Solution Proposal Functional Programming --Sheet 7 Proof: Let S be an arbitrary chain of D 1 , let f : D 1 # D 2 and g : D 2 # D 3 be 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). Collections: Computer Technologies and Information Sciences