Metamath Proof Explorer


Theorem cdlemkyuu

Description: cdlemkyu with some hypotheses eliminated. TODO: Clean all this up. (Contributed by NM, 21-Jul-2013)

Ref Expression
Hypotheses cdlemk5.b B = Base K
cdlemk5.l ˙ = K
cdlemk5.j ˙ = join K
cdlemk5.m ˙ = meet K
cdlemk5.a A = Atoms K
cdlemk5.h H = LHyp K
cdlemk5.t T = LTrn K W
cdlemk5.r R = trL K W
cdlemk5.z Z = P ˙ R b ˙ N P ˙ R b F -1
cdlemk5.y Y = P ˙ R g ˙ Z ˙ R g b -1
cdlemk5c.s S = f T ι i T | i P = P ˙ R f ˙ N P ˙ R f F -1
cdlemk5a.u2 C = e T ι j T | j P = P ˙ R e ˙ S b P ˙ R e b -1
Assertion cdlemkyuu K HL W H F T F I B G T G I B N T P A ¬ P ˙ W R F = R N b T b I B R b R F R b R G G / g Y = C G P

Proof

Step Hyp Ref Expression
1 cdlemk5.b B = Base K
2 cdlemk5.l ˙ = K
3 cdlemk5.j ˙ = join K
4 cdlemk5.m ˙ = meet K
5 cdlemk5.a A = Atoms K
6 cdlemk5.h H = LHyp K
7 cdlemk5.t T = LTrn K W
8 cdlemk5.r R = trL K W
9 cdlemk5.z Z = P ˙ R b ˙ N P ˙ R b F -1
10 cdlemk5.y Y = P ˙ R g ˙ Z ˙ R g b -1
11 cdlemk5c.s S = f T ι i T | i P = P ˙ R f ˙ N P ˙ R f F -1
12 cdlemk5a.u2 C = e T ι j T | j P = P ˙ R e ˙ S b P ˙ R e b -1
13 eqid d T , e T ι j T | j P = P ˙ R e ˙ S d P ˙ R e d -1 = d T , e T ι j T | j P = P ˙ R e ˙ S d P ˙ R e d -1
14 eqid S b = S b
15 1 2 3 4 5 6 7 8 9 10 11 13 14 12 cdlemkyu K HL W H F T F I B G T G I B N T P A ¬ P ˙ W R F = R N b T b I B R b R F R b R G G / g Y = C G P