Metamath Proof Explorer


Theorem lsatcv0eq

Description: If the sum of two atoms cover the zero subspace, they are equal. ( atcv0eq analog.) (Contributed by NM, 10-Jan-2015)

Ref Expression
Hypotheses lsatcv0eq.o ⊢ 0 ˙ = 0 W
lsatcv0eq.p ⊢ ⊕ ˙ = LSSum ⁡ W
lsatcv0eq.a ⊢ A = LSAtoms ⁡ W
lsatcv0eq.c ⊢ C = ⋖ L ⁡ W
lsatcv0eq.w ⊢ φ → W ∈ LVec
lsatcv0eq.q ⊢ φ → Q ∈ A
lsatcv0eq.r ⊢ φ → R ∈ A
Assertion lsatcv0eq ⊢ φ → 0 ˙ C Q ⊕ ˙ R ↔ Q = R

Proof

Step Hyp Ref Expression
1 lsatcv0eq.o ⊢ 0 ˙ = 0 W
2 lsatcv0eq.p ⊢ ⊕ ˙ = LSSum ⁡ W
3 lsatcv0eq.a ⊢ A = LSAtoms ⁡ W
4 lsatcv0eq.c ⊢ C = ⋖ L ⁡ W
5 lsatcv0eq.w ⊢ φ → W ∈ LVec
6 lsatcv0eq.q ⊢ φ → Q ∈ A
7 lsatcv0eq.r ⊢ φ → R ∈ A
8 1 3 5 6 7 lsatnem0 ⊢ φ → Q ≠ R ↔ Q ∩ R = 0 ˙
9 eqid ⊢ LSubSp ⁡ W = LSubSp ⁡ W
10 lveclmod ⊢ W ∈ LVec → W ∈ LMod
11 5 10 syl ⊢ φ → W ∈ LMod
12 9 3 11 6 lsatlssel ⊢ φ → Q ∈ LSubSp ⁡ W
13 9 2 1 3 4 5 12 7 lcvp ⊢ φ → Q ∩ R = 0 ˙ ↔ Q C Q ⊕ ˙ R
14 1 3 4 5 6 lsatcv0 ⊢ φ → 0 ˙ C Q
15 14 biantrurd ⊢ φ → Q C Q ⊕ ˙ R ↔ 0 ˙ C Q ∧ Q C Q ⊕ ˙ R
16 8 13 15 3bitrd ⊢ φ → Q ≠ R ↔ 0 ˙ C Q ∧ Q C Q ⊕ ˙ R
17 5 adantr ⊢ φ ∧ 0 ˙ C Q ∧ Q C Q ⊕ ˙ R → W ∈ LVec
18 1 9 lsssn0 ⊢ W ∈ LMod → 0 ˙ ∈ LSubSp ⁡ W
19 11 18 syl ⊢ φ → 0 ˙ ∈ LSubSp ⁡ W
20 19 adantr ⊢ φ ∧ 0 ˙ C Q ∧ Q C Q ⊕ ˙ R → 0 ˙ ∈ LSubSp ⁡ W
21 12 adantr ⊢ φ ∧ 0 ˙ C Q ∧ Q C Q ⊕ ˙ R → Q ∈ LSubSp ⁡ W
22 9 3 11 7 lsatlssel ⊢ φ → R ∈ LSubSp ⁡ W
23 9 2 lsmcl ⊢ W ∈ LMod ∧ Q ∈ LSubSp ⁡ W ∧ R ∈ LSubSp ⁡ W → Q ⊕ ˙ R ∈ LSubSp ⁡ W
24 11 12 22 23 syl3anc ⊢ φ → Q ⊕ ˙ R ∈ LSubSp ⁡ W
25 24 adantr ⊢ φ ∧ 0 ˙ C Q ∧ Q C Q ⊕ ˙ R → Q ⊕ ˙ R ∈ LSubSp ⁡ W
26 simprl ⊢ φ ∧ 0 ˙ C Q ∧ Q C Q ⊕ ˙ R → 0 ˙ C Q
27 simprr ⊢ φ ∧ 0 ˙ C Q ∧ Q C Q ⊕ ˙ R → Q C Q ⊕ ˙ R
28 9 4 17 20 21 25 26 27 lcvntr ⊢ φ ∧ 0 ˙ C Q ∧ Q C Q ⊕ ˙ R → ¬ 0 ˙ C Q ⊕ ˙ R
29 28 ex ⊢ φ → 0 ˙ C Q ∧ Q C Q ⊕ ˙ R → ¬ 0 ˙ C Q ⊕ ˙ R
30 16 29 sylbid ⊢ φ → Q ≠ R → ¬ 0 ˙ C Q ⊕ ˙ R
31 30 necon4ad ⊢ φ → 0 ˙ C Q ⊕ ˙ R → Q = R
32 9 lsssssubg ⊢ W ∈ LMod → LSubSp ⁡ W ⊆ SubGrp ⁡ W
33 11 32 syl ⊢ φ → LSubSp ⁡ W ⊆ SubGrp ⁡ W
34 33 12 sseldd ⊢ φ → Q ∈ SubGrp ⁡ W
35 2 lsmidm ⊢ Q ∈ SubGrp ⁡ W → Q ⊕ ˙ Q = Q
36 34 35 syl ⊢ φ → Q ⊕ ˙ Q = Q
37 14 36 breqtrrd ⊢ φ → 0 ˙ C Q ⊕ ˙ Q
38 oveq2 ⊢ Q = R → Q ⊕ ˙ Q = Q ⊕ ˙ R
39 38 breq2d ⊢ Q = R → 0 ˙ C Q ⊕ ˙ Q ↔ 0 ˙ C Q ⊕ ˙ R
40 37 39 syl5ibcom ⊢ φ → Q = R → 0 ˙ C Q ⊕ ˙ R
41 31 40 impbid ⊢ φ → 0 ˙ C Q ⊕ ˙ R ↔ Q = R