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Metamath Proof Explorer

Theorem 3polN

Description: Triple polarity cancels to a single polarity. (Contributed by NM, 6-Mar-2012) (New usage is discouraged.)

Ref Expression
Hypotheses 2polss.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
2polss.p ⊢ ⊥ = ( ⊥𝑃 ‘ 𝐾 )
Assertion 3polN ( ( 𝐾 ∈ HL ∧ 𝑆 ⊆ 𝐴 ) → ( ⊥ ‘ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ) = ( ⊥ ‘ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 2polss.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
2 2polss.p ⊢ ⊥ = ( ⊥𝑃 ‘ 𝐾 )
3 hlclat ⊢ ( 𝐾 ∈ HL → 𝐾 ∈ CLat )
4 eqid ⊢ ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
5 4 1 atssbase ⊢ 𝐴 ⊆ ( Base ‘ 𝐾 )
6 sstr ⊢ ( ( 𝑆 ⊆ 𝐴 ∧ 𝐴 ⊆ ( Base ‘ 𝐾 ) ) → 𝑆 ⊆ ( Base ‘ 𝐾 ) )
7 5 6 mpan2 ⊢ ( 𝑆 ⊆ 𝐴 → 𝑆 ⊆ ( Base ‘ 𝐾 ) )
8 eqid ⊢ ( lub ‘ 𝐾 ) = ( lub ‘ 𝐾 )
9 4 8 clatlubcl ⊢ ( ( 𝐾 ∈ CLat ∧ 𝑆 ⊆ ( Base ‘ 𝐾 ) ) → ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ∈ ( Base ‘ 𝐾 ) )
10 3 7 9 syl2an ⊢ ( ( 𝐾 ∈ HL ∧ 𝑆 ⊆ 𝐴 ) → ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ∈ ( Base ‘ 𝐾 ) )
11 eqid ⊢ ( oc ‘ 𝐾 ) = ( oc ‘ 𝐾 )
12 eqid ⊢ ( pmap ‘ 𝐾 ) = ( pmap ‘ 𝐾 )
13 4 11 12 2 polpmapN ⊢ ( ( 𝐾 ∈ HL ∧ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ∈ ( Base ‘ 𝐾 ) ) → ( ⊥ ‘ ( ( pmap ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) ) = ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) ) )
14 10 13 syldan ⊢ ( ( 𝐾 ∈ HL ∧ 𝑆 ⊆ 𝐴 ) → ( ⊥ ‘ ( ( pmap ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) ) = ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) ) )
15 8 1 12 2 2polvalN ⊢ ( ( 𝐾 ∈ HL ∧ 𝑆 ⊆ 𝐴 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) = ( ( pmap ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) )
16 15 fveq2d ⊢ ( ( 𝐾 ∈ HL ∧ 𝑆 ⊆ 𝐴 ) → ( ⊥ ‘ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ) = ( ⊥ ‘ ( ( pmap ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) ) )
17 8 11 1 12 2 polval2N ⊢ ( ( 𝐾 ∈ HL ∧ 𝑆 ⊆ 𝐴 ) → ( ⊥ ‘ 𝑆 ) = ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) ) )
18 14 16 17 3eqtr4d ⊢ ( ( 𝐾 ∈ HL ∧ 𝑆 ⊆ 𝐴 ) → ( ⊥ ‘ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ) = ( ⊥ ‘ 𝑆 ) )