Metamath Proof Explorer


Theorem dvh3dimatN

Description: There is an atom that is outside the subspace sum of 2 others. (Contributed by NM, 25-Apr-2015) (New usage is discouraged.)

Ref Expression
Hypotheses dvh4dimat.h 𝐻 = ( LHyp ‘ 𝐾 )
dvh4dimat.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
dvh3dimat.s = ( LSSum ‘ 𝑈 )
dvh3dimat.a 𝐴 = ( LSAtoms ‘ 𝑈 )
dvh3dimat.k ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) )
dvh3dimat.p ( 𝜑𝑃𝐴 )
dvh3dimat.q ( 𝜑𝑄𝐴 )
Assertion dvh3dimatN ( 𝜑 → ∃ 𝑠𝐴 ¬ 𝑠 ⊆ ( 𝑃 𝑄 ) )

Proof

Step Hyp Ref Expression
1 dvh4dimat.h 𝐻 = ( LHyp ‘ 𝐾 )
2 dvh4dimat.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
3 dvh3dimat.s = ( LSSum ‘ 𝑈 )
4 dvh3dimat.a 𝐴 = ( LSAtoms ‘ 𝑈 )
5 dvh3dimat.k ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) )
6 dvh3dimat.p ( 𝜑𝑃𝐴 )
7 dvh3dimat.q ( 𝜑𝑄𝐴 )
8 1 2 3 4 5 6 6 7 dvh4dimat ( 𝜑 → ∃ 𝑠𝐴 ¬ 𝑠 ⊆ ( ( 𝑃 𝑃 ) 𝑄 ) )
9 1 2 5 dvhlmod ( 𝜑𝑈 ∈ LMod )
10 eqid ( LSubSp ‘ 𝑈 ) = ( LSubSp ‘ 𝑈 )
11 10 4 9 6 lsatlssel ( 𝜑𝑃 ∈ ( LSubSp ‘ 𝑈 ) )
12 10 lsssubg ( ( 𝑈 ∈ LMod ∧ 𝑃 ∈ ( LSubSp ‘ 𝑈 ) ) → 𝑃 ∈ ( SubGrp ‘ 𝑈 ) )
13 9 11 12 syl2anc ( 𝜑𝑃 ∈ ( SubGrp ‘ 𝑈 ) )
14 3 lsmidm ( 𝑃 ∈ ( SubGrp ‘ 𝑈 ) → ( 𝑃 𝑃 ) = 𝑃 )
15 13 14 syl ( 𝜑 → ( 𝑃 𝑃 ) = 𝑃 )
16 15 oveq1d ( 𝜑 → ( ( 𝑃 𝑃 ) 𝑄 ) = ( 𝑃 𝑄 ) )
17 16 sseq2d ( 𝜑 → ( 𝑠 ⊆ ( ( 𝑃 𝑃 ) 𝑄 ) ↔ 𝑠 ⊆ ( 𝑃 𝑄 ) ) )
18 17 notbid ( 𝜑 → ( ¬ 𝑠 ⊆ ( ( 𝑃 𝑃 ) 𝑄 ) ↔ ¬ 𝑠 ⊆ ( 𝑃 𝑄 ) ) )
19 18 rexbidv ( 𝜑 → ( ∃ 𝑠𝐴 ¬ 𝑠 ⊆ ( ( 𝑃 𝑃 ) 𝑄 ) ↔ ∃ 𝑠𝐴 ¬ 𝑠 ⊆ ( 𝑃 𝑄 ) ) )
20 8 19 mpbid ( 𝜑 → ∃ 𝑠𝐴 ¬ 𝑠 ⊆ ( 𝑃 𝑄 ) )