Metamath Proof Explorer


Theorem lsatn0

Description: A 1-dim subspace (atom) of a left module or left vector space is nonzero. ( atne0 analog.) (Contributed by NM, 25-Aug-2014)

Ref Expression
Hypotheses lsatn0.o ⊢ 0 ˙ = 0 W
lsatn0.a ⊢ A = LSAtoms ⁡ W
lsatn0.w ⊢ φ → W ∈ LMod
lsatn0.u ⊢ φ → U ∈ A
Assertion lsatn0 ⊢ φ → U ≠ 0 ˙

Proof

Step Hyp Ref Expression
1 lsatn0.o ⊢ 0 ˙ = 0 W
2 lsatn0.a ⊢ A = LSAtoms ⁡ W
3 lsatn0.w ⊢ φ → W ∈ LMod
4 lsatn0.u ⊢ φ → U ∈ A
5 eqid ⊢ Base W = Base W
6 eqid ⊢ LSpan ⁡ W = LSpan ⁡ W
7 5 6 1 2 islsat ⊢ W ∈ LMod → U ∈ A ↔ ∃ v ∈ Base W ∖ 0 ˙ U = LSpan ⁡ W ⁡ v
8 3 7 syl ⊢ φ → U ∈ A ↔ ∃ v ∈ Base W ∖ 0 ˙ U = LSpan ⁡ W ⁡ v
9 4 8 mpbid ⊢ φ → ∃ v ∈ Base W ∖ 0 ˙ U = LSpan ⁡ W ⁡ v
10 eldifsn ⊢ v ∈ Base W ∖ 0 ˙ ↔ v ∈ Base W ∧ v ≠ 0 ˙
11 5 1 6 lspsneq0 ⊢ W ∈ LMod ∧ v ∈ Base W → LSpan ⁡ W ⁡ v = 0 ˙ ↔ v = 0 ˙
12 3 11 sylan ⊢ φ ∧ v ∈ Base W → LSpan ⁡ W ⁡ v = 0 ˙ ↔ v = 0 ˙
13 12 biimpd ⊢ φ ∧ v ∈ Base W → LSpan ⁡ W ⁡ v = 0 ˙ → v = 0 ˙
14 13 necon3d ⊢ φ ∧ v ∈ Base W → v ≠ 0 ˙ → LSpan ⁡ W ⁡ v ≠ 0 ˙
15 14 expimpd ⊢ φ → v ∈ Base W ∧ v ≠ 0 ˙ → LSpan ⁡ W ⁡ v ≠ 0 ˙
16 10 15 biimtrid ⊢ φ → v ∈ Base W ∖ 0 ˙ → LSpan ⁡ W ⁡ v ≠ 0 ˙
17 neeq1 ⊢ U = LSpan ⁡ W ⁡ v → U ≠ 0 ˙ ↔ LSpan ⁡ W ⁡ v ≠ 0 ˙
18 17 biimprcd ⊢ LSpan ⁡ W ⁡ v ≠ 0 ˙ → U = LSpan ⁡ W ⁡ v → U ≠ 0 ˙
19 16 18 syl6 ⊢ φ → v ∈ Base W ∖ 0 ˙ → U = LSpan ⁡ W ⁡ v → U ≠ 0 ˙
20 19 rexlimdv ⊢ φ → ∃ v ∈ Base W ∖ 0 ˙ U = LSpan ⁡ W ⁡ v → U ≠ 0 ˙
21 9 20 mpd ⊢ φ → U ≠ 0 ˙