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Klaus Madlener sheet 6 Exercise 1: Let≤⊆Term(F, V)×Term(F, V) be defined as: s.tiff

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WS 2011-2012 21.11.2011 Exercises to the Lecture FSVT

Prof. Dr. Klaus Madlener sheet 6

Exercise 1:

Let≤⊆Term(F, V)×Term(F, V) be defined as:

s.tiff. there is a substitutionσ with tσ(s) stiff.s.tand t.s

s < t iff.s.tand s6≈t Show:

1. <is the strict part of a well-founded partial order. Which elements are comparable in this order?

2. st holds iff. a permutationξ exists with sξ(t) (variable renaming).

Exercise 2:

This exercise is on an alternative specification of the integers INTEGER = (sig, E) with

sig= (int,0, succ, pred, add),

E ={succ(pred(x)) =x, pred(succ(x)) =x, add(0, y) =y, add(succ(x), y) =succ(add(x, y))}

1. Show, that (Z,0,+1,−1,+) is initial in Alg(INTEGER).

2. Structurize this specification using the specification INT. Show that INTEGER is an enrichment of INT.

3. Extend INTEGER by a function absolute with the properties of the absolute value function onZ. Show that this is an enrichment of INT.

Exercise 3:

Let INT2 be the specification of integers from example 7.9 of the lecture. We combine INT2 with BOOL and (({},{<}), E) to obtain a specification INT3, where

E ={<(0,succ(x)) = true, <(pred(x),0) = true, <(0,pred(x)) = false, <(succ(x),0) = false, <(pred(x),pred(y)) =<(x, y), <(succ(x),succ(y)) =<(x, y)}

1. Check, whether TINT3 |bool= Bool. Why would this be important? Hint: Look at

<(succ(pred(x)),pred(succ(y))).

2. Show that INT3 can not be fixed by additional equations.

3. Find further problems of INT3.

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4. Make a suggestion for a specification INT4, such thatTINT4|int = Z,TINT4|bool = Bool and < is properly defined by its equations. Hint: Consider further function symbols.

Exercise 4:

Let sig1 = ({NAT,EVEN},{0,1, S, f},{0 :→ NAT,1 :→ EVEN, S : NAT → NAT, f : NAT→EVEN}). Further, let the sig1-Algebra A1 be defined by:

A1,NAT =N, A1,EVEN = 2N∪ {1},0A1 = 0,1A1 = 1, SA1(x) =x+ 1, fA1(x) =

( x, ifx even 1, else Prove:

1. There is no specification spec1 = (sig1, E1) with finite E1, such that Tspec1 ∼=A1. 2. There is a specification spec2 = (sig2, E2) with sig1sig2, E2 finite, such that

Tspec2|sig1 ∼=A1.

Delivery: until 27.11.2011,

by E-Mail to huechting@informatik.uni-kl.de

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