4 let R be num->num->bool;
6 assume !x y z. R x y /\ R y z ==> R x z [2];
7 thus (!m n. m <= n ==> R m n) <=> (!n. R n (SUC n))
9 now [3] // back direction first
10 assume !n. R n (SUC n);
12 !d. R m (m + d) ==> R m (m + SUC d) [4] by 2,ADD_CLAUSES;
13 R m (m + 0) by 1,ADD_CLAUSES;
14 !d. R m (m + d) by 4,INDUCT_TAC;
15 thus m <= n ==> R m n by LE_EXISTS;