Metamath Proof Explorer


Theorem cdlemkfid2N

Description: Lemma for cdlemkfid3N . (Contributed by NM, 29-Jul-2013) (New usage is discouraged.)

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
Assertion cdlemkfid2N K HL W H F = N F T F I B b T R b R F P A ¬ P ˙ W Z = b 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 simp1r K HL W H F = N F T F I B b T R b R F P A ¬ P ˙ W F = N
11 10 fveq1d K HL W H F = N F T F I B b T R b R F P A ¬ P ˙ W F P = N P
12 11 oveq1d K HL W H F = N F T F I B b T R b R F P A ¬ P ˙ W F P ˙ R b F -1 = N P ˙ R b F -1
13 12 oveq2d K HL W H F = N F T F I B b T R b R F P A ¬ P ˙ W P ˙ R b ˙ F P ˙ R b F -1 = P ˙ R b ˙ N P ˙ R b F -1
14 1 2 3 4 5 6 7 8 cdlemkfid1N K HL W H F T F I B b T R b R F P A ¬ P ˙ W P ˙ R b ˙ F P ˙ R b F -1 = b P
15 14 3adant1r K HL W H F = N F T F I B b T R b R F P A ¬ P ˙ W P ˙ R b ˙ F P ˙ R b F -1 = b P
16 13 15 eqtr3d K HL W H F = N F T F I B b T R b R F P A ¬ P ˙ W P ˙ R b ˙ N P ˙ R b F -1 = b P
17 9 16 syl5eq K HL W H F = N F T F I B b T R b R F P A ¬ P ˙ W Z = b P