Metamath Proof Explorer


Theorem cdleme50trn12

Description: Part of proof that F is a translation. Combine R .<_ ( P .\/ Q ) and -. R .<_ ( P .\/ Q ) cases. TODO: fix comment. (Contributed by NM, 10-Apr-2013)

Ref Expression
Hypotheses cdlemef50.b ⊢ B = Base K
cdlemef50.l ⊢ ≤ ˙ = ≤ K
cdlemef50.j ⊢ ∨ ˙ = join ⁡ K
cdlemef50.m ⊢ ∧ ˙ = meet ⁡ K
cdlemef50.a ⊢ A = Atoms ⁡ K
cdlemef50.h ⊢ H = LHyp ⁡ K
cdlemef50.u ⊢ U = P ∨ ˙ Q ∧ ˙ W
cdlemef50.d ⊢ D = t ∨ ˙ U ∧ ˙ Q ∨ ˙ P ∨ ˙ t ∧ ˙ W
cdlemefs50.e ⊢ E = P ∨ ˙ Q ∧ ˙ D ∨ ˙ s ∨ ˙ t ∧ ˙ W
cdlemef50.f ⊢ F = x ∈ B ⟼ if P ≠ Q ∧ ¬ x ≤ ˙ W ι z ∈ B | ∀ s ∈ A ¬ s ≤ ˙ W ∧ s ∨ ˙ x ∧ ˙ W = x → z = if s ≤ ˙ P ∨ ˙ Q ι y ∈ B | ∀ t ∈ A ¬ t ≤ ˙ W ∧ ¬ t ≤ ˙ P ∨ ˙ Q → y = E ⦋ s / t⦌ D ∨ ˙ x ∧ ˙ W x
Assertion cdleme50trn12 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ P ≠ Q ∧ R ∈ A ∧ ¬ R ≤ ˙ W → R ∨ ˙ F ⁡ R ∧ ˙ W = U

Proof

Step Hyp Ref Expression
1 cdlemef50.b ⊢ B = Base K
2 cdlemef50.l ⊢ ≤ ˙ = ≤ K
3 cdlemef50.j ⊢ ∨ ˙ = join ⁡ K
4 cdlemef50.m ⊢ ∧ ˙ = meet ⁡ K
5 cdlemef50.a ⊢ A = Atoms ⁡ K
6 cdlemef50.h ⊢ H = LHyp ⁡ K
7 cdlemef50.u ⊢ U = P ∨ ˙ Q ∧ ˙ W
8 cdlemef50.d ⊢ D = t ∨ ˙ U ∧ ˙ Q ∨ ˙ P ∨ ˙ t ∧ ˙ W
9 cdlemefs50.e ⊢ E = P ∨ ˙ Q ∧ ˙ D ∨ ˙ s ∨ ˙ t ∧ ˙ W
10 cdlemef50.f ⊢ F = x ∈ B ⟼ if P ≠ Q ∧ ¬ x ≤ ˙ W ι z ∈ B | ∀ s ∈ A ¬ s ≤ ˙ W ∧ s ∨ ˙ x ∧ ˙ W = x → z = if s ≤ ˙ P ∨ ˙ Q ι y ∈ B | ∀ t ∈ A ¬ t ≤ ˙ W ∧ ¬ t ≤ ˙ P ∨ ˙ Q → y = E ⦋ s / t⦌ D ∨ ˙ x ∧ ˙ W x
11 1 2 3 4 5 6 7 8 9 10 cdleme50trn2 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ P ≠ Q ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q → R ∨ ˙ F ⁡ R ∧ ˙ W = U
12 11 3expa ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ P ≠ Q ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ R ≤ ˙ P ∨ ˙ Q → R ∨ ˙ F ⁡ R ∧ ˙ W = U
13 1 2 3 4 5 6 7 8 9 10 cdleme50trn1 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ P ≠ Q ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ ¬ R ≤ ˙ P ∨ ˙ Q → R ∨ ˙ F ⁡ R ∧ ˙ W = U
14 13 3expa ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ P ≠ Q ∧ R ∈ A ∧ ¬ R ≤ ˙ W ∧ ¬ R ≤ ˙ P ∨ ˙ Q → R ∨ ˙ F ⁡ R ∧ ˙ W = U
15 12 14 pm2.61dan ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ P ≠ Q ∧ R ∈ A ∧ ¬ R ≤ ˙ W → R ∨ ˙ F ⁡ R ∧ ˙ W = U