Metamath Proof Explorer


Theorem cdlemk19w

Description: Use a fixed element to eliminate P in cdlemk19u . (Contributed by NM, 1-Aug-2013)

Ref Expression
Hypotheses cdlemk6.b B = Base K
cdlemk6.j ˙ = join K
cdlemk6.m ˙ = meet K
cdlemk6.o ˙ = oc K
cdlemk6.a A = Atoms K
cdlemk6.h H = LHyp K
cdlemk6.t T = LTrn K W
cdlemk6.r R = trL K W
cdlemk6.p P = ˙ W
cdlemk6.z Z = P ˙ R b ˙ N P ˙ R b F -1
cdlemk6.y Y = P ˙ R g ˙ Z ˙ R g b -1
cdlemk6.x X = ι z T | b T b I B R b R F R b R g z P = Y
cdlemk6.u U = g T if F = N g X
Assertion cdlemk19w K HL W H F T N T R F = R N U F = N

Proof

Step Hyp Ref Expression
1 cdlemk6.b B = Base K
2 cdlemk6.j ˙ = join K
3 cdlemk6.m ˙ = meet K
4 cdlemk6.o ˙ = oc K
5 cdlemk6.a A = Atoms K
6 cdlemk6.h H = LHyp K
7 cdlemk6.t T = LTrn K W
8 cdlemk6.r R = trL K W
9 cdlemk6.p P = ˙ W
10 cdlemk6.z Z = P ˙ R b ˙ N P ˙ R b F -1
11 cdlemk6.y Y = P ˙ R g ˙ Z ˙ R g b -1
12 cdlemk6.x X = ι z T | b T b I B R b R F R b R g z P = Y
13 cdlemk6.u U = g T if F = N g X
14 3simpb K HL W H F T N T R F = R N K HL W H R F = R N
15 simp2 K HL W H F T N T R F = R N F T N T
16 eqid K = K
17 16 4 5 6 lhpocnel K HL W H ˙ W A ¬ ˙ W K W
18 17 3ad2ant1 K HL W H F T N T R F = R N ˙ W A ¬ ˙ W K W
19 9 eleq1i P A ˙ W A
20 9 breq1i P K W ˙ W K W
21 20 notbii ¬ P K W ¬ ˙ W K W
22 19 21 anbi12i P A ¬ P K W ˙ W A ¬ ˙ W K W
23 18 22 sylibr K HL W H F T N T R F = R N P A ¬ P K W
24 1 16 2 3 5 6 7 8 10 11 12 13 cdlemk19u K HL W H R F = R N F T N T P A ¬ P K W U F = N
25 14 15 23 24 syl3anc K HL W H F T N T R F = R N U F = N