Metamath Proof Explorer


Theorem dihjat5N

Description: Transfer lattice join with atom to subspace sum. (Contributed by NM, 25-Apr-2015) (New usage is discouraged.)

Ref Expression
Hypotheses dihjat5.b B = Base K
dihjat5.h H = LHyp K
dihjat5.j ˙ = join K
dihjat5.a A = Atoms K
dihjat5.u U = DVecH K W
dihjat5.s ˙ = LSSum U
dihjat5.i I = DIsoH K W
dihjat5.k φ K HL W H
dihjat5.x φ X B
dihjat5.p φ P A
Assertion dihjat5N φ X ˙ P = I -1 I X ˙ I P

Proof

Step Hyp Ref Expression
1 dihjat5.b B = Base K
2 dihjat5.h H = LHyp K
3 dihjat5.j ˙ = join K
4 dihjat5.a A = Atoms K
5 dihjat5.u U = DVecH K W
6 dihjat5.s ˙ = LSSum U
7 dihjat5.i I = DIsoH K W
8 dihjat5.k φ K HL W H
9 dihjat5.x φ X B
10 dihjat5.p φ P A
11 1 2 3 4 5 6 7 8 9 10 dihjat3 φ I X ˙ P = I X ˙ I P
12 eqid LSAtoms U = LSAtoms U
13 1 2 7 dihcl K HL W H X B I X ran I
14 8 9 13 syl2anc φ I X ran I
15 4 2 5 7 12 dihatlat K HL W H P A I P LSAtoms U
16 8 10 15 syl2anc φ I P LSAtoms U
17 2 7 5 6 12 8 14 16 dihsmatrn φ I X ˙ I P ran I
18 2 7 dihcnvid2 K HL W H I X ˙ I P ran I I I -1 I X ˙ I P = I X ˙ I P
19 8 17 18 syl2anc φ I I -1 I X ˙ I P = I X ˙ I P
20 11 19 eqtr4d φ I X ˙ P = I I -1 I X ˙ I P
21 8 simpld φ K HL
22 21 hllatd φ K Lat
23 1 4 atbase P A P B
24 10 23 syl φ P B
25 1 3 latjcl K Lat X B P B X ˙ P B
26 22 9 24 25 syl3anc φ X ˙ P B
27 1 2 7 dihcnvcl K HL W H I X ˙ I P ran I I -1 I X ˙ I P B
28 8 17 27 syl2anc φ I -1 I X ˙ I P B
29 1 2 7 dih11 K HL W H X ˙ P B I -1 I X ˙ I P B I X ˙ P = I I -1 I X ˙ I P X ˙ P = I -1 I X ˙ I P
30 8 26 28 29 syl3anc φ I X ˙ P = I I -1 I X ˙ I P X ˙ P = I -1 I X ˙ I P
31 20 30 mpbid φ X ˙ P = I -1 I X ˙ I P