Metamath Proof Explorer


Theorem polcon3N

Description: Contraposition law for polarity. Remark in Holland95 p. 223. (Contributed by NM, 23-Mar-2012) (New usage is discouraged.)

Ref Expression
Hypotheses 2polss.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
2polss.p ⊢ ⊥ = ( ⊥𝑃 ‘ 𝐾 )
Assertion polcon3N ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ⊆ 𝑌 ) → ( ⊥ ‘ 𝑌 ) ⊆ ( ⊥ ‘ 𝑋 ) )

Proof

Step Hyp Ref Expression
1 2polss.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
2 2polss.p ⊢ ⊥ = ( ⊥𝑃 ‘ 𝐾 )
3 simp3 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ⊆ 𝑌 ) → 𝑋 ⊆ 𝑌 )
4 iinss1 ⊢ ( 𝑋 ⊆ 𝑌 → ∩ 𝑝 ∈ 𝑌 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) ⊆ ∩ 𝑝 ∈ 𝑋 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) )
5 sslin ⊢ ( ∩ 𝑝 ∈ 𝑌 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) ⊆ ∩ 𝑝 ∈ 𝑋 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) → ( 𝐴 ∩ ∩ 𝑝 ∈ 𝑌 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) ) ⊆ ( 𝐴 ∩ ∩ 𝑝 ∈ 𝑋 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) ) )
6 3 4 5 3syl ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ⊆ 𝑌 ) → ( 𝐴 ∩ ∩ 𝑝 ∈ 𝑌 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) ) ⊆ ( 𝐴 ∩ ∩ 𝑝 ∈ 𝑋 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) ) )
7 eqid ⊢ ( oc ‘ 𝐾 ) = ( oc ‘ 𝐾 )
8 eqid ⊢ ( pmap ‘ 𝐾 ) = ( pmap ‘ 𝐾 )
9 7 1 8 2 polvalN ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ) → ( ⊥ ‘ 𝑌 ) = ( 𝐴 ∩ ∩ 𝑝 ∈ 𝑌 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) ) )
10 9 3adant3 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ⊆ 𝑌 ) → ( ⊥ ‘ 𝑌 ) = ( 𝐴 ∩ ∩ 𝑝 ∈ 𝑌 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) ) )
11 simp1 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ⊆ 𝑌 ) → 𝐾 ∈ HL )
12 simp2 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ⊆ 𝑌 ) → 𝑌 ⊆ 𝐴 )
13 3 12 sstrd ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ⊆ 𝑌 ) → 𝑋 ⊆ 𝐴 )
14 7 1 8 2 polvalN ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ) → ( ⊥ ‘ 𝑋 ) = ( 𝐴 ∩ ∩ 𝑝 ∈ 𝑋 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) ) )
15 11 13 14 syl2anc ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ⊆ 𝑌 ) → ( ⊥ ‘ 𝑋 ) = ( 𝐴 ∩ ∩ 𝑝 ∈ 𝑋 ( ( pmap ‘ 𝐾 ) ‘ ( ( oc ‘ 𝐾 ) ‘ 𝑝 ) ) ) )
16 6 10 15 3sstr4d ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ⊆ 𝑌 ) → ( ⊥ ‘ 𝑌 ) ⊆ ( ⊥ ‘ 𝑋 ) )