X = | x | A x ^ O => X = -x

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X=|x| A x ^ O => X = x .

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These are both clearly true, thus completing the verification. Solution 10.11. The precondition is

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x =X A y = Y .

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The postcondition is

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We consider two cases: x ^ y and y ^ x. hi the first case, there is nothing to do. In the second case, an interchange of x and y establishes the postcondition. Thus

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298 the program is

{ x =X A y =Y }

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Solutions to Exercises

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if x ^ y skip

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D y ^ x x,y := y,x

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A (x = X vy = X) A (x = Y vy = Y) } .

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Use of the assignment axiom and the skip rule in order to check the correctness yields the verification conditions:

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A (x = X v y = X) A (x = Y v y = Y) and

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x = X A y = Y A y ^ x => y ^ x A (y = X v x = X) A (y = Yvx = Y) . Both are clearly true. Solution 10.12. Using the conditional rule, the requirements are D

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k , x ,y := k-1 ,a,b and

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{ 0 < k = K A x f c x y = CA even.k } k,x,y := k+2,c,d { 0^k<K A xkxy = C } . Applying the assignment axiom, the requirement on a and b is 0 < k = K A xkxy = C => C K f c - l < K A ak~lxb = C .

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Now,

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0<k = K A X k x y = C => { heading for introducing 'k-1* we use the fact that

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0 < k = K => k = (k-l) + l A

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0 < k - l < K A x(k~l)+1xy = C = { property of powers } ( K k - l < K A xk~lxxxy = C .

Solutions to Exercises

So, suitable values for a and b are a = x and b - x x y. (Note the implicit use of the associativity of multiplication!) Again applying the assignment axiom, the requirement on c and d is 0 < k = K A xkxy = C A even.k => ( K k - - 2 < K A ck+2xd = C . 0<k = K A xkxy = C A even.k => { heading for introducing 'fc-s-2' we use the fact that 0<k = K A even.k => k = ( k ^ - 2 ) x 2 A 0 < k - 2 < K } O s C k + 2 < K AX ( f c - 2 ) x 2 x y = c { property of powers } (Kk4-2<K A (x'2)k-2xy = C . So, suitable values for c and d are c = x2 and d = y. In summary, taking account of the fact that x is unchanged when the assignment k := k-1 is chosen, and y is unchanged when the assignment k := k-=-2 is chosen, the program we have constructed is {0<k=K A X k x y = C } if true k,y n even.k - k,x

{ (Kk<K A

Now,

:= k-l,xxy := k4-2,x 2

Solution 10.21. Using Exercise 10.12, we have { 0 < k - = - 2 < K A even.k A xkxy = C } k,x := k+2,x2 So the statement 52 is k , x := k^2 , x2 and the assertion P is 0 < k-^-2 < K. Now, satisfying the postcondition even.k is achieved by doing nothing in the case that k is already even, and subtracting one in the case that k is odd. So, again making use of Exercise 10.12, we postulate that statement SI is the statement if even.k - skip D odd.k > k,y :- k-l,xxy

fi .

300 We have to check that it meets its specification, i.e. we have

Solutions to Exercises

if even.k - skip n odd.k > k,y := k-l,xxy

{ ( K k - ^ 2 < K A even.k A xkxy = C } . Using the conditional rule and the assignment axiom, this is the case if 0 < k = K A even.k A xkxy = C => 0 < k + 2 < K A even.k A xkxy = C

0 < k = K A odd.k A xkxy = C => (K(fc-lH2<K A even.(fc-l) A xk~lxxxy = C .

Simple properties of arithmetic show that this is indeed the case. The complete program is, thus,

if even.k skip

D odd.k k,y := k-l,xxy

{ (Kk-=-2<K A even.k A xkxy = C }

k,x := k+2,x2

{ (Kfc<K A xkxy = C } .

Solution 11.1. (a) 6, (b) 5, (c) 3, (d) 0. (There are no integers i and j such that ( K i < j X 2 A odd.i A odd.j.) n Solution 11.2. (a) 4 (there is only one natural number k such that k2 = 4). (b) 8 (there are two integers k such that k2 = 4). D Solution 11.3. (a), (b) The one occurrence of T is free, all other occurrences of variables are bound, (c) There are no free occurrences of variables, (d) The occurrences of 'm' and 'n' are free, (e) The occurrences of 'm' and 'n' are free, as is the rightmost occurrence of '/. n