Metamath Proof Explorer


Theorem pexmidlem5N

Description: Lemma for pexmidN . (Contributed by NM, 2-Feb-2012) (New usage is discouraged.)

Ref Expression
Hypotheses pexmidlem.l = ( le ‘ 𝐾 )
pexmidlem.j = ( join ‘ 𝐾 )
pexmidlem.a 𝐴 = ( Atoms ‘ 𝐾 )
pexmidlem.p + = ( +𝑃𝐾 )
pexmidlem.o = ( ⊥𝑃𝐾 )
pexmidlem.m 𝑀 = ( 𝑋 + { 𝑝 } )
Assertion pexmidlem5N ( ( ( 𝐾 ∈ HL ∧ 𝑋𝐴𝑝𝐴 ) ∧ ( 𝑋 ≠ ∅ ∧ ¬ 𝑝 ∈ ( 𝑋 + ( 𝑋 ) ) ) ) → ( ( 𝑋 ) ∩ 𝑀 ) = ∅ )

Proof

Step Hyp Ref Expression
1 pexmidlem.l = ( le ‘ 𝐾 )
2 pexmidlem.j = ( join ‘ 𝐾 )
3 pexmidlem.a 𝐴 = ( Atoms ‘ 𝐾 )
4 pexmidlem.p + = ( +𝑃𝐾 )
5 pexmidlem.o = ( ⊥𝑃𝐾 )
6 pexmidlem.m 𝑀 = ( 𝑋 + { 𝑝 } )
7 n0 ( ( ( 𝑋 ) ∩ 𝑀 ) ≠ ∅ ↔ ∃ 𝑞 𝑞 ∈ ( ( 𝑋 ) ∩ 𝑀 ) )
8 1 2 3 4 5 6 pexmidlem4N ( ( ( 𝐾 ∈ HL ∧ 𝑋𝐴𝑝𝐴 ) ∧ ( 𝑋 ≠ ∅ ∧ 𝑞 ∈ ( ( 𝑋 ) ∩ 𝑀 ) ) ) → 𝑝 ∈ ( 𝑋 + ( 𝑋 ) ) )
9 8 expr ( ( ( 𝐾 ∈ HL ∧ 𝑋𝐴𝑝𝐴 ) ∧ 𝑋 ≠ ∅ ) → ( 𝑞 ∈ ( ( 𝑋 ) ∩ 𝑀 ) → 𝑝 ∈ ( 𝑋 + ( 𝑋 ) ) ) )
10 9 exlimdv ( ( ( 𝐾 ∈ HL ∧ 𝑋𝐴𝑝𝐴 ) ∧ 𝑋 ≠ ∅ ) → ( ∃ 𝑞 𝑞 ∈ ( ( 𝑋 ) ∩ 𝑀 ) → 𝑝 ∈ ( 𝑋 + ( 𝑋 ) ) ) )
11 7 10 syl5bi ( ( ( 𝐾 ∈ HL ∧ 𝑋𝐴𝑝𝐴 ) ∧ 𝑋 ≠ ∅ ) → ( ( ( 𝑋 ) ∩ 𝑀 ) ≠ ∅ → 𝑝 ∈ ( 𝑋 + ( 𝑋 ) ) ) )
12 11 necon1bd ( ( ( 𝐾 ∈ HL ∧ 𝑋𝐴𝑝𝐴 ) ∧ 𝑋 ≠ ∅ ) → ( ¬ 𝑝 ∈ ( 𝑋 + ( 𝑋 ) ) → ( ( 𝑋 ) ∩ 𝑀 ) = ∅ ) )
13 12 impr ( ( ( 𝐾 ∈ HL ∧ 𝑋𝐴𝑝𝐴 ) ∧ ( 𝑋 ≠ ∅ ∧ ¬ 𝑝 ∈ ( 𝑋 + ( 𝑋 ) ) ) ) → ( ( 𝑋 ) ∩ 𝑀 ) = ∅ )