Solutions to Exercises in .NET

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Solutions to Exercises
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reflexivity of equality }
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Induction Step:
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range splitting, arithmetic } ) + (n+l) 2 - |(n+l)(n+2)(2n ^k^nik2} = n(n+l)(2n + assume that
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arithmetic and reflexivity of equality }
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(b) Basis:
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one-point rule to simplify the summation, arithmetic }
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arithmetic and reflexivity of equality }
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Induction Step:
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range splitting, arithmetic } assume that
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l ^ k ^ n : k 3 ) + (n+l) 3 = |(n+l) 2 (n+2) 2 ( Z k : l ^ k ^ n : k 3 ) = |n 2 (n+l) 2 . |n 2 (n+l) 2 + (n+1) 3 = |(n+l) 2 (n+2) 2 arithmetic and reflexivity of equality }
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Solutions to Exercises
(c) Basis:
v o+i
x-l one-point rule to simplify the summation,
arithmetic }
x * i, arithmetic and reflexivity of equality } true . Induction Step:
x-l range splitting, arithmetic
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assume that
x-l -1 x-l arithmetic and reflexivity of equality }
n+l + X
true (d) Basis:
1 + Oxx
arithmetic }
true Induction Step: l + (n+l)xx transitivity of ^ (heading towards using the induction hypothesis, we introduce the middle term
nxx) ) }
nxx) ^ 1 + (n+l)xx assume that (l+x) n ^ 1 + nxx . x > - 1, so 1 +x > 0. Multiplication by a positive
Solutions to Exercises
number is monotonic. That is, nxx) ^ l + (n+l)xx { { true . (e) Basis: : -r i- \ - -^kx(k+l)/ 0+1 empty-range rule to simplify the summation, arithmetic on the right side } true . Induction Step: kx(k+l) range splitting }
arithmetic
1 + x + nxx + nxx ^ 1 + x + nxx nxx2 ^ 0. Addition is monotonic. }
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assume that
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kx(k+l)/ n n+1 n+1 (n+l)x(n+2) n+2 { arithmetic and reflexMty of equality }
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(f) Basis: empty-range rule to simplify the summation, arithmetic }
0 = 0 {
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reflexivity of equality }
312 Induction Step: 'zfcil^fc { range splitting }
Solutions to Exercises
2n+i
.
assume that
n+2 2n {
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n+l _ n+3 ~~ 2 n+1 ~~ 2 n+1 arithmetic and reflexivity of equality }
Solution 12.13. Basis: the basis is the case n = 1. This is because there are individual definitions of F.O and F.I. We have F.(l + l ) x F . ( l - l ) - (F.I) 2
IxO-l
definition } arithmetic } arithmetic }
{ -1
(-D 1 .
Induction Step: care must be taken in the induction step when expanding the definition of F. Assuming that n ^ 1, the definitions of F. (n+2) and F.(n+l) are given by the rule F.(k+2) = F.(fc+l) + F.fe, for all k, k ^ 0. (This is not the case for the definition of F.n.) { { arithmetic } definition: F.(k+2) = F.(k+l) + F.k applied to the cases k = n and k = n-1. } (F. (n+l) + F.n) x F.n - (F.n + F.(n-l)) xF.(n+l) { arithmetic }
F.(n+2)xF.n - (F.(n+l)) 2
Solutions to Exercises
(F.n)2 - F.(n-l)x { -(-l) n { (-l)
assume that
F.(n+l)xF.(n-l) - (F.n)2 = (-1)" . } arithmetic and reflexivity of equality } . D D
The proof uses simple induction only. Solution 12.14. The first step is invalid in the case that n = 0.
Solution 13.13. To verify the correctness of the initialization, we must verify the Hoare triple
fc,5 := JV.O
{ (Kk^iV A sxXk = (Zi:k^i By the assignment axiom, this reduces to
=> CKAKAT A OxXN =
This is true by virtue of the empty-range rule (since N < i < N is false) and simple properties of arithmetic. The termination condition is valid if ( K f c ^ N A sxXk = (I.i:k^i<N:a[i]xXi) => 5 = ( E i : Q ^ i < N : a [ i ] x X i } . D A -.(fe>0)
This is clearly true as 0 < k < N A ->(fe > 0) reduces to 0 = k < N, Solution 13.15. The invariant is established by the assignment
k,y,z := M,l,X
and, when k = 0, we have
y = XM
independently of the value of z. As the reader will have seen in solving Exercise 10.21, we also have { 0 < k = K A yxzk = XM }