Metamath Proof Explorer

Theorem osumcllem10N

Description: Lemma for osumclN . Contradict osumcllem9N . (Contributed by NM, 25-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 osumcllem10N ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → 𝑀 ≠ 𝑋 )

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 simp11 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → 𝐾 ∈ HL )
10 simp2 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → 𝑝 ∈ 𝐴 )
11 10 snssd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → { 𝑝 } ⊆ 𝐴 )
12 simp12 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → 𝑋 ⊆ 𝐴 )
13 3 4 sspadd2 ⊢ ( ( 𝐾 ∈ HL ∧ { 𝑝 } ⊆ 𝐴 ∧ 𝑋 ⊆ 𝐴 ) → { 𝑝 } ⊆ ( 𝑋 + { 𝑝 } ) )
14 9 11 12 13 syl3anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → { 𝑝 } ⊆ ( 𝑋 + { 𝑝 } ) )
15 vex ⊢ 𝑝 ∈ V
16 15 snss ⊢ ( 𝑝 ∈ ( 𝑋 + { 𝑝 } ) ↔ { 𝑝 } ⊆ ( 𝑋 + { 𝑝 } ) )
17 14 16 sylibr ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → 𝑝 ∈ ( 𝑋 + { 𝑝 } ) )
18 17 7 eleqtrrdi ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → 𝑝 ∈ 𝑀 )
19 3 4 sspadd1 ⊢ ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) → 𝑋 ⊆ ( 𝑋 + 𝑌 ) )
20 19 3ad2ant1 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → 𝑋 ⊆ ( 𝑋 + 𝑌 ) )
21 simp3 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) )
22 20 21 ssneldd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → ¬ 𝑝 ∈ 𝑋 )
23 nelne1 ⊢ ( ( 𝑝 ∈ 𝑀 ∧ ¬ 𝑝 ∈ 𝑋 ) → 𝑀 ≠ 𝑋 )
24 18 22 23 syl2anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ) ∧ 𝑝 ∈ 𝐴 ∧ ¬ 𝑝 ∈ ( 𝑋 + 𝑌 ) ) → 𝑀 ≠ 𝑋 )