Quelle PresburgerEx.thy Sprache: Isabelle

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

section \<open>Some examples for Presburger Arithmetic\<close>

theory PresburgerEx
imports Main
begin

lemma "\m n ja ia. \\ m \ j; \ (n::nat) \ i; (e::nat) \ 0; Suc j \ ja\ \ \m. \ja ia. m \ ja \ (if j = ja \ i = ia then e else 0) = 0" by presburger
lemma "(0::nat) < emBits mod 8 \ 8 + emBits div 8 * 8 - emBits = 8 - emBits mod 8"
by presburger
lemma "(0::nat) < emBits mod 8 \ 8 + emBits div 8 * 8 - emBits = 8 - emBits mod 8"
by presburger

theorem "(\(y::int). 3 dvd y) ==> \(x::int). b < x --> a \ 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 "\(x::nat). \(y::nat). (0::nat) \ 5 --> y = 5 + x "
  by presburger

text\<open>Slow: about 7 seconds on a 1.6GHz machine.\<close>
theorem "\(x::nat). \(y::nat). y = 5 + x | x div 6 + 1= 2"
  by presburger

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

theorem "\(x::int) y. x < y --> 2 * x + 1 < 2 * y"
  by presburger

theorem "\(x::int) y. 2 * x + 1 \ 2 * y"
  by presburger

theorem "\(x::int) y. 0 < x & 0 \ y & 3 * x - 5 * y = 1"
  by presburger

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

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

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

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

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

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

theorem "\(x::int). (2 dvd x) = (\(y::int). x = 2*y)"
  by presburger

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

theorem "~ (\(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 "~ (\(i::int). 4 \ i --> (\x y. 0 \ x & 0 \ y & 3 * x + 5 * y = i))"
  by presburger

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

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

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

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

text\<open>This following theorem proves that all solutions to the
recurrence relation $x_{i+2} = |x_{i+1}| - x_i$ are periodic with
period 9.  The example was brought to our attention by John
Harrison. It does does not require Presburger arithmetic but merely
quantifier-free linear arithmetic and holds for the rationals as well.

Warning: it takes (in 2006) over 4.2 minutes!\<close>

lemma "\ x3 = \x2\ - x1; x4 = \x3\ - x2; x5 = \x4\ - x3;
         x6 = \<bar>x5\<bar> - x4; x7 = \<bar>x6\<bar> - x5; x8 = \<bar>x7\<bar> - x6;
         x9 = \<bar>x8\<bar> - x7; x10 = \<bar>x9\<bar> - x8; x11 = \<bar>x10\<bar> - x9 \<rbrakk>
\<Longrightarrow> x1 = x10 & x2 = (x11::int)"
by arith

end

quality94%

¤ Dauer der Verarbeitung: 0.3 Sekunden ¤

Wurzel

Suchen

Beweissystem der NASA

Beweissystem Isabelle

NIST Cobol Testsuite

Cephes Mathematical Library

Wiener Entwicklungsmethode

Haftungshinweis

Die Informationen auf dieser Webseite wurden nach bestem Wissen sorgfältig zusammengestellt. Es wird jedoch weder Vollständigkeit, noch Richtigkeit, noch Qualität der bereit gestellten Informationen zugesichert.

Bemerkung:

Die farbliche Syntaxdarstellung ist noch experimentell.