number-theory

126 8 Counting a type's inhabitantsProof By 8A13(i), Habs(T) = 0 e:> #(Nhabs(T)) = #(T) = 0. And by 8A10(ii) and(iii), #(T) = 0 a #(Long(T)) = 0.8D6 Theorem (Counting all normal inhabitants) (Ben-Yelles 1979 Ch. 3.) Whengiven a type T the algorithm in 8D7 below outputs #(T) and an enumeration of the setNhabs(T) of all the /3-normal inhabitants of T.8D7 Counting Algorithm for Nhabs(r) If T is an atom then Nhabs(T) is empty by8B5. If r is composite, apply Algorithm 8D5 to count Long(T).Case 1: If Long(T) = 0 then Nhabs(T) = 0. [By 8A8.1]Case II: If Long(T) # 0 then Nhabs(T) is counted and enumerated by counting andenumerating Long(T) and enumerating the n family of each member of Long(T).[By 8A10 these rl-families are finite and their union is Nhabs(T).]Before ending this section we shall digress from the main theme to look at aninteresting application of the counting algorithm to types that contain only oneatom. Such types have been studied and used as tools by several workers, forexample see Statman 1979a, 1979b and 1980, Dekkers 1988, and Zaionc 1987b and1990.8D8 Definition A type T will be called monatomic if IITII = 1, i.e. if only one atomoccurs in T.8D9 Theorem (Ben-Yelles 1979 §3.25.) Let r be a monatomic type with the formT -TI ->...Tm->a (m > 0). Then(i) if at least one Ti is composite, #(T) is either oo or 0;(ii) if TI - ... - Tm = a then #(T) = M.Proof Part (ii) is just 8B4. To prove (i), assume Nhabs(T) is finite and non-emptyand assume r has formT =with m >_ 1 and at least one premise composite. By 8A8 Long(T) is also finite andhas a member MT. By 8D2, Depth(MT) < fITli = 1, so MT must have formMT = AXT IXT'" ' xTI1 m P(1T;,i,m+,;-'a(m;j > 0).For each j < m, choose distinct new variables yl,... , ym J and defineP.+,i-T+i,l..T 1,i,m;' j . XaY1 Ym; p