Metamath Proof Explorer


Theorem lhpmat

Description: An element covered by the lattice unit, when conjoined with an atom not under it, equals the lattice zero. (Contributed by NM, 6-Jun-2012)

Ref Expression
Hypotheses lhpmat.l = ( le ‘ 𝐾 )
lhpmat.m = ( meet ‘ 𝐾 )
lhpmat.z 0 = ( 0. ‘ 𝐾 )
lhpmat.a 𝐴 = ( Atoms ‘ 𝐾 )
lhpmat.h 𝐻 = ( LHyp ‘ 𝐾 )
Assertion lhpmat ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ) → ( 𝑃 𝑊 ) = 0 )

Proof

Step Hyp Ref Expression
1 lhpmat.l = ( le ‘ 𝐾 )
2 lhpmat.m = ( meet ‘ 𝐾 )
3 lhpmat.z 0 = ( 0. ‘ 𝐾 )
4 lhpmat.a 𝐴 = ( Atoms ‘ 𝐾 )
5 lhpmat.h 𝐻 = ( LHyp ‘ 𝐾 )
6 simprr ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ) → ¬ 𝑃 𝑊 )
7 hlatl ( 𝐾 ∈ HL → 𝐾 ∈ AtLat )
8 7 ad2antrr ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ) → 𝐾 ∈ AtLat )
9 simprl ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ) → 𝑃𝐴 )
10 eqid ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
11 10 5 lhpbase ( 𝑊𝐻𝑊 ∈ ( Base ‘ 𝐾 ) )
12 11 ad2antlr ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ) → 𝑊 ∈ ( Base ‘ 𝐾 ) )
13 10 1 2 3 4 atnle ( ( 𝐾 ∈ AtLat ∧ 𝑃𝐴𝑊 ∈ ( Base ‘ 𝐾 ) ) → ( ¬ 𝑃 𝑊 ↔ ( 𝑃 𝑊 ) = 0 ) )
14 8 9 12 13 syl3anc ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ) → ( ¬ 𝑃 𝑊 ↔ ( 𝑃 𝑊 ) = 0 ) )
15 6 14 mpbid ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑃𝐴 ∧ ¬ 𝑃 𝑊 ) ) → ( 𝑃 𝑊 ) = 0 )