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