Metamath Proof Explorer


Theorem dihord2

Description: Part of proof after Lemma N of Crawley p. 122. Reverse ordering property. TODO: do we need -. X .<_ W and -. Y .<_ W ? (Contributed by NM, 4-Mar-2014)

Ref Expression
Hypotheses dihord2.b ⊢ B = Base K
dihord2.l ⊢ ≤ ˙ = ≤ K
dihord2.j ⊢ ∨ ˙ = join ⁡ K
dihord2.m ⊢ ∧ ˙ = meet ⁡ K
dihord2.a ⊢ A = Atoms ⁡ K
dihord2.h ⊢ H = LHyp ⁡ K
dihord2.i ⊢ I = DIsoB ⁡ K ⁡ W
dihord2.J ⊢ J = DIsoC ⁡ K ⁡ W
dihord2.u ⊢ U = DVecH ⁡ K ⁡ W
dihord2.s ⊢ ⊕ ˙ = LSSum ⁡ U
Assertion dihord2 ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ N ∈ A ∧ ¬ N ≤ ˙ W ∧ X ∈ B ∧ Y ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = X ∧ N ∨ ˙ Y ∧ ˙ W = Y ∧ J ⁡ Q ⊕ ˙ I ⁡ X ∧ ˙ W ⊆ J ⁡ N ⊕ ˙ I ⁡ Y ∧ ˙ W → X ≤ ˙ Y

Proof

Step Hyp Ref Expression
1 dihord2.b ⊢ B = Base K
2 dihord2.l ⊢ ≤ ˙ = ≤ K
3 dihord2.j ⊢ ∨ ˙ = join ⁡ K
4 dihord2.m ⊢ ∧ ˙ = meet ⁡ K
5 dihord2.a ⊢ A = Atoms ⁡ K
6 dihord2.h ⊢ H = LHyp ⁡ K
7 dihord2.i ⊢ I = DIsoB ⁡ K ⁡ W
8 dihord2.J ⊢ J = DIsoC ⁡ K ⁡ W
9 dihord2.u ⊢ U = DVecH ⁡ K ⁡ W
10 dihord2.s ⊢ ⊕ ˙ = LSSum ⁡ U
11 eqid ⊢ LTrn ⁡ K ⁡ W = LTrn ⁡ K ⁡ W
12 eqid ⊢ trL ⁡ K ⁡ W = trL ⁡ K ⁡ W
13 eqid ⊢ h ∈ LTrn ⁡ K ⁡ W ⟼ I ↾ B = h ∈ LTrn ⁡ K ⁡ W ⟼ I ↾ B
14 eqid ⊢ oc ⁡ K ⁡ W = oc ⁡ K ⁡ W
15 eqid ⊢ TEndo ⁡ K ⁡ W = TEndo ⁡ K ⁡ W
16 eqid ⊢ + U = + U
17 eqid ⊢ ι h ∈ LTrn ⁡ K ⁡ W | h ⁡ oc ⁡ K ⁡ W = N = ι h ∈ LTrn ⁡ K ⁡ W | h ⁡ oc ⁡ K ⁡ W = N
18 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 dihord2pre2 ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ N ∈ A ∧ ¬ N ≤ ˙ W ∧ X ∈ B ∧ Y ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = X ∧ N ∨ ˙ Y ∧ ˙ W = Y ∧ J ⁡ Q ⊕ ˙ I ⁡ X ∧ ˙ W ⊆ J ⁡ N ⊕ ˙ I ⁡ Y ∧ ˙ W → Q ∨ ˙ X ∧ ˙ W ≤ ˙ N ∨ ˙ Y ∧ ˙ W
19 simp31 ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ N ∈ A ∧ ¬ N ≤ ˙ W ∧ X ∈ B ∧ Y ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = X ∧ N ∨ ˙ Y ∧ ˙ W = Y ∧ J ⁡ Q ⊕ ˙ I ⁡ X ∧ ˙ W ⊆ J ⁡ N ⊕ ˙ I ⁡ Y ∧ ˙ W → Q ∨ ˙ X ∧ ˙ W = X
20 simp32 ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ N ∈ A ∧ ¬ N ≤ ˙ W ∧ X ∈ B ∧ Y ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = X ∧ N ∨ ˙ Y ∧ ˙ W = Y ∧ J ⁡ Q ⊕ ˙ I ⁡ X ∧ ˙ W ⊆ J ⁡ N ⊕ ˙ I ⁡ Y ∧ ˙ W → N ∨ ˙ Y ∧ ˙ W = Y
21 18 19 20 3brtr3d ⊢ K ∈ HL ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W ∧ N ∈ A ∧ ¬ N ≤ ˙ W ∧ X ∈ B ∧ Y ∈ B ∧ Q ∨ ˙ X ∧ ˙ W = X ∧ N ∨ ˙ Y ∧ ˙ W = Y ∧ J ⁡ Q ⊕ ˙ I ⁡ X ∧ ˙ W ⊆ J ⁡ N ⊕ ˙ I ⁡ Y ∧ ˙ W → X ≤ ˙ Y