src/HOL/ex/PresburgerEx.thy
 author bulwahn Fri Oct 21 11:17:14 2011 +0200 (2011-10-21) changeset 45231 d85a2fdc586c parent 29705 a1ecdd8cf81c child 58889 5b7a9633cfa8 permissions -rw-r--r--
replacing code_inline by code_unfold, removing obsolete code_unfold, code_inline del now that the ancient code generator is removed
1 (*  Title:      HOL/ex/PresburgerEx.thy
2     Author:     Amine Chaieb, TU Muenchen
3 *)
5 header {* Some examples for Presburger Arithmetic *}
7 theory PresburgerEx
8 imports Presburger
9 begin
11 lemma "\<And>m n ja ia. \<lbrakk>\<not> m \<le> j; \<not> (n::nat) \<le> i; (e::nat) \<noteq> 0; Suc j \<le> ja\<rbrakk> \<Longrightarrow> \<exists>m. \<forall>ja ia. m \<le> ja \<longrightarrow> (if j = ja \<and> i = ia then e else 0) = 0" by presburger
12 lemma "(0::nat) < emBits mod 8 \<Longrightarrow> 8 + emBits div 8 * 8 - emBits = 8 - emBits mod 8"
13 by presburger
14 lemma "(0::nat) < emBits mod 8 \<Longrightarrow> 8 + emBits div 8 * 8 - emBits = 8 - emBits mod 8"
15 by presburger
17 theorem "(\<forall>(y::int). 3 dvd y) ==> \<forall>(x::int). b < x --> a \<le> x"
18   by presburger
20 theorem "!! (y::int) (z::int) (n::int). 3 dvd z ==> 2 dvd (y::int) ==>
21   (\<exists>(x::int).  2*x =  y) & (\<exists>(k::int). 3*k = z)"
22   by presburger
24 theorem "!! (y::int) (z::int) n. Suc(n::nat) < 6 ==>  3 dvd z ==>
25   2 dvd (y::int) ==> (\<exists>(x::int).  2*x =  y) & (\<exists>(k::int). 3*k = z)"
26   by presburger
28 theorem "\<forall>(x::nat). \<exists>(y::nat). (0::nat) \<le> 5 --> y = 5 + x "
29   by presburger
31 text{*Slow: about 7 seconds on a 1.6GHz machine.*}
32 theorem "\<forall>(x::nat). \<exists>(y::nat). y = 5 + x | x div 6 + 1= 2"
33   by presburger
35 theorem "\<exists>(x::int). 0 < x"
36   by presburger
38 theorem "\<forall>(x::int) y. x < y --> 2 * x + 1 < 2 * y"
39   by presburger
41 theorem "\<forall>(x::int) y. 2 * x + 1 \<noteq> 2 * y"
42   by presburger
44 theorem "\<exists>(x::int) y. 0 < x  & 0 \<le> y  & 3 * x - 5 * y = 1"
45   by presburger
47 theorem "~ (\<exists>(x::int) (y::int) (z::int). 4*x + (-6::int)*y = 1)"
48   by presburger
50 theorem "\<forall>(x::int). b < x --> a \<le> x"
51   apply (presburger elim)
52   oops
54 theorem "~ (\<exists>(x::int). False)"
55   by presburger
57 theorem "\<forall>(x::int). (a::int) < 3 * x --> b < 3 * x"
58   apply (presburger elim)
59   oops
61 theorem "\<forall>(x::int). (2 dvd x) --> (\<exists>(y::int). x = 2*y)"
62   by presburger
64 theorem "\<forall>(x::int). (2 dvd x) --> (\<exists>(y::int). x = 2*y)"
65   by presburger
67 theorem "\<forall>(x::int). (2 dvd x) = (\<exists>(y::int). x = 2*y)"
68   by presburger
70 theorem "\<forall>(x::int). ((2 dvd x) = (\<forall>(y::int). x \<noteq> 2*y + 1))"
71   by presburger
73 theorem "~ (\<forall>(x::int).
74             ((2 dvd x) = (\<forall>(y::int). x \<noteq> 2*y+1) |
75              (\<exists>(q::int) (u::int) i. 3*i + 2*q - u < 17)
76              --> 0 < x | ((~ 3 dvd x) &(x + 8 = 0))))"
77   by presburger
79 theorem "~ (\<forall>(i::int). 4 \<le> i --> (\<exists>x y. 0 \<le> x & 0 \<le> y & 3 * x + 5 * y = i))"
80   by presburger
82 theorem "\<forall>(i::int). 8 \<le> i --> (\<exists>x y. 0 \<le> x & 0 \<le> y & 3 * x + 5 * y = i)"
83   by presburger
85 theorem "\<exists>(j::int). \<forall>i. j \<le> i --> (\<exists>x y. 0 \<le> x & 0 \<le> y & 3 * x + 5 * y = i)"
86   by presburger
88 theorem "~ (\<forall>j (i::int). j \<le> i --> (\<exists>x y. 0 \<le> x & 0 \<le> y & 3 * x + 5 * y = i))"
89   by presburger
91 text{*Slow: about 5 seconds on a 1.6GHz machine.*}
92 theorem "(\<exists>m::nat. n = 2 * m) --> (n + 1) div 2 = n div 2"
93   by presburger
95 text{* This following theorem proves that all solutions to the
96 recurrence relation \$x_{i+2} = |x_{i+1}| - x_i\$ are periodic with
97 period 9.  The example was brought to our attention by John
98 Harrison. It does does not require Presburger arithmetic but merely
99 quantifier-free linear arithmetic and holds for the rationals as well.
101 Warning: it takes (in 2006) over 4.2 minutes! *}
103 lemma "\<lbrakk> x3 = abs x2 - x1; x4 = abs x3 - x2; x5 = abs x4 - x3;
104          x6 = abs x5 - x4; x7 = abs x6 - x5; x8 = abs x7 - x6;
105          x9 = abs x8 - x7; x10 = abs x9 - x8; x11 = abs x10 - x9 \<rbrakk>
106  \<Longrightarrow> x1 = x10 & x2 = (x11::int)"
107 by arith
109 end