Metamath Proof Explorer


Theorem cdleml8

Description: Part of proof of Lemma L of Crawley p. 120. TODO: fix comment. (Contributed by NM, 11-Aug-2013)

Ref Expression
Hypotheses cdleml6.b B = Base K
cdleml6.j ˙ = join K
cdleml6.m ˙ = meet K
cdleml6.h H = LHyp K
cdleml6.t T = LTrn K W
cdleml6.r R = trL K W
cdleml6.p Q = oc K W
cdleml6.z Z = Q ˙ R b ˙ h Q ˙ R b s h -1
cdleml6.y Y = Q ˙ R g ˙ Z ˙ R g b -1
cdleml6.x X = ι z T | b T b I B R b R s h R b R g z Q = Y
cdleml6.u U = g T if s h = h g X
cdleml6.e E = TEndo K W
cdleml6.o 0 ˙ = f T I B
Assertion cdleml8 K HL W H h T h I B s E s 0 ˙ U s = I T

Proof

Step Hyp Ref Expression
1 cdleml6.b B = Base K
2 cdleml6.j ˙ = join K
3 cdleml6.m ˙ = meet K
4 cdleml6.h H = LHyp K
5 cdleml6.t T = LTrn K W
6 cdleml6.r R = trL K W
7 cdleml6.p Q = oc K W
8 cdleml6.z Z = Q ˙ R b ˙ h Q ˙ R b s h -1
9 cdleml6.y Y = Q ˙ R g ˙ Z ˙ R g b -1
10 cdleml6.x X = ι z T | b T b I B R b R s h R b R g z Q = Y
11 cdleml6.u U = g T if s h = h g X
12 cdleml6.e E = TEndo K W
13 cdleml6.o 0 ˙ = f T I B
14 simp1 K HL W H h T h I B s E s 0 ˙ K HL W H
15 1 2 3 4 5 6 7 8 9 10 11 12 13 cdleml6 K HL W H h T s E s 0 ˙ U E U s h = h
16 15 3adant2r K HL W H h T h I B s E s 0 ˙ U E U s h = h
17 16 simpld K HL W H h T h I B s E s 0 ˙ U E
18 simp3l K HL W H h T h I B s E s 0 ˙ s E
19 4 12 tendococl K HL W H U E s E U s E
20 14 17 18 19 syl3anc K HL W H h T h I B s E s 0 ˙ U s E
21 4 5 12 tendoidcl K HL W H I T E
22 21 3ad2ant1 K HL W H h T h I B s E s 0 ˙ I T E
23 1 2 3 4 5 6 7 8 9 10 11 12 13 cdleml7 K HL W H h T s E s 0 ˙ U s h = I T h
24 23 3adant2r K HL W H h T h I B s E s 0 ˙ U s h = I T h
25 simp2 K HL W H h T h I B s E s 0 ˙ h T h I B
26 1 4 5 12 tendocan K HL W H U s E I T E U s h = I T h h T h I B U s = I T
27 14 20 22 24 25 26 syl131anc K HL W H h T h I B s E s 0 ˙ U s = I T