Metamath Proof Explorer


Theorem nonbooli

Description: A Hilbert lattice with two or more dimensions fails the distributive law and therefore cannot be a Boolean algebra. This counterexample demonstrates a condition where ( ( H i^i F ) vH ( H i^i G ) ) = 0H but ( H i^i ( F vH G ) ) =/= 0H . The antecedent specifies that the vectors A and B are nonzero and non-colinear. The last three hypotheses assign one-dimensional subspaces to F , G , and H . (Contributed by NM, 1-Nov-2005) (New usage is discouraged.)

Ref Expression
Hypotheses nonbool.1 ⊢ A ∈ ℋ
nonbool.2 ⊢ B ∈ ℋ
nonbool.3 ⊢ F = span ⁡ A
nonbool.4 ⊢ G = span ⁡ B
nonbool.5 ⊢ H = span ⁡ A + ℎ B
Assertion nonbooli ⊢ ¬ A ∈ G ∨ B ∈ F → H ∩ F ∨ ℋ G ≠ H ∩ F ∨ ℋ H ∩ G

Proof

Step Hyp Ref Expression
1 nonbool.1 ⊢ A ∈ ℋ
2 nonbool.2 ⊢ B ∈ ℋ
3 nonbool.3 ⊢ F = span ⁡ A
4 nonbool.4 ⊢ G = span ⁡ B
5 nonbool.5 ⊢ H = span ⁡ A + ℎ B
6 1 2 hvaddcli ⊢ A + ℎ B ∈ ℋ
7 spansnid ⊢ A + ℎ B ∈ ℋ → A + ℎ B ∈ span ⁡ A + ℎ B
8 6 7 ax-mp ⊢ A + ℎ B ∈ span ⁡ A + ℎ B
9 8 5 eleqtrri ⊢ A + ℎ B ∈ H
10 1 spansnchi ⊢ span ⁡ A ∈ C ℋ
11 10 chshii ⊢ span ⁡ A ∈ S ℋ
12 3 11 eqeltri ⊢ F ∈ S ℋ
13 2 spansnchi ⊢ span ⁡ B ∈ C ℋ
14 13 chshii ⊢ span ⁡ B ∈ S ℋ
15 4 14 eqeltri ⊢ G ∈ S ℋ
16 12 15 shsleji ⊢ F + ℋ G ⊆ F ∨ ℋ G
17 spansnid ⊢ A ∈ ℋ → A ∈ span ⁡ A
18 1 17 ax-mp ⊢ A ∈ span ⁡ A
19 18 3 eleqtrri ⊢ A ∈ F
20 spansnid ⊢ B ∈ ℋ → B ∈ span ⁡ B
21 2 20 ax-mp ⊢ B ∈ span ⁡ B
22 21 4 eleqtrri ⊢ B ∈ G
23 12 15 shsvai ⊢ A ∈ F ∧ B ∈ G → A + ℎ B ∈ F + ℋ G
24 19 22 23 mp2an ⊢ A + ℎ B ∈ F + ℋ G
25 16 24 sselii ⊢ A + ℎ B ∈ F ∨ ℋ G
26 elin ⊢ A + ℎ B ∈ H ∩ F ∨ ℋ G ↔ A + ℎ B ∈ H ∧ A + ℎ B ∈ F ∨ ℋ G
27 9 25 26 mpbir2an ⊢ A + ℎ B ∈ H ∩ F ∨ ℋ G
28 eleq2 ⊢ H ∩ F ∨ ℋ G = 0 ℋ → A + ℎ B ∈ H ∩ F ∨ ℋ G ↔ A + ℎ B ∈ 0 ℋ
29 27 28 mpbii ⊢ H ∩ F ∨ ℋ G = 0 ℋ → A + ℎ B ∈ 0 ℋ
30 elch0 ⊢ A + ℎ B ∈ 0 ℋ ↔ A + ℎ B = 0 ℎ
31 29 30 sylib ⊢ H ∩ F ∨ ℋ G = 0 ℋ → A + ℎ B = 0 ℎ
32 ch0 ⊢ span ⁡ A ∈ C ℋ → 0 ℎ ∈ span ⁡ A
33 10 32 ax-mp ⊢ 0 ℎ ∈ span ⁡ A
34 31 33 eqeltrdi ⊢ H ∩ F ∨ ℋ G = 0 ℋ → A + ℎ B ∈ span ⁡ A
35 3 eleq2i ⊢ B ∈ F ↔ B ∈ span ⁡ A
36 sumspansn ⊢ A ∈ ℋ ∧ B ∈ ℋ → A + ℎ B ∈ span ⁡ A ↔ B ∈ span ⁡ A
37 1 2 36 mp2an ⊢ A + ℎ B ∈ span ⁡ A ↔ B ∈ span ⁡ A
38 35 37 bitr4i ⊢ B ∈ F ↔ A + ℎ B ∈ span ⁡ A
39 34 38 sylibr ⊢ H ∩ F ∨ ℋ G = 0 ℋ → B ∈ F
40 39 con3i ⊢ ¬ B ∈ F → ¬ H ∩ F ∨ ℋ G = 0 ℋ
41 40 adantl ⊢ ¬ A ∈ G ∧ ¬ B ∈ F → ¬ H ∩ F ∨ ℋ G = 0 ℋ
42 5 3 ineq12i ⊢ H ∩ F = span ⁡ A + ℎ B ∩ span ⁡ A
43 6 1 spansnm0i ⊢ ¬ A + ℎ B ∈ span ⁡ A → span ⁡ A + ℎ B ∩ span ⁡ A = 0 ℋ
44 38 43 sylnbi ⊢ ¬ B ∈ F → span ⁡ A + ℎ B ∩ span ⁡ A = 0 ℋ
45 42 44 eqtrid ⊢ ¬ B ∈ F → H ∩ F = 0 ℋ
46 5 4 ineq12i ⊢ H ∩ G = span ⁡ A + ℎ B ∩ span ⁡ B
47 sumspansn ⊢ B ∈ ℋ ∧ A ∈ ℋ → B + ℎ A ∈ span ⁡ B ↔ A ∈ span ⁡ B
48 2 1 47 mp2an ⊢ B + ℎ A ∈ span ⁡ B ↔ A ∈ span ⁡ B
49 1 2 hvcomi ⊢ A + ℎ B = B + ℎ A
50 49 eleq1i ⊢ A + ℎ B ∈ span ⁡ B ↔ B + ℎ A ∈ span ⁡ B
51 4 eleq2i ⊢ A ∈ G ↔ A ∈ span ⁡ B
52 48 50 51 3bitr4ri ⊢ A ∈ G ↔ A + ℎ B ∈ span ⁡ B
53 6 2 spansnm0i ⊢ ¬ A + ℎ B ∈ span ⁡ B → span ⁡ A + ℎ B ∩ span ⁡ B = 0 ℋ
54 52 53 sylnbi ⊢ ¬ A ∈ G → span ⁡ A + ℎ B ∩ span ⁡ B = 0 ℋ
55 46 54 eqtrid ⊢ ¬ A ∈ G → H ∩ G = 0 ℋ
56 45 55 oveqan12rd ⊢ ¬ A ∈ G ∧ ¬ B ∈ F → H ∩ F ∨ ℋ H ∩ G = 0 ℋ ∨ ℋ 0 ℋ
57 h0elch ⊢ 0 ℋ ∈ C ℋ
58 57 chj0i ⊢ 0 ℋ ∨ ℋ 0 ℋ = 0 ℋ
59 56 58 eqtrdi ⊢ ¬ A ∈ G ∧ ¬ B ∈ F → H ∩ F ∨ ℋ H ∩ G = 0 ℋ
60 eqeq2 ⊢ H ∩ F ∨ ℋ H ∩ G = 0 ℋ → H ∩ F ∨ ℋ G = H ∩ F ∨ ℋ H ∩ G ↔ H ∩ F ∨ ℋ G = 0 ℋ
61 60 notbid ⊢ H ∩ F ∨ ℋ H ∩ G = 0 ℋ → ¬ H ∩ F ∨ ℋ G = H ∩ F ∨ ℋ H ∩ G ↔ ¬ H ∩ F ∨ ℋ G = 0 ℋ
62 61 biimparc ⊢ ¬ H ∩ F ∨ ℋ G = 0 ℋ ∧ H ∩ F ∨ ℋ H ∩ G = 0 ℋ → ¬ H ∩ F ∨ ℋ G = H ∩ F ∨ ℋ H ∩ G
63 41 59 62 syl2anc ⊢ ¬ A ∈ G ∧ ¬ B ∈ F → ¬ H ∩ F ∨ ℋ G = H ∩ F ∨ ℋ H ∩ G
64 ioran ⊢ ¬ A ∈ G ∨ B ∈ F ↔ ¬ A ∈ G ∧ ¬ B ∈ F
65 df-ne ⊢ H ∩ F ∨ ℋ G ≠ H ∩ F ∨ ℋ H ∩ G ↔ ¬ H ∩ F ∨ ℋ G = H ∩ F ∨ ℋ H ∩ G
66 63 64 65 3imtr4i ⊢ ¬ A ∈ G ∨ B ∈ F → H ∩ F ∨ ℋ G ≠ H ∩ F ∨ ℋ H ∩ G