U G Base W
28 9 4 2 lspcl W LMod G Base W M G L
29 5 27 28 syl2anc W LMod U L G U M G L
30 4 2 lspssp W LMod U L G U M G U
31 1 4 16 lsslss W LMod U L M G LSubSp X M G L M G U
32 31 3adant3 W LMod U L G U M G LSubSp X M G L M G U
33 29 30 32 mpbir2and W LMod U L G U M G LSubSp X
34 9 2 lspssid W LMod G Base W G M G
35 5 27 34 syl2anc W LMod U L G U G M G
36 16 3 lspssp X LMod M G LSubSp X G M G N G M G
37 7 33 35 36 syl3anc W LMod U L G U N G M G
38 26 37 eqssd W LMod U L G U M G = N G