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Metamath Proof Explorer

Theorem 4atlem0be

Description: Lemma for 4at . (Contributed by NM, 10-Jul-2012)

Ref Expression
Hypotheses 4at.l ⊢ ≤ = ( le ‘ 𝐾 )
4at.j ⊢ ∨ = ( join ‘ 𝐾 )
4at.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
Assertion 4atlem0be ( ( 𝐾 ∈ HL ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ ¬ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑃 ≠ 𝑅 )

Proof

Step Hyp Ref Expression
1 4at.l ⊢ ≤ = ( le ‘ 𝐾 )
2 4at.j ⊢ ∨ = ( join ‘ 𝐾 )
3 4at.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
4 simp1 ⊢ ( ( 𝐾 ∈ HL ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ ¬ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝐾 ∈ HL )
5 simp23 ⊢ ( ( 𝐾 ∈ HL ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ ¬ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑅 ∈ 𝐴 )
6 simp21 ⊢ ( ( 𝐾 ∈ HL ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ ¬ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑃 ∈ 𝐴 )
7 simp22 ⊢ ( ( 𝐾 ∈ HL ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ ¬ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑄 ∈ 𝐴 )
8 simp3 ⊢ ( ( 𝐾 ∈ HL ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ ¬ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → ¬ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) )
9 1 2 3 atnlej1 ⊢ ( ( 𝐾 ∈ HL ∧ ( 𝑅 ∈ 𝐴 ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) ∧ ¬ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑅 ≠ 𝑃 )
10 9 necomd ⊢ ( ( 𝐾 ∈ HL ∧ ( 𝑅 ∈ 𝐴 ∧ 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ) ∧ ¬ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑃 ≠ 𝑅 )
11 4 5 6 7 8 10 syl131anc ⊢ ( ( 𝐾 ∈ HL ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ ¬ 𝑅 ≤ ( 𝑃 ∨ 𝑄 ) ) → 𝑃 ≠ 𝑅 )