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By David A. Schmidt

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Xn ) = e does not permit f itself to appear in e. There is good reason: a recursive definition may not uniquely define a function. Here is an example: q(x) = x equals zero→ one [] q(x plus one) This specification apparently defines a function in IN → IN_| . 3 Recursive Function Definitions f1 (x) = one if x = zero otherwise −| f2 (x) = one if x = zero two otherwise 45 f3 (x) = one and there are infinitely many others. Routine substitution verifies that f3 is a meaning of q: for any n∈ Nat, n equals zero → one [] f3 (n plus one) = n equals zero → one [] one by the definition of f3 = one by the definition of the choice function = f3 (n) Similar derivations also show that f1 and f2 are meanings of q.

Identity: f : R→ R is the identity function for R iff for all x∈R, f(x) = x. inverse: for some f : R→ S, if f is one-one and onto, then the function g : S→ R, defined as g(y) = x iff f(x) = y is called the inverse function of f. Function g is denoted by f−1 . Functions are used to define many interesting relationships between sets. The most important relationship is called an isomorphism: two sets R and S are isomorphic if there exist a pair of functions f : R→ S and g : S → R such that g ° f is the identity function for R and f ° g is the identity function for S.

3. 5, simplify the following (note that we use identifiers m,n∈Nat, t∈Tr, p∈Tr×Tr, r∈Tr+Nat, and x,y∈Nat_| ): a. n) ) b. (λp. (snd p, fst p))(true, (two equals one)) c. ((λr. cases r of isTr(t) → (λm . zero) ' [] isNat(n) → (λm. n) end)(inNat(two)) )(one) d. cases (false → inNat(one) [] inTr(false)) of isTr(t) → true or t [] isNat(n) → false end e. y(x))(one)(λn. n plus two) f. ((λ _ n. [ zero |→ n](λm. zero))(two)) zero g. (λx. (λm. m equals zero → x [] one)(two))(−| ) h. (λ _ m. one)(true → −| [] zero) i.

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