Metamath Proof Explorer


Theorem frlmsslss

Description: A subset of a free module obtained by restricting the support set is a submodule. J is the set of forbidden unit vectors. (Contributed by Stefan O'Rear, 4-Feb-2015)

Ref Expression
Hypotheses frlmsslss.y Y = R freeLMod I
frlmsslss.u U = LSubSp Y
frlmsslss.b B = Base Y
frlmsslss.z 0 ˙ = 0 R
frlmsslss.c C = x B | x J = J × 0 ˙
Assertion frlmsslss R Ring I V J I C U

Proof

Step Hyp Ref Expression
1 frlmsslss.y Y = R freeLMod I
2 frlmsslss.u U = LSubSp Y
3 frlmsslss.b B = Base Y
4 frlmsslss.z 0 ˙ = 0 R
5 frlmsslss.c C = x B | x J = J × 0 ˙
6 simp1 R Ring I V J I R Ring
7 simp2 R Ring I V J I I V
8 simp3 R Ring I V J I J I
9 7 8 ssexd R Ring I V J I J V
10 eqid R freeLMod J = R freeLMod J
11 10 4 frlm0 R Ring J V J × 0 ˙ = 0 R freeLMod J
12 6 9 11 syl2anc R Ring I V J I J × 0 ˙ = 0 R freeLMod J
13 12 eqeq2d R Ring I V J I x J = J × 0 ˙ x J = 0 R freeLMod J
14 13 rabbidv R Ring I V J I x B | x J = J × 0 ˙ = x B | x J = 0 R freeLMod J
15 5 14 eqtrid R Ring I V J I C = x B | x J = 0 R freeLMod J
16 eqid Base R freeLMod J = Base R freeLMod J
17 eqid x B x J = x B x J
18 1 10 3 16 17 frlmsplit2 R Ring I V J I x B x J Y LMHom R freeLMod J
19 fvex 0 R freeLMod J V
20 17 mptiniseg 0 R freeLMod J V x B x J -1 0 R freeLMod J = x B | x J = 0 R freeLMod J
21 19 20 ax-mp x B x J -1 0 R freeLMod J = x B | x J = 0 R freeLMod J
22 21 eqcomi x B | x J = 0 R freeLMod J = x B x J -1 0 R freeLMod J
23 eqid 0 R freeLMod J = 0 R freeLMod J
24 22 23 2 lmhmkerlss x B x J Y LMHom R freeLMod J x B | x J = 0 R freeLMod J U
25 18 24 syl R Ring I V J I x B | x J = 0 R freeLMod J U
26 15 25 eqeltrd R Ring I V J I C U