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
nonbool.2 B
nonbool.3 F = span A
nonbool.4 G = span B
nonbool.5 H = span A + B
Assertion nonbooli ¬ A G B F H F G H F H G

Proof

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