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; Book fibonacci
; Contributed by Matyas Sustik
; Date created 2000-05-08
; $Id: fibonacci.lisp,v 1.3 2001/09/05 19:12:21 matyas Exp $
;
; Matt Kaufmann has simplified and rearranged several proofs. Thank
; you! See his comments inserted before the theorems.
(in-package "ACL2")
(include-book "int-division")
(include-book "grcd")
(defmacro np (x)
(list 'and (list 'integerp x) (list '<= 0 x)))
(defmacro pintp (x)
(list 'and (list 'integerp x) (list '< 0 x)))
; This is the efficient definition.
(defun fib-general (x y n)
(cond ((zp n) 0)
((equal n 0) 0)
((equal n 1) x)
((equal n 2) y)
((< 2 n) (fib-general y (+ x y) (+ -1 n)))
(t nil)))
; This is better for proofs.
(defun fib (n)
(cond ((zp n) 0)
((equal n 0) 0)
((equal n 1) 1)
(t (+ (fib (- n 1)) (fib (- n 2))))))
(defthm fib-general-at-3
(equal (fib-general x y 3) (+ x y))
:hints (("Goal" :expand (fib-general x y 3))))
(defthm fib-general-recursive-equation ; requires top-with-meta
(implies (and (integerp n)
(< 2 n))
(equal (+ (fib-general x y (+ -1 n))
(fib-general x y (+ -2 n)))
(fib-general x y n))))
; The function fib is really a special case of fib-general:
(defthm fib-is-special-case-of-fib-general
(equal (fib n)
(fib-general 1 1 n))
:hints (("Goal" :induct (fib n)))
:rule-classes nil)
; The fib-general is more efficient but proofs are easier on fib.
; We use fib from now on.
(defthm fib-identity-base-case ; This is needed for the next theorem.
(implies (and (integerp n) (< 1 n))
(equal (fib n)
(+ (fib (- n 1))
(fib (- n 2))))))
(defthm fib-identity
(implies (and (pintp n)
(pintp k)
(< k n))
(equal (fib n)
(+ (* (fib k)
(fib (+ n (- k) 1)))
(* (fib (+ k -1))
(fib (+ n (- k)))))))
:rule-classes nil)
; MattK: Moved fib-identity-base-case below the proof of fib-neighbours-lemma
; in order to avoid the need for one of the :use hints.
(in-theory (disable fib))
(defthm fib-neighbours-lemma
(implies (and (integerp k)
(< 1 k))
(equal (grcd (fib k)
(fib (+ -1 k)))
(grcd (fib (+ -1 k))
(fib (+ -2 k)))))
:hints (("Goal"
:use ((:instance grcd-addition-lemma
(a (fib (+ -1 k)))
(b (fib (+ -2 k)))
(c 1))))))
(in-theory (disable fib-identity-base-case))
; MattK: Trivial mod, probably unnecessary, but result is simpler.
(local
(defun ind-int->=-2 (n)
(if (and (integerp n)
(<= 2 n))
(ind-int->=-2 (+ -1 n))
t)))
(defthm fib-neighbours-are-relative-primes
(implies (pintp k)
(equal (grcd (fib k)
(fib (+ -1 k)))
1))
:hints (("Goal"
:induct (ind-int->=-2 k))))
; (F(k); F(n-k)*F(k-1)) = (F(k); F(n-k))
; MattK: I rearranged the terms of the following in order to make it a better
; rewrite rule. In order for the left side to match, we need to be sure it's
; in a form that is likely to be matched during the proof of the next lemma.
; The documentation for loop-stopper hints that (- x) and x are ordered the
; same when directly under a call of +, and that variables closer to the start
; of the alphabet are ordered before those later in the alphabet. Those facts
; are relevant below since there are rewrite rules for commutativity of + and *
; that have loop stoppers. This is all pretty subtle, and of course there is
; nothing really wrong with just giving a :use hint, as was done originally.
(defthm grcd-fib-recursion-lemma-1
(implies (and (pintp n)
(pintp k)
(<= k n))
(equal (grcd (fib k)
(* (fib (+ -1 k))
(fib (+ (- k) n))))
(grcd (fib k)
(fib (- n k)))))
:hints (("Goal"
:use ((:instance grcd-multiplication-with-relative-prime
(a (fib k))
(b (fib (- n k)))
(c (fib (+ -1 k))))))))
; (F(k); F(n-k)*F(k-1)+F(n-k+1)*F(k)) = (F(k); F(n-k))
; MattK: Just for fun, I restated this lemma in a form that would apply in the
; proof of grcd-fib-recursion. I came up with this restatement by looking at
; the failed proof of grcd-fib-recursion. when this lemma was not supplied ina
; :use hint.
(defthm grcd-fib-recursion-lemma-2
(implies (and (pintp n)
(pintp k)
(<= k n)
(equal (fib n)
(+ (* (fib k)
(fib (+ 1 (- k) n)))
(* (fib (+ -1 k))
(fib (+ (- k) n))))))
(equal (grcd (fib k)
(fib n))
(grcd (fib k)
(fib (- n k)))))
:hints (("Goal"
:use ((:instance grcd-addition-lemma
(a (fib k))
(b (* (fib (- n k))
(fib (+ -1 k))))
(c (fib (+ 1 n (- k)))))))))
; MattK: The following is used automatically in grcd-fib-recursion and
; main-grcd-fib, thus saving two :use hints. Except, we also need grcd-0 for
; the proof of grcd-fib-recursion, which is OK since we will need it in
; main-grcd-fib, anyhow.
(defthm grcd-subtract
(implies (and (integerp k) (integerp n))
(equal (grcd k (+ (- k) n))
(grcd k n)))
:hints (("Goal"
:use ((:instance grcd-addition-lemma
(a k)
(b n)
(c (- 1)))))))
; We already have a similar grcd-with-0 but that has (grcd a 0) in it.
; ACL2 does not seem to use commutativity with constants in this case.
(defthm grcd-0
(implies (np a)
(equal (grcd 0 a) a))
:hints (("Goal" :in-theory (enable grcd euclid-alg-nat))))
; This says that the Euclidean algorithm for Fibonacci numbers can be
; thought of as operating on the indices.
(defthm grcd-fib-recursion
(implies (and (pintp n)
(pintp k)
(<= k n))
(equal (grcd (fib k)
(fib n))
(grcd (fib k)
(fib (- n k)))))
:hints (("Goal" :use fib-identity)))
(in-theory (enable Euclid-alg-nat))
(defthm main-grcd-fib
(implies (and (pintp n)
(pintp k))
(equal (grcd (fib k)
(fib n))
(fib (grcd k n))))
:hints (("Goal"
:induct (Euclid-alg-nat k n))))
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