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 = ( ⊥𝑃𝐾 )
Assertion 3polN ( ( 𝐾 ∈ HL ∧ 𝑆𝐴 ) → ( ‘ ( ‘ ( 𝑆 ) ) ) = ( 𝑆 ) )

Proof

Step Hyp Ref Expression
1 2polss.a 𝐴 = ( Atoms ‘ 𝐾 )
2 2polss.p = ( ⊥𝑃𝐾 )
3 hlclat ( 𝐾 ∈ HL → 𝐾 ∈ CLat )
4 eqid ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
5 4 1 atssbase 𝐴 ⊆ ( Base ‘ 𝐾 )
6 sstr ( ( 𝑆𝐴𝐴 ⊆ ( Base ‘ 𝐾 ) ) → 𝑆 ⊆ ( Base ‘ 𝐾 ) )
7 5 6 mpan2 ( 𝑆𝐴𝑆 ⊆ ( 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 ∧ 𝑆𝐴 ) → ( ( 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 ∧ 𝑆𝐴 ) → ( ‘ ( ( pmap ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) ) = ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) ) )
15 8 1 12 2 2polvalN ( ( 𝐾 ∈ HL ∧ 𝑆𝐴 ) → ( ‘ ( 𝑆 ) ) = ( ( pmap ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) )
16 15 fveq2d ( ( 𝐾 ∈ HL ∧ 𝑆𝐴 ) → ( ‘ ( ‘ ( 𝑆 ) ) ) = ( ‘ ( ( pmap ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) ) )
17 8 11 1 12 2 polval2N ( ( 𝐾 ∈ HL ∧ 𝑆𝐴 ) → ( 𝑆 ) = ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ ( ( lub ‘ 𝐾 ) ‘ 𝑆 ) ) ) )
18 14 16 17 3eqtr4d ( ( 𝐾 ∈ HL ∧ 𝑆𝐴 ) → ( ‘ ( ‘ ( 𝑆 ) ) ) = ( 𝑆 ) )