Metamath Proof Explorer


Theorem djh01

Description: Closed subspace join with zero. (Contributed by NM, 9-Aug-2014)

Ref Expression
Hypotheses djh01.h ⊢ H = LHyp ⁡ K
djh01.u ⊢ U = DVecH ⁡ K ⁡ W
djh01.o ⊢ 0 ˙ = 0 U
djh01.i ⊢ I = DIsoH ⁡ K ⁡ W
djh01.j ⊢ ∨ ˙ = joinH ⁡ K ⁡ W
djh01.k ⊢ φ → K ∈ HL ∧ W ∈ H
djh01.x ⊢ φ → X ∈ ran ⁡ I
Assertion djh01 ⊢ φ → X ∨ ˙ 0 ˙ = X

Proof

Step Hyp Ref Expression
1 djh01.h ⊢ H = LHyp ⁡ K
2 djh01.u ⊢ U = DVecH ⁡ K ⁡ W
3 djh01.o ⊢ 0 ˙ = 0 U
4 djh01.i ⊢ I = DIsoH ⁡ K ⁡ W
5 djh01.j ⊢ ∨ ˙ = joinH ⁡ K ⁡ W
6 djh01.k ⊢ φ → K ∈ HL ∧ W ∈ H
7 djh01.x ⊢ φ → X ∈ ran ⁡ I
8 eqid ⊢ join ⁡ K = join ⁡ K
9 1 4 2 3 dih0rn ⊢ K ∈ HL ∧ W ∈ H → 0 ˙ ∈ ran ⁡ I
10 6 9 syl ⊢ φ → 0 ˙ ∈ ran ⁡ I
11 8 1 4 5 6 7 10 djhjlj ⊢ φ → X ∨ ˙ 0 ˙ = I ⁡ I -1 ⁡ X join ⁡ K I -1 ⁡ 0 ˙
12 eqid ⊢ 0. ⁡ K = 0. ⁡ K
13 1 12 4 2 3 dih0cnv ⊢ K ∈ HL ∧ W ∈ H → I -1 ⁡ 0 ˙ = 0. ⁡ K
14 6 13 syl ⊢ φ → I -1 ⁡ 0 ˙ = 0. ⁡ K
15 14 oveq2d ⊢ φ → I -1 ⁡ X join ⁡ K I -1 ⁡ 0 ˙ = I -1 ⁡ X join ⁡ K 0. ⁡ K
16 6 simpld ⊢ φ → K ∈ HL
17 hlol ⊢ K ∈ HL → K ∈ OL
18 16 17 syl ⊢ φ → K ∈ OL
19 eqid ⊢ Base K = Base K
20 19 1 4 dihcnvcl ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ ran ⁡ I → I -1 ⁡ X ∈ Base K
21 6 7 20 syl2anc ⊢ φ → I -1 ⁡ X ∈ Base K
22 19 8 12 olj01 ⊢ K ∈ OL ∧ I -1 ⁡ X ∈ Base K → I -1 ⁡ X join ⁡ K 0. ⁡ K = I -1 ⁡ X
23 18 21 22 syl2anc ⊢ φ → I -1 ⁡ X join ⁡ K 0. ⁡ K = I -1 ⁡ X
24 15 23 eqtrd ⊢ φ → I -1 ⁡ X join ⁡ K I -1 ⁡ 0 ˙ = I -1 ⁡ X
25 24 fveq2d ⊢ φ → I ⁡ I -1 ⁡ X join ⁡ K I -1 ⁡ 0 ˙ = I ⁡ I -1 ⁡ X
26 1 4 dihcnvid2 ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ ran ⁡ I → I ⁡ I -1 ⁡ X = X
27 6 7 26 syl2anc ⊢ φ → I ⁡ I -1 ⁡ X = X
28 11 25 27 3eqtrd ⊢ φ → X ∨ ˙ 0 ˙ = X