(* Let f be a function over natural numbers such that, for all n f(f(n)) < f(n+1) Show that f(n)=n for all n. Inspired by a Dafny example (see http://searchco.de/codesearch/view/28108482) Original reference is Edsger W. Dijkstra: Heuristics for a Calculational Proof. Inf. Process. Lett. (IPL) 53(3):141-143 (1995) *) theory Puzzle use export int.Int function f int: int axiom H1: forall n: int. 0 <= n -> 0 <= f n axiom H2: forall n: int. 0 <= n -> f (f n) < f (n+1) end theory Step1 (* k <= f(n+k) by induction over k *) use Puzzle predicate p (k: int) = forall n: int. 0 <= n -> k <= f (n+k) clone int.SimpleInduction as I1 with predicate p = p, lemma base, lemma induction_step end theory Solution use Puzzle use Step1 lemma L3: forall n: int. 0 <= n -> n <= f n && f n <= f (f n) lemma L4: forall n: int. 0 <= n -> f n < f (n+1) (* so f is increasing *) predicate p' (k: int) = forall n m: int. 0 <= n <= m <= k -> f n <= f m clone int.SimpleInduction as I2 with predicate p = p', lemma base, lemma induction_step lemma L5: forall n m: int. 0 <= n <= m -> f n <= f m lemma L6: forall n: int. 0 <= n -> f n < n+1 goal G: forall n: int. 0 <= n -> f n = n end