(*原始帰納的関数*) (*零関数*) let zero x = 0;; (*後者関数*) let suc x = x + 1;; (*射影関数*) let p n i = function x -> x.(i - 1);; (*前者関数*) let pred = function 0 -> zero () | x -> p 2 1 [|(x - 1); pred (x - 1)|];; (*加算*) let rec add …