src/HOL/ex/PresburgerEx.thy
author wenzelm
Sun Nov 02 18:21:45 2014 +0100 (2014-11-02)
changeset 58889 5b7a9633cfa8
parent 29705 a1ecdd8cf81c
child 61343 5b5656a63bd6
permissions -rw-r--r--
modernized header uniformly as section;
     1 (*  Title:      HOL/ex/PresburgerEx.thy
     2     Author:     Amine Chaieb, TU Muenchen
     3 *)
     4 
     5 section {* Some examples for Presburger Arithmetic *}
     6 
     7 theory PresburgerEx
     8 imports Presburger
     9 begin
    10 
    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
    16 
    17 theorem "(\<forall>(y::int). 3 dvd y) ==> \<forall>(x::int). b < x --> a \<le> x"
    18   by presburger
    19 
    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
    23 
    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
    27 
    28 theorem "\<forall>(x::nat). \<exists>(y::nat). (0::nat) \<le> 5 --> y = 5 + x "
    29   by presburger
    30 
    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
    34 
    35 theorem "\<exists>(x::int). 0 < x"
    36   by presburger
    37 
    38 theorem "\<forall>(x::int) y. x < y --> 2 * x + 1 < 2 * y"
    39   by presburger
    40  
    41 theorem "\<forall>(x::int) y. 2 * x + 1 \<noteq> 2 * y"
    42   by presburger
    43  
    44 theorem "\<exists>(x::int) y. 0 < x  & 0 \<le> y  & 3 * x - 5 * y = 1"
    45   by presburger
    46 
    47 theorem "~ (\<exists>(x::int) (y::int) (z::int). 4*x + (-6::int)*y = 1)"
    48   by presburger
    49 
    50 theorem "\<forall>(x::int). b < x --> a \<le> x"
    51   apply (presburger elim)
    52   oops
    53 
    54 theorem "~ (\<exists>(x::int). False)"
    55   by presburger
    56 
    57 theorem "\<forall>(x::int). (a::int) < 3 * x --> b < 3 * x"
    58   apply (presburger elim)
    59   oops
    60 
    61 theorem "\<forall>(x::int). (2 dvd x) --> (\<exists>(y::int). x = 2*y)"
    62   by presburger 
    63 
    64 theorem "\<forall>(x::int). (2 dvd x) --> (\<exists>(y::int). x = 2*y)"
    65   by presburger 
    66 
    67 theorem "\<forall>(x::int). (2 dvd x) = (\<exists>(y::int). x = 2*y)"
    68   by presburger 
    69 
    70 theorem "\<forall>(x::int). ((2 dvd x) = (\<forall>(y::int). x \<noteq> 2*y + 1))"
    71   by presburger 
    72 
    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
    78  
    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
    81 
    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
    84 
    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
    87 
    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
    90 
    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
    94 
    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.
   100 
   101 Warning: it takes (in 2006) over 4.2 minutes! *}
   102 
   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
   108 
   109 end