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Metamath Proof Explorer

Theorem 2polssN

Description: A set of atoms is a subset of its double polarity. (Contributed by NM, 29-Jan-2012) (New usage is discouraged.)

Ref Expression
Hypotheses 2polss.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
2polss.p ⊢ ⊥ = ( ⊥𝑃 ‘ 𝐾 )
Assertion 2polssN ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → 𝑋 ⊆ ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) )

Proof

Step Hyp Ref Expression
1 2polss.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
2 2polss.p ⊢ ⊥ = ( ⊥𝑃 ‘ 𝐾 )
3 hlclat ⊢ ( 𝐾 ∈ HL → 𝐾 ∈ CLat )
4 3 ad3antrrr ⊢ ( ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ) ∧ 𝑝 ∈ 𝑋 ) → 𝐾 ∈ CLat )
5 simpr ⊢ ( ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ) ∧ 𝑝 ∈ 𝑋 ) → 𝑝 ∈ 𝑋 )
6 simpllr ⊢ ( ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ) ∧ 𝑝 ∈ 𝑋 ) → 𝑋 ⊆ 𝐴 )
7 eqid ⊢ ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
8 7 1 atssbase ⊢ 𝐴 ⊆ ( Base ‘ 𝐾 )
9 6 8 sstrdi ⊢ ( ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ) ∧ 𝑝 ∈ 𝑋 ) → 𝑋 ⊆ ( Base ‘ 𝐾 ) )
10 eqid ⊢ ( le ‘ 𝐾 ) = ( le ‘ 𝐾 )
11 eqid ⊢ ( lub ‘ 𝐾 ) = ( lub ‘ 𝐾 )
12 7 10 11 lubel ⊢ ( ( 𝐾 ∈ CLat ∧ 𝑝 ∈ 𝑋 ∧ 𝑋 ⊆ ( Base ‘ 𝐾 ) ) → 𝑝 ( le ‘ 𝐾 ) ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) )
13 4 5 9 12 syl3anc ⊢ ( ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ) ∧ 𝑝 ∈ 𝑋 ) → 𝑝 ( le ‘ 𝐾 ) ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) )
14 13 ex ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ) → ( 𝑝 ∈ 𝑋 → 𝑝 ( le ‘ 𝐾 ) ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) ) )
15 14 ss2rabdv ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → { 𝑝 ∈ 𝐴 ∣ 𝑝 ∈ 𝑋 } ⊆ { 𝑝 ∈ 𝐴 ∣ 𝑝 ( le ‘ 𝐾 ) ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) } )
16 sseqin2 ⊢ ( 𝑋 ⊆ 𝐴 ↔ ( 𝐴 ∩ 𝑋 ) = 𝑋 )
17 16 biimpi ⊢ ( 𝑋 ⊆ 𝐴 → ( 𝐴 ∩ 𝑋 ) = 𝑋 )
18 17 adantl ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( 𝐴 ∩ 𝑋 ) = 𝑋 )
19 dfin5 ⊢ ( 𝐴 ∩ 𝑋 ) = { 𝑝 ∈ 𝐴 ∣ 𝑝 ∈ 𝑋 }
20 18 19 eqtr3di ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → 𝑋 = { 𝑝 ∈ 𝐴 ∣ 𝑝 ∈ 𝑋 } )
21 eqid ⊢ ( pmap ‘ 𝐾 ) = ( pmap ‘ 𝐾 )
22 11 1 21 2 2polvalN ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = ( ( pmap ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) ) )
23 sstr ⊢ ( ( 𝑋 ⊆ 𝐴 ∧ 𝐴 ⊆ ( Base ‘ 𝐾 ) ) → 𝑋 ⊆ ( Base ‘ 𝐾 ) )
24 8 23 mpan2 ⊢ ( 𝑋 ⊆ 𝐴 → 𝑋 ⊆ ( Base ‘ 𝐾 ) )
25 7 11 clatlubcl ⊢ ( ( 𝐾 ∈ CLat ∧ 𝑋 ⊆ ( Base ‘ 𝐾 ) ) → ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) ∈ ( Base ‘ 𝐾 ) )
26 3 24 25 syl2an ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) ∈ ( Base ‘ 𝐾 ) )
27 7 10 1 21 pmapval ⊢ ( ( 𝐾 ∈ HL ∧ ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) ∈ ( Base ‘ 𝐾 ) ) → ( ( pmap ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) ) = { 𝑝 ∈ 𝐴 ∣ 𝑝 ( le ‘ 𝐾 ) ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) } )
28 26 27 syldan ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ( pmap ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) ) = { 𝑝 ∈ 𝐴 ∣ 𝑝 ( le ‘ 𝐾 ) ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) } )
29 22 28 eqtrd ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = { 𝑝 ∈ 𝐴 ∣ 𝑝 ( le ‘ 𝐾 ) ( ( lub ‘ 𝐾 ) ‘ 𝑋 ) } )
30 15 20 29 3sstr4d ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → 𝑋 ⊆ ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) )