Metamath Proof Explorer


Theorem cdlemednuN

Description: Part of proof of Lemma E in Crawley p. 113. Utility lemma. D represents s_2. (Contributed by NM, 18-Nov-2012) (New usage is discouraged.)

Ref Expression
Hypotheses cdlemeda.l ⊢ ≤ ˙ = ≤ K
cdlemeda.j ⊢ ∨ ˙ = join ⁡ K
cdlemeda.m ⊢ ∧ ˙ = meet ⁡ K
cdlemeda.a ⊢ A = Atoms ⁡ K
cdlemeda.h ⊢ H = LHyp ⁡ K
cdlemeda.d ⊢ D = R ∨ ˙ S ∧ ˙ W
cdlemednu.u ⊢ U = P ∨ ˙ Q ∧ ˙ W
Assertion cdlemednuN ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → D ≠ U

Proof

Step Hyp Ref Expression
1 cdlemeda.l ⊢ ≤ ˙ = ≤ K
2 cdlemeda.j ⊢ ∨ ˙ = join ⁡ K
3 cdlemeda.m ⊢ ∧ ˙ = meet ⁡ K
4 cdlemeda.a ⊢ A = Atoms ⁡ K
5 cdlemeda.h ⊢ H = LHyp ⁡ K
6 cdlemeda.d ⊢ D = R ∨ ˙ S ∧ ˙ W
7 cdlemednu.u ⊢ U = P ∨ ˙ Q ∧ ˙ W
8 1 2 3 4 5 6 cdlemednpq ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → ¬ D ≤ ˙ P ∨ ˙ Q
9 simp1l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → K ∈ HL
10 simp1r ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → W ∈ H
11 simp21 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → P ∈ A
12 simp22 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → Q ∈ A
13 1 2 3 4 5 7 cdlemeulpq ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A → U ≤ ˙ P ∨ ˙ Q
14 9 10 11 12 13 syl22anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → U ≤ ˙ P ∨ ˙ Q
15 breq1 ⊢ D = U → D ≤ ˙ P ∨ ˙ Q ↔ U ≤ ˙ P ∨ ˙ Q
16 14 15 syl5ibrcom ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → D = U → D ≤ ˙ P ∨ ˙ Q
17 16 necon3bd ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → ¬ D ≤ ˙ P ∨ ˙ Q → D ≠ U
18 8 17 mpd ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ S ∈ A ∧ ¬ S ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q ∧ ¬ S ≤ ˙ P ∨ ˙ Q → D ≠ U