Metamath Proof Explorer


Theorem lcfrlem24

Description: Lemma for lcfr . (Contributed by NM, 24-Feb-2015)

Ref Expression
Hypotheses lcfrlem17.h ⊢ H = LHyp ⁡ K
lcfrlem17.o ⊢ ⊥ ˙ = ocH ⁡ K ⁡ W
lcfrlem17.u ⊢ U = DVecH ⁡ K ⁡ W
lcfrlem17.v ⊢ V = Base U
lcfrlem17.p ⊢ + ˙ = + U
lcfrlem17.z ⊢ 0 ˙ = 0 U
lcfrlem17.n ⊢ N = LSpan ⁡ U
lcfrlem17.a ⊢ A = LSAtoms ⁡ U
lcfrlem17.k ⊢ φ → K ∈ HL ∧ W ∈ H
lcfrlem17.x ⊢ φ → X ∈ V ∖ 0 ˙
lcfrlem17.y ⊢ φ → Y ∈ V ∖ 0 ˙
lcfrlem17.ne ⊢ φ → N ⁡ X ≠ N ⁡ Y
lcfrlem22.b ⊢ B = N ⁡ X Y ∩ ⊥ ˙ ⁡ X + ˙ Y
lcfrlem24.t ⊢ · ˙ = ⋅ U
lcfrlem24.s ⊢ S = Scalar ⁡ U
lcfrlem24.q ⊢ Q = 0 S
lcfrlem24.r ⊢ R = Base S
lcfrlem24.j ⊢ J = x ∈ V ∖ 0 ˙ ⟼ v ∈ V ⟼ ι k ∈ R | ∃ w ∈ ⊥ ˙ ⁡ x v = w + ˙ k · ˙ x
lcfrlem24.ib ⊢ φ → I ∈ B
lcfrlem24.l ⊢ L = LKer ⁡ U
Assertion lcfrlem24 ⊢ φ → ⊥ ˙ ⁡ X Y = L ⁡ J ⁡ X ∩ L ⁡ J ⁡ Y

Proof

Step Hyp Ref Expression
1 lcfrlem17.h ⊢ H = LHyp ⁡ K
2 lcfrlem17.o ⊢ ⊥ ˙ = ocH ⁡ K ⁡ W
3 lcfrlem17.u ⊢ U = DVecH ⁡ K ⁡ W
4 lcfrlem17.v ⊢ V = Base U
5 lcfrlem17.p ⊢ + ˙ = + U
6 lcfrlem17.z ⊢ 0 ˙ = 0 U
7 lcfrlem17.n ⊢ N = LSpan ⁡ U
8 lcfrlem17.a ⊢ A = LSAtoms ⁡ U
9 lcfrlem17.k ⊢ φ → K ∈ HL ∧ W ∈ H
10 lcfrlem17.x ⊢ φ → X ∈ V ∖ 0 ˙
11 lcfrlem17.y ⊢ φ → Y ∈ V ∖ 0 ˙
12 lcfrlem17.ne ⊢ φ → N ⁡ X ≠ N ⁡ Y
13 lcfrlem22.b ⊢ B = N ⁡ X Y ∩ ⊥ ˙ ⁡ X + ˙ Y
14 lcfrlem24.t ⊢ · ˙ = ⋅ U
15 lcfrlem24.s ⊢ S = Scalar ⁡ U
16 lcfrlem24.q ⊢ Q = 0 S
17 lcfrlem24.r ⊢ R = Base S
18 lcfrlem24.j ⊢ J = x ∈ V ∖ 0 ˙ ⟼ v ∈ V ⟼ ι k ∈ R | ∃ w ∈ ⊥ ˙ ⁡ x v = w + ˙ k · ˙ x
19 lcfrlem24.ib ⊢ φ → I ∈ B
20 lcfrlem24.l ⊢ L = LKer ⁡ U
21 1 2 3 4 5 6 7 8 9 10 11 12 lcfrlem18 ⊢ φ → ⊥ ˙ ⁡ X Y = ⊥ ˙ ⁡ X ∩ ⊥ ˙ ⁡ Y
22 eqid ⊢ LFnl ⁡ U = LFnl ⁡ U
23 eqid ⊢ LDual ⁡ U = LDual ⁡ U
24 eqid ⊢ 0 LDual ⁡ U = 0 LDual ⁡ U
25 eqid ⊢ f ∈ LFnl ⁡ U | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f = f ∈ LFnl ⁡ U | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f
26 1 2 3 4 5 14 15 17 6 22 20 23 24 25 18 9 10 lcfrlem11 ⊢ φ → L ⁡ J ⁡ X = ⊥ ˙ ⁡ X
27 1 2 3 4 5 14 15 17 6 22 20 23 24 25 18 9 11 lcfrlem11 ⊢ φ → L ⁡ J ⁡ Y = ⊥ ˙ ⁡ Y
28 26 27 ineq12d ⊢ φ → L ⁡ J ⁡ X ∩ L ⁡ J ⁡ Y = ⊥ ˙ ⁡ X ∩ ⊥ ˙ ⁡ Y
29 21 28 eqtr4d ⊢ φ → ⊥ ˙ ⁡ X Y = L ⁡ J ⁡ X ∩ L ⁡ J ⁡ Y