Metamath Proof Explorer


Theorem dvh2dimatN

Description: Given an atom, there exists another. (Contributed by NM, 25-Apr-2015) (New usage is discouraged.)

Ref Expression
Hypotheses dvh4dimat.h ⊢ H = LHyp ⁡ K
dvh4dimat.u ⊢ U = DVecH ⁡ K ⁡ W
dvh2dimat.a ⊢ A = LSAtoms ⁡ U
dvh2dimat.k ⊢ φ → K ∈ HL ∧ W ∈ H
dvh2dimat.p ⊢ φ → P ∈ A
Assertion dvh2dimatN ⊢ φ → ∃ s ∈ A s ≠ P

Proof

Step Hyp Ref Expression
1 dvh4dimat.h ⊢ H = LHyp ⁡ K
2 dvh4dimat.u ⊢ U = DVecH ⁡ K ⁡ W
3 dvh2dimat.a ⊢ A = LSAtoms ⁡ U
4 dvh2dimat.k ⊢ φ → K ∈ HL ∧ W ∈ H
5 dvh2dimat.p ⊢ φ → P ∈ A
6 eqid ⊢ LSSum ⁡ U = LSSum ⁡ U
7 1 2 6 3 4 5 5 dvh3dimatN ⊢ φ → ∃ s ∈ A ¬ s ⊆ P LSSum ⁡ U P
8 1 2 4 dvhlmod ⊢ φ → U ∈ LMod
9 eqid ⊢ LSubSp ⁡ U = LSubSp ⁡ U
10 9 3 8 5 lsatlssel ⊢ φ → P ∈ LSubSp ⁡ U
11 9 lsssubg ⊢ U ∈ LMod ∧ P ∈ LSubSp ⁡ U → P ∈ SubGrp ⁡ U
12 8 10 11 syl2anc ⊢ φ → P ∈ SubGrp ⁡ U
13 6 lsmidm ⊢ P ∈ SubGrp ⁡ U → P LSSum ⁡ U P = P
14 12 13 syl ⊢ φ → P LSSum ⁡ U P = P
15 14 sseq2d ⊢ φ → s ⊆ P LSSum ⁡ U P ↔ s ⊆ P
16 15 adantr ⊢ φ ∧ s ∈ A → s ⊆ P LSSum ⁡ U P ↔ s ⊆ P
17 1 2 4 dvhlvec ⊢ φ → U ∈ LVec
18 17 adantr ⊢ φ ∧ s ∈ A → U ∈ LVec
19 simpr ⊢ φ ∧ s ∈ A → s ∈ A
20 5 adantr ⊢ φ ∧ s ∈ A → P ∈ A
21 3 18 19 20 lsatcmp ⊢ φ ∧ s ∈ A → s ⊆ P ↔ s = P
22 16 21 bitrd ⊢ φ ∧ s ∈ A → s ⊆ P LSSum ⁡ U P ↔ s = P
23 22 necon3bbid ⊢ φ ∧ s ∈ A → ¬ s ⊆ P LSSum ⁡ U P ↔ s ≠ P
24 23 rexbidva ⊢ φ → ∃ s ∈ A ¬ s ⊆ P LSSum ⁡ U P ↔ ∃ s ∈ A s ≠ P
25 7 24 mpbid ⊢ φ → ∃ s ∈ A s ≠ P