section ‹Some examples for Presburger Arithmetic›
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) ==>
(∃(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 = ¦x2¦ - x1; x4 = ¦x3¦ - x2; x5 = ¦x4¦ - x3;
x6 = ¦x5¦ - x4; x7 = ¦x6¦ - x5; x8 = ¦x7¦ - x6;
x9 = ¦x8¦ - x7; x10 = ¦x9¦ - x8; x11 = ¦x10¦ - x9 ⟧
⟹ x1 = x10 & x2 = (x11::int)"
by arith
end