Metamath Proof Explorer


Theorem cdlemk42yN

Description: Part of proof of Lemma K of Crawley p. 118. TODO: fix comment. (Contributed by NM, 20-Jul-2013) (New usage is discouraged.)

Ref Expression
Hypotheses cdlemk5.b ⊢ B = Base K
cdlemk5.l ⊢ ≤ ˙ = ≤ K
cdlemk5.j ⊢ ∨ ˙ = join ⁡ K
cdlemk5.m ⊢ ∧ ˙ = meet ⁡ K
cdlemk5.a ⊢ A = Atoms ⁡ K
cdlemk5.h ⊢ H = LHyp ⁡ K
cdlemk5.t ⊢ T = LTrn ⁡ K ⁡ W
cdlemk5.r ⊢ R = trL ⁡ K ⁡ W
cdlemk5.z ⊢ Z = P ∨ ˙ R ⁡ b ∧ ˙ N ⁡ P ∨ ˙ R ⁡ b ∘ F -1
cdlemk5.y ⊢ Y = P ∨ ˙ R ⁡ g ∧ ˙ Z ∨ ˙ R ⁡ g ∘ b -1
cdlemk5.x ⊢ X = ι z ∈ T | ∀ b ∈ T b ≠ I ↾ B ∧ R ⁡ b ≠ R ⁡ F ∧ R ⁡ b ≠ R ⁡ g → z ⁡ P = Y
Assertion cdlemk42yN ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ F ≠ I ↾ B ∧ G ∈ T ∧ G ≠ I ↾ B ∧ N ∈ T ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ R ⁡ F = R ⁡ N ∧ b ∈ T ∧ b ≠ I ↾ B ∧ R ⁡ b ≠ R ⁡ F ∧ R ⁡ b ≠ R ⁡ G → ⦋ G / g⦌ X ⁡ P = P ∨ ˙ R ⁡ G ∧ ˙ Z ∨ ˙ R ⁡ G ∘ b -1

Proof

Step Hyp Ref Expression
1 cdlemk5.b ⊢ B = Base K
2 cdlemk5.l ⊢ ≤ ˙ = ≤ K
3 cdlemk5.j ⊢ ∨ ˙ = join ⁡ K
4 cdlemk5.m ⊢ ∧ ˙ = meet ⁡ K
5 cdlemk5.a ⊢ A = Atoms ⁡ K
6 cdlemk5.h ⊢ H = LHyp ⁡ K
7 cdlemk5.t ⊢ T = LTrn ⁡ K ⁡ W
8 cdlemk5.r ⊢ R = trL ⁡ K ⁡ W
9 cdlemk5.z ⊢ Z = P ∨ ˙ R ⁡ b ∧ ˙ N ⁡ P ∨ ˙ R ⁡ b ∘ F -1
10 cdlemk5.y ⊢ Y = P ∨ ˙ R ⁡ g ∧ ˙ Z ∨ ˙ R ⁡ g ∘ b -1
11 cdlemk5.x ⊢ X = ι z ∈ T | ∀ b ∈ T b ≠ I ↾ B ∧ R ⁡ b ≠ R ⁡ F ∧ R ⁡ b ≠ R ⁡ g → z ⁡ P = Y
12 1 2 3 4 5 6 7 8 9 10 11 cdlemk42 ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ F ≠ I ↾ B ∧ G ∈ T ∧ G ≠ I ↾ B ∧ N ∈ T ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ R ⁡ F = R ⁡ N ∧ b ∈ T ∧ b ≠ I ↾ B ∧ R ⁡ b ≠ R ⁡ F ∧ R ⁡ b ≠ R ⁡ G → ⦋ G / g⦌ X ⁡ P = ⦋ G / g⦌ Y
13 simp13l ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ F ≠ I ↾ B ∧ G ∈ T ∧ G ≠ I ↾ B ∧ N ∈ T ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ R ⁡ F = R ⁡ N ∧ b ∈ T ∧ b ≠ I ↾ B ∧ R ⁡ b ≠ R ⁡ F ∧ R ⁡ b ≠ R ⁡ G → G ∈ T
14 10 cdlemk41 ⊢ G ∈ T → ⦋ G / g⦌ Y = P ∨ ˙ R ⁡ G ∧ ˙ Z ∨ ˙ R ⁡ G ∘ b -1
15 13 14 syl ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ F ≠ I ↾ B ∧ G ∈ T ∧ G ≠ I ↾ B ∧ N ∈ T ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ R ⁡ F = R ⁡ N ∧ b ∈ T ∧ b ≠ I ↾ B ∧ R ⁡ b ≠ R ⁡ F ∧ R ⁡ b ≠ R ⁡ G → ⦋ G / g⦌ Y = P ∨ ˙ R ⁡ G ∧ ˙ Z ∨ ˙ R ⁡ G ∘ b -1
16 12 15 eqtrd ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ F ≠ I ↾ B ∧ G ∈ T ∧ G ≠ I ↾ B ∧ N ∈ T ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ R ⁡ F = R ⁡ N ∧ b ∈ T ∧ b ≠ I ↾ B ∧ R ⁡ b ≠ R ⁡ F ∧ R ⁡ b ≠ R ⁡ G → ⦋ G / g⦌ X ⁡ P = P ∨ ˙ R ⁡ G ∧ ˙ Z ∨ ˙ R ⁡ G ∘ b -1