Metamath Proof Explorer


Theorem cdlemg4a

Description: TODO: FIX COMMENT If fg(p) = p, then tr f = tr g. (Contributed by NM, 23-Apr-2013)

Ref Expression
Hypotheses cdlemg4.l ˙ = K
cdlemg4.a A = Atoms K
cdlemg4.h H = LHyp K
cdlemg4.t T = LTrn K W
cdlemg4.r R = trL K W
Assertion cdlemg4a K HL W H P A ¬ P ˙ W F T G T F G P = P R F = R G

Proof

Step Hyp Ref Expression
1 cdlemg4.l ˙ = K
2 cdlemg4.a A = Atoms K
3 cdlemg4.h H = LHyp K
4 cdlemg4.t T = LTrn K W
5 cdlemg4.r R = trL K W
6 simp3 K HL W H P A ¬ P ˙ W F T G T F G P = P F G P = P
7 6 oveq2d K HL W H P A ¬ P ˙ W F T G T F G P = P G P join K F G P = G P join K P
8 simp1l K HL W H P A ¬ P ˙ W F T G T F G P = P K HL
9 simp1 K HL W H P A ¬ P ˙ W F T G T F G P = P K HL W H
10 simp23 K HL W H P A ¬ P ˙ W F T G T F G P = P G T
11 simp21 K HL W H P A ¬ P ˙ W F T G T F G P = P P A ¬ P ˙ W
12 1 2 3 4 ltrnel K HL W H G T P A ¬ P ˙ W G P A ¬ G P ˙ W
13 12 simpld K HL W H G T P A ¬ P ˙ W G P A
14 9 10 11 13 syl3anc K HL W H P A ¬ P ˙ W F T G T F G P = P G P A
15 simp21l K HL W H P A ¬ P ˙ W F T G T F G P = P P A
16 eqid join K = join K
17 16 2 hlatjcom K HL G P A P A G P join K P = P join K G P
18 8 14 15 17 syl3anc K HL W H P A ¬ P ˙ W F T G T F G P = P G P join K P = P join K G P
19 7 18 eqtrd K HL W H P A ¬ P ˙ W F T G T F G P = P G P join K F G P = P join K G P
20 19 oveq1d K HL W H P A ¬ P ˙ W F T G T F G P = P G P join K F G P meet K W = P join K G P meet K W
21 simp22 K HL W H P A ¬ P ˙ W F T G T F G P = P F T
22 9 10 11 12 syl3anc K HL W H P A ¬ P ˙ W F T G T F G P = P G P A ¬ G P ˙ W
23 eqid meet K = meet K
24 1 16 23 2 3 4 5 trlval2 K HL W H F T G P A ¬ G P ˙ W R F = G P join K F G P meet K W
25 9 21 22 24 syl3anc K HL W H P A ¬ P ˙ W F T G T F G P = P R F = G P join K F G P meet K W
26 1 16 23 2 3 4 5 trlval2 K HL W H G T P A ¬ P ˙ W R G = P join K G P meet K W
27 9 10 11 26 syl3anc K HL W H P A ¬ P ˙ W F T G T F G P = P R G = P join K G P meet K W
28 20 25 27 3eqtr4d K HL W H P A ¬ P ˙ W F T G T F G P = P R F = R G