horizon := 1;; thm `; let R be num->num->bool; assume !x. R x x [1]; assume !x y z. R x y /\ R y z ==> R x z [2]; thus (!m n. m <= n ==> R m n) <=> (!n. R n (SUC n)) proof now [3] // back direction first assume !n. R n (SUC n); let m n be num; !d. R m (m + d) ==> R m (m + SUC d) [4] by 2,ADD_CLAUSES; R m (m + 0) by 1,ADD_CLAUSES; !d. R m (m + d) by 4,INDUCT_TAC; thus m <= n ==> R m n by LE_EXISTS; end; !n. n <= SUC n; qed by 3`;;