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Solution Proposal Functional Programming Sheet 12 (a) Consider Y = f.(x.f (x x)) (x.f (x x)). Then we have
 

Summary: Solution Proposal Functional Programming ­ Sheet 12
Exercise 1
(a) Consider Y = f.(x.f (x x)) (x.f (x x)). Then we have:
Y z
= (f.(x.f (x x)) (x.f (x x))) z
(x.z (x x)) (x.z (x x))
z ((x.z (x x)) (x.z (x x)))
z ((f.(x.f (x x)) (x.f (x x))) z)
= z (Y z)
(b) As z ((x.z (x x)) (x.z (x x))) reduces with to z (z ((x.z (x x)) (x.z (x x)))),
there is no chance to ever reach z ((f.(x.f (x x)) (x.f (x x))) z). The reason is
that a term z (z (...)) can never reduce to a term z ((f. ...) ...) with .
Note that one can apply -reduction not only on the outermost application of a
-abstraction to a term (i.e., (f. ...) z), but also on the application (x. ...) (...)
further inside the term. Nevertheless, there similar reasoning allows to conclude that
z (Y z) cannot be reached.
Exercise 2
(a) W(A0, x. if x x)
W(A0 + {x :: b1}, if x x)
W(A0 + {x :: b1}, if x)

  

Source: Ábrahám, Erika - Fachgruppe Informatik, Rheinisch Westfälische Technische Hochschule Aachen (RWTH)

 

Collections: Computer Technologies and Information Sciences