Metamath Proof Explorer


Theorem cdlemkuv2-3N

Description: Part of proof of Lemma K of Crawley p. 118. Line 16 on p. 119 for i = 1, where sigma_2 (p) is Y , f_1 is D , and k_1 is O . (Contributed by NM, 6-Jul-2013) (New usage is discouraged.)

Ref Expression
Hypotheses cdlemk3.b B = Base K
cdlemk3.l ˙ = K
cdlemk3.j ˙ = join K
cdlemk3.m ˙ = meet K
cdlemk3.a A = Atoms K
cdlemk3.h H = LHyp K
cdlemk3.t T = LTrn K W
cdlemk3.r R = trL K W
cdlemk3.s S = f T ι i T | i P = P ˙ R f ˙ N P ˙ R f F -1
cdlemk3.u1 Y = d T , e T ι j T | j P = P ˙ R e ˙ S d P ˙ R e d -1
Assertion cdlemkuv2-3N K HL W H R F = R N G T F T D T N T R D R F R D R G F I B G I B D I B P A ¬ P ˙ W D Y G P = P ˙ R G ˙ S D P ˙ R G D -1

Proof

Step Hyp Ref Expression
1 cdlemk3.b B = Base K
2 cdlemk3.l ˙ = K
3 cdlemk3.j ˙ = join K
4 cdlemk3.m ˙ = meet K
5 cdlemk3.a A = Atoms K
6 cdlemk3.h H = LHyp K
7 cdlemk3.t T = LTrn K W
8 cdlemk3.r R = trL K W
9 cdlemk3.s S = f T ι i T | i P = P ˙ R f ˙ N P ˙ R f F -1
10 cdlemk3.u1 Y = d T , e T ι j T | j P = P ˙ R e ˙ S d P ˙ R e d -1
11 simp22 K HL W H R F = R N G T F T D T N T R D R F R D R G F I B G I B D I B P A ¬ P ˙ W D T
12 simp13 K HL W H R F = R N G T F T D T N T R D R F R D R G F I B G I B D I B P A ¬ P ˙ W G T
13 eqid S D = S D
14 eqid e T ι j T | j P = P ˙ R e ˙ S D P ˙ R e D -1 = e T ι j T | j P = P ˙ R e ˙ S D P ˙ R e D -1
15 1 2 3 4 5 6 7 8 9 10 13 14 cdlemkuu D T G T D Y G = e T ι j T | j P = P ˙ R e ˙ S D P ˙ R e D -1 G
16 11 12 15 syl2anc K HL W H R F = R N G T F T D T N T R D R F R D R G F I B G I B D I B P A ¬ P ˙ W D Y G = e T ι j T | j P = P ˙ R e ˙ S D P ˙ R e D -1 G
17 16 fveq1d K HL W H R F = R N G T F T D T N T R D R F R D R G F I B G I B D I B P A ¬ P ˙ W D Y G P = e T ι j T | j P = P ˙ R e ˙ S D P ˙ R e D -1 G P
18 1 2 3 4 5 6 7 8 9 13 14 cdlemkuv2 K HL W H R F = R N G T F T D T N T R D R F R D R G F I B G I B D I B P A ¬ P ˙ W e T ι j T | j P = P ˙ R e ˙ S D P ˙ R e D -1 G P = P ˙ R G ˙ S D P ˙ R G D -1
19 17 18 eqtrd K HL W H R F = R N G T F T D T N T R D R F R D R G F I B G I B D I B P A ¬ P ˙ W D Y G P = P ˙ R G ˙ S D P ˙ R G D -1