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 + = ( +𝑃𝐾 )
osumcllem.o = ( ⊥𝑃𝐾 )
osumcllem.c 𝐶 = ( PSubCl ‘ 𝐾 )
osumcllem.m 𝑀 = ( 𝑋 + { 𝑝 } )
osumcllem.u 𝑈 = ( ‘ ( ‘ ( 𝑋 + 𝑌 ) ) )
Assertion osumcllem3N ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 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 + = ( +𝑃𝐾 )
5 osumcllem.o = ( ⊥𝑃𝐾 )
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 ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → 𝐾 ∈ HL )
11 simp3 ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → 𝑋 ⊆ ( 𝑌 ) )
12 3 6 psubclssatN ( ( 𝐾 ∈ HL ∧ 𝑌𝐶 ) → 𝑌𝐴 )
13 12 3adant3 ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → 𝑌𝐴 )
14 3 5 polssatN ( ( 𝐾 ∈ HL ∧ 𝑌𝐴 ) → ( 𝑌 ) ⊆ 𝐴 )
15 10 13 14 syl2anc ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → ( 𝑌 ) ⊆ 𝐴 )
16 11 15 sstrd ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → 𝑋𝐴 )
17 3 4 5 poldmj1N ( ( 𝐾 ∈ HL ∧ 𝑋𝐴𝑌𝐴 ) → ( ‘ ( 𝑋 + 𝑌 ) ) = ( ( 𝑋 ) ∩ ( 𝑌 ) ) )
18 10 16 13 17 syl3anc ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → ( ‘ ( 𝑋 + 𝑌 ) ) = ( ( 𝑋 ) ∩ ( 𝑌 ) ) )
19 incom ( ( 𝑋 ) ∩ ( 𝑌 ) ) = ( ( 𝑌 ) ∩ ( 𝑋 ) )
20 18 19 eqtrdi ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → ( ‘ ( 𝑋 + 𝑌 ) ) = ( ( 𝑌 ) ∩ ( 𝑋 ) ) )
21 20 fveq2d ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → ( ‘ ( ‘ ( 𝑋 + 𝑌 ) ) ) = ( ‘ ( ( 𝑌 ) ∩ ( 𝑋 ) ) ) )
22 8 21 syl5eq ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → 𝑈 = ( ‘ ( ( 𝑌 ) ∩ ( 𝑋 ) ) ) )
23 22 ineq1d ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → ( 𝑈 ∩ ( 𝑋 ) ) = ( ( ‘ ( ( 𝑌 ) ∩ ( 𝑋 ) ) ) ∩ ( 𝑋 ) ) )
24 3 5 polcon2N ( ( 𝐾 ∈ HL ∧ 𝑌𝐴𝑋 ⊆ ( 𝑌 ) ) → 𝑌 ⊆ ( 𝑋 ) )
25 13 24 syld3an2 ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → 𝑌 ⊆ ( 𝑋 ) )
26 3 5 poml5N ( ( 𝐾 ∈ HL ∧ 𝑋𝐴𝑌 ⊆ ( 𝑋 ) ) → ( ( ‘ ( ( 𝑌 ) ∩ ( 𝑋 ) ) ) ∩ ( 𝑋 ) ) = ( ‘ ( 𝑌 ) ) )
27 10 16 25 26 syl3anc ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → ( ( ‘ ( ( 𝑌 ) ∩ ( 𝑋 ) ) ) ∩ ( 𝑋 ) ) = ( ‘ ( 𝑌 ) ) )
28 5 6 psubcli2N ( ( 𝐾 ∈ HL ∧ 𝑌𝐶 ) → ( ‘ ( 𝑌 ) ) = 𝑌 )
29 28 3adant3 ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → ( ‘ ( 𝑌 ) ) = 𝑌 )
30 23 27 29 3eqtrd ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → ( 𝑈 ∩ ( 𝑋 ) ) = 𝑌 )
31 9 30 syl5eq ( ( 𝐾 ∈ HL ∧ 𝑌𝐶𝑋 ⊆ ( 𝑌 ) ) → ( ( 𝑋 ) ∩ 𝑈 ) = 𝑌 )