 
Summary: Solution Proposal Functional Programming  Sheet 7
Exercise 1
Proof: Let S be an arbitrary chain of D 1 , let f : D 1 # D 2 and g : D 2 # D 3 be
continuous functions. We then have:
(g # f)(#S) by definition of composition
= g(f(#S)) f is continuous
= g(#f(S)) g is continuous
= #g(f(S)) by definition of composition
= #(g # f)(S)
Therefore, also the composition (g # f) : D 1 # D 3 is continuous. #
Exercise 2
(a) Claim: For g 1 , g 2 # #Z# # Z# #, we have:
g 1 # #Z ##Z# # g 2 # # sum (g 1 ) # #Z ##Z# # # sum (g 2 )
Proof: Let g 1 , g 2 # #Z# # Z# # with g 1 # #Z ##Z# # g 2 . By definition of # #Z ##Z# # ,
one can show # sum (g 1 ) # #Z ##Z# # # sum (g 2 ) by showing that
#x # Z# : (# sum (g 1 ))(x) # Z# (# sum (g 2 ))(x)
holds. There are three cases:
(1) x = #. Then (# sum (g 1 ))(x) = g(#  1) +# = # # Z# (# sum (g 2 ))(x) for all
g 2 .
(2) x # 0. Then (# sum (g 1 ))(x) = 1 # Z# 1 = (# sum (g 2 ))(x).
