src/HOL/ex/PresburgerEx.thy

Merged in license change from Isabelle2004

(*  Title:      HOL/ex/PresburgerEx.thy
    ID:         $Id$
    Author:     Amine Chaieb, TU Muenchen

Some examples for Presburger Arithmetic
*)

theory PresburgerEx = Main:

theorem "(\<forall>(y::int). 3 dvd y) ==> \<forall>(x::int). b < x --> a \<le> x"
  by presburger

theorem "!! (y::int) (z::int) (n::int). 3 dvd z ==> 2 dvd (y::int) ==>
  (\<exists>(x::int).  2*x =  y) & (\<exists>(k::int). 3*k = z)"
  by presburger

theorem "!! (y::int) (z::int) n. Suc(n::nat) < 6 ==>  3 dvd z ==>
  2 dvd (y::int) ==> (\<exists>(x::int).  2*x =  y) & (\<exists>(k::int). 3*k = z)"
  by presburger

theorem "\<forall>(x::nat). \<exists>(y::nat). (0::nat) \<le> 5 --> y = 5 + x ";
  by presburger

text{*Very slow: about 55 seconds on a 1.8GHz machine.*}
theorem "\<forall>(x::nat). \<exists>(y::nat). y = 5 + x | x div 6 + 1= 2";
  by presburger

theorem "\<exists>(x::int). 0 < x" by presburger

theorem "\<forall>(x::int) y. x < y --> 2 * x + 1 < 2 * y" by presburger
 
theorem "\<forall>(x::int) y. 2 * x + 1 \<noteq> 2 * y" by presburger
 
theorem
   "\<exists>(x::int) y. 0 < x  & 0 \<le> y  & 3 * x - 5 * y = 1" by presburger

theorem "~ (\<exists>(x::int) (y::int) (z::int). 4*x + (-6::int)*y = 1)"
  by presburger

theorem "\<forall>(x::int). b < x --> a \<le> x"
  apply (presburger (no_quantify))
  oops

theorem "\<forall>(x::int). b < x --> a \<le> x"
  apply (presburger (no_quantify))
  oops

theorem "~ (\<exists>(x::int). False)"
  by presburger

theorem "\<forall>(x::int). (a::int) < 3 * x --> b < 3 * x"
  apply (presburger (no_quantify))
  oops

theorem "\<forall>(x::int). (2 dvd x) --> (\<exists>(y::int). x = 2*y)" by presburger 

theorem "\<forall>(x::int). (2 dvd x) --> (\<exists>(y::int). x = 2*y)" by presburger 

theorem "\<forall>(x::int). (2 dvd x) = (\<exists>(y::int). x = 2*y)" by presburger 
  
theorem "\<forall>(x::int). ((2 dvd x) = (\<forall>(y::int). x \<noteq> 2*y + 1))" by presburger 

theorem "\<forall>(x::int). ((2 dvd x) = (\<forall>(y::int). x \<noteq> 2*y + 1))" by presburger 

theorem "~ (\<forall>(x::int). 
            ((2 dvd x) = (\<forall>(y::int). x \<noteq> 2*y+1) | 
             (\<exists>(q::int) (u::int) i. 3*i + 2*q - u < 17)
             --> 0 < x | ((~ 3 dvd x) &(x + 8 = 0))))"
  by presburger
 
theorem 
   "~ (\<forall>(i::int). 4 \<le> i --> (\<exists>x y. 0 \<le> x & 0 \<le> y & 3 * x + 5 * y = i))"
  by presburger

theorem
    "\<forall>(i::int). 8 \<le> i --> (\<exists>x y. 0 \<le> x & 0 \<le> y & 3 * x + 5 * y = i)" 
  by presburger
   
theorem
   "\<exists>(j::int). \<forall>i. j \<le> i --> (\<exists>x y. 0 \<le> x & 0 \<le> y & 3 * x + 5 * y = i)"
  by presburger

theorem
   "~ (\<forall>j (i::int). j \<le> i --> (\<exists>x y. 0 \<le> x & 0 \<le> y & 3 * x + 5 * y = i))"
  by presburger

text{*Very slow: about 80 seconds on a 1.8GHz machine.*}
theorem "(\<exists>m::nat. n = 2 * m) --> (n + 1) div 2 = n div 2" by presburger

theorem "(\<exists>m::int. n = 2 * m) --> (n + 1) div 2 = n div 2" by presburger

end