Metamath Proof Explorer


Theorem islss

Description: The predicate "is a subspace" (of a left module or left vector space). (Contributed by NM, 8-Dec-2013) (Revised by Mario Carneiro, 8-Jan-2015)

Ref Expression
Hypotheses lssset.f F = Scalar W
lssset.b B = Base F
lssset.v V = Base W
lssset.p + ˙ = + W
lssset.t · ˙ = W
lssset.s S = LSubSp W
Assertion islss U S U V U x B a U b U x · ˙ a + ˙ b U

Proof

Step Hyp Ref Expression
1 lssset.f F = Scalar W
2 lssset.b B = Base F
3 lssset.v V = Base W
4 lssset.p + ˙ = + W
5 lssset.t · ˙ = W
6 lssset.s S = LSubSp W
7 elfvex U LSubSp W W V
8 7 6 eleq2s U S W V
9 fvprc ¬ W V Base W =
10 3 9 eqtrid ¬ W V V =
11 10 sseq2d ¬ W V U V U
12 11 biimpcd U V ¬ W V U
13 ss0 U U =
14 12 13 syl6 U V ¬ W V U =
15 14 necon1ad U V U W V
16 15 imp U V U W V
17 16 3adant3 U V U x B a U b U x · ˙ a + ˙ b U W V
18 1 2 3 4 5 6 lssset W V S = s 𝒫 V | x B a s b s x · ˙ a + ˙ b s
19 18 eleq2d W V U S U s 𝒫 V | x B a s b s x · ˙ a + ˙ b s
20 eldifsn U 𝒫 V U 𝒫 V U
21 3 fvexi V V
22 21 elpw2 U 𝒫 V U V
23 22 anbi1i U 𝒫 V U U V U
24 20 23 bitri U 𝒫 V U V U
25 24 anbi1i U 𝒫 V x B a U b U x · ˙ a + ˙ b U U V U x B a U b U x · ˙ a + ˙ b U
26 eleq2 s = U x · ˙ a + ˙ b s x · ˙ a + ˙ b U
27 26 raleqbi1dv s = U b s x · ˙ a + ˙ b s b U x · ˙ a + ˙ b U
28 27 raleqbi1dv s = U a s b s x · ˙ a + ˙ b s a U b U x · ˙ a + ˙ b U
29 28 ralbidv s = U x B a s b s x · ˙ a + ˙ b s x B a U b U x · ˙ a + ˙ b U
30 29 elrab U s 𝒫 V | x B a s b s x · ˙ a + ˙ b s U 𝒫 V x B a U b U x · ˙ a + ˙ b U
31 df-3an U V U x B a U b U x · ˙ a + ˙ b U U V U x B a U b U x · ˙ a + ˙ b U
32 25 30 31 3bitr4i U s 𝒫 V | x B a s b s x · ˙ a + ˙ b s U V U x B a U b U x · ˙ a + ˙ b U
33 19 32 bitrdi W V U S U V U x B a U b U x · ˙ a + ˙ b U
34 8 17 33 pm5.21nii U S U V U x B a U b U x · ˙ a + ˙ b U