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Metamath Proof Explorer

Theorem osumcllem3N

Description: Lemma for osumclN . (Contributed by NM, 23-Mar-2012) (New usage is discouraged.)

Ref Expression
Hypotheses osumcllem.l ⊢ ≤ = ( le ‘ 𝐾 )
osumcllem.j ⊢ ∨ = ( join ‘ 𝐾 )
osumcllem.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
osumcllem.p ⊢ + = ( +𝑃 ‘ 𝐾 )
osumcllem.o ⊢ ⊥ = ( ⊥𝑃 ‘ 𝐾 )
osumcllem.c ⊢ 𝐶 = ( PSubCl ‘ 𝐾 )
osumcllem.m ⊢ 𝑀 = ( 𝑋 + { 𝑝 } )
osumcllem.u ⊢ 𝑈 = ( ⊥ ‘ ( ⊥ ‘ ( 𝑋 + 𝑌 ) ) )
Assertion osumcllem3N ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → ( ( ⊥ ‘ 𝑋 ) ∩ 𝑈 ) = 𝑌 )

Proof

Step Hyp Ref Expression
1 osumcllem.l ⊢ ≤ = ( le ‘ 𝐾 )
2 osumcllem.j ⊢ ∨ = ( join ‘ 𝐾 )
3 osumcllem.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
4 osumcllem.p ⊢ + = ( +𝑃 ‘ 𝐾 )
5 osumcllem.o ⊢ ⊥ = ( ⊥𝑃 ‘ 𝐾 )
6 osumcllem.c ⊢ 𝐶 = ( PSubCl ‘ 𝐾 )
7 osumcllem.m ⊢ 𝑀 = ( 𝑋 + { 𝑝 } )
8 osumcllem.u ⊢ 𝑈 = ( ⊥ ‘ ( ⊥ ‘ ( 𝑋 + 𝑌 ) ) )
9 incom ⊢ ( ( ⊥ ‘ 𝑋 ) ∩ 𝑈 ) = ( 𝑈 ∩ ( ⊥ ‘ 𝑋 ) )
10 simp1 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → 𝐾 ∈ HL )
11 simp3 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) )
12 3 6 psubclssatN ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ) → 𝑌 ⊆ 𝐴 )
13 12 3adant3 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → 𝑌 ⊆ 𝐴 )
14 3 5 polssatN ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ) → ( ⊥ ‘ 𝑌 ) ⊆ 𝐴 )
15 10 13 14 syl2anc ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → ( ⊥ ‘ 𝑌 ) ⊆ 𝐴 )
16 11 15 sstrd ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → 𝑋 ⊆ 𝐴 )
17 3 4 5 poldmj1N ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) → ( ⊥ ‘ ( 𝑋 + 𝑌 ) ) = ( ( ⊥ ‘ 𝑋 ) ∩ ( ⊥ ‘ 𝑌 ) ) )
18 10 16 13 17 syl3anc ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → ( ⊥ ‘ ( 𝑋 + 𝑌 ) ) = ( ( ⊥ ‘ 𝑋 ) ∩ ( ⊥ ‘ 𝑌 ) ) )
19 incom ⊢ ( ( ⊥ ‘ 𝑋 ) ∩ ( ⊥ ‘ 𝑌 ) ) = ( ( ⊥ ‘ 𝑌 ) ∩ ( ⊥ ‘ 𝑋 ) )
20 18 19 eqtrdi ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → ( ⊥ ‘ ( 𝑋 + 𝑌 ) ) = ( ( ⊥ ‘ 𝑌 ) ∩ ( ⊥ ‘ 𝑋 ) ) )
21 20 fveq2d ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → ( ⊥ ‘ ( ⊥ ‘ ( 𝑋 + 𝑌 ) ) ) = ( ⊥ ‘ ( ( ⊥ ‘ 𝑌 ) ∩ ( ⊥ ‘ 𝑋 ) ) ) )
22 8 21 eqtrid ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → 𝑈 = ( ⊥ ‘ ( ( ⊥ ‘ 𝑌 ) ∩ ( ⊥ ‘ 𝑋 ) ) ) )
23 22 ineq1d ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → ( 𝑈 ∩ ( ⊥ ‘ 𝑋 ) ) = ( ( ⊥ ‘ ( ( ⊥ ‘ 𝑌 ) ∩ ( ⊥ ‘ 𝑋 ) ) ) ∩ ( ⊥ ‘ 𝑋 ) ) )
24 3 5 polcon2N ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → 𝑌 ⊆ ( ⊥ ‘ 𝑋 ) )
25 13 24 syld3an2 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → 𝑌 ⊆ ( ⊥ ‘ 𝑋 ) )
26 3 5 poml5N ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ ( ⊥ ‘ 𝑋 ) ) → ( ( ⊥ ‘ ( ( ⊥ ‘ 𝑌 ) ∩ ( ⊥ ‘ 𝑋 ) ) ) ∩ ( ⊥ ‘ 𝑋 ) ) = ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) )
27 10 16 25 26 syl3anc ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → ( ( ⊥ ‘ ( ( ⊥ ‘ 𝑌 ) ∩ ( ⊥ ‘ 𝑋 ) ) ) ∩ ( ⊥ ‘ 𝑋 ) ) = ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) )
28 5 6 psubcli2N ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) = 𝑌 )
29 28 3adant3 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) = 𝑌 )
30 23 27 29 3eqtrd ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → ( 𝑈 ∩ ( ⊥ ‘ 𝑋 ) ) = 𝑌 )
31 9 30 eqtrid ⊢ ( ( 𝐾 ∈ HL ∧ 𝑌 ∈ 𝐶 ∧ 𝑋 ⊆ ( ⊥ ‘ 𝑌 ) ) → ( ( ⊥ ‘ 𝑋 ) ∩ 𝑈 ) = 𝑌 )