(* Title: HOL/ex/PresburgerEx.thy

Author: Amine Chaieb, TU Muenchen

*)

header {* Some examples for Presburger Arithmetic *}

theory PresburgerEx

imports Presburger

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) ==>

(∃(x::int). 2*x = y) & (∃(k::int). 3*k = z)"

by presburger

theorem "!! (y::int) (z::int) n. Suc(n::nat) < 6 ==> 3 dvd z ==>

2 dvd (y::int) ==> (∃(x::int). 2*x = y) & (∃(k::int). 3*k = z)"

by presburger

theorem "∀(x::nat). ∃(y::nat). (0::nat) ≤ 5 --> y = 5 + x "

by presburger

text{*Slow: about 7 seconds on a 1.6GHz machine.*}

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) = (∀(y::int). x ≠ 2*y+1) |

(∃(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

text{*Slow: about 5 seconds on a 1.6GHz machine.*}

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

by presburger

text{* 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! *}

lemma "[| x3 = abs x2 - x1; x4 = abs x3 - x2; x5 = abs x4 - x3;

x6 = abs x5 - x4; x7 = abs x6 - x5; x8 = abs x7 - x6;

x9 = abs x8 - x7; x10 = abs x9 - x8; x11 = abs x10 - x9 |]

==> x1 = x10 & x2 = (x11::int)"

by arith

end