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Metamath Proof Explorer

Theorem 2polvalN

Description: Value of double polarity. (Contributed by NM, 25-Jan-2012) (New usage is discouraged.)

Ref Expression
Hypotheses 2polval.u ⊢ 𝑈 = ( lub ‘ 𝐾 )
2polval.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
2polval.m ⊢ 𝑀 = ( pmap ‘ 𝐾 )
2polval.p ⊢ ⊥ = ( ⊥𝑃 ‘ 𝐾 )
Assertion 2polvalN ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = ( 𝑀 ‘ ( 𝑈 ‘ 𝑋 ) ) )

Proof

Step Hyp Ref Expression
1 2polval.u ⊢ 𝑈 = ( lub ‘ 𝐾 )
2 2polval.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
3 2polval.m ⊢ 𝑀 = ( pmap ‘ 𝐾 )
4 2polval.p ⊢ ⊥ = ( ⊥𝑃 ‘ 𝐾 )
5 eqid ⊢ ( oc ‘ 𝐾 ) = ( oc ‘ 𝐾 )
6 1 5 2 3 4 polval2N ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ⊥ ‘ 𝑋 ) = ( 𝑀 ‘ ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ) )
7 6 fveq2d ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = ( ⊥ ‘ ( 𝑀 ‘ ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ) ) )
8 hlop ⊢ ( 𝐾 ∈ HL → 𝐾 ∈ OP )
9 8 adantr ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → 𝐾 ∈ OP )
10 hlclat ⊢ ( 𝐾 ∈ HL → 𝐾 ∈ CLat )
11 eqid ⊢ ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
12 11 2 atssbase ⊢ 𝐴 ⊆ ( Base ‘ 𝐾 )
13 sstr ⊢ ( ( 𝑋 ⊆ 𝐴 ∧ 𝐴 ⊆ ( Base ‘ 𝐾 ) ) → 𝑋 ⊆ ( Base ‘ 𝐾 ) )
14 12 13 mpan2 ⊢ ( 𝑋 ⊆ 𝐴 → 𝑋 ⊆ ( Base ‘ 𝐾 ) )
15 11 1 clatlubcl ⊢ ( ( 𝐾 ∈ CLat ∧ 𝑋 ⊆ ( Base ‘ 𝐾 ) ) → ( 𝑈 ‘ 𝑋 ) ∈ ( Base ‘ 𝐾 ) )
16 10 14 15 syl2an ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( 𝑈 ‘ 𝑋 ) ∈ ( Base ‘ 𝐾 ) )
17 11 5 opoccl ⊢ ( ( 𝐾 ∈ OP ∧ ( 𝑈 ‘ 𝑋 ) ∈ ( Base ‘ 𝐾 ) ) → ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ∈ ( Base ‘ 𝐾 ) )
18 9 16 17 syl2anc ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ∈ ( Base ‘ 𝐾 ) )
19 11 5 3 4 polpmapN ⊢ ( ( 𝐾 ∈ HL ∧ ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ∈ ( Base ‘ 𝐾 ) ) → ( ⊥ ‘ ( 𝑀 ‘ ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ) ) = ( 𝑀 ‘ ( ( oc ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ) ) )
20 18 19 syldan ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ⊥ ‘ ( 𝑀 ‘ ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ) ) = ( 𝑀 ‘ ( ( oc ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ) ) )
21 11 5 opococ ⊢ ( ( 𝐾 ∈ OP ∧ ( 𝑈 ‘ 𝑋 ) ∈ ( Base ‘ 𝐾 ) ) → ( ( oc ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ) = ( 𝑈 ‘ 𝑋 ) )
22 9 16 21 syl2anc ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ( oc ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ) = ( 𝑈 ‘ 𝑋 ) )
23 22 fveq2d ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( 𝑀 ‘ ( ( oc ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ ( 𝑈 ‘ 𝑋 ) ) ) ) = ( 𝑀 ‘ ( 𝑈 ‘ 𝑋 ) ) )
24 7 20 23 3eqtrd ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = ( 𝑀 ‘ ( 𝑈 ‘ 𝑋 ) ) )