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(f) q^r . (g) P (h) true .
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Solution 5.18.
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(a) p = q . (b) p*q . (c) p = q=r = s .
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272 Solution 5.19. - true { law ->p = p = false with p := true } true = false = { false . law true = p = p with p := false }
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Solutions to Exercises
Solution 5.20. { law -ip = p=false with p := -ip } ->p = false { law ->p = p = false with p := p and symmetry of equivalence }
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Solution 5.21. The three most important examples are (p (q r)) = (p = q = r) , ((p q) r) = (p = q = r) , (p q) = (a r) = p = r . The first two establish that inequivalence is associative. Solution 5.22. The process of decryption after encryption computes But, is associative } D
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false b { definition of ^ } false = ->b { definition of negation: (5.15) } b .
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Solutions to Exercises
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Solution 5.23. A's statement is B = ->A. So, what we are given is A=B=^A . This simplifies to ~>B as follows. A=B=^A = = {
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rearranging terms } law -ip = p = false with p := A } law ->p = p = false with p := B and rearranging } D
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So, B is a knave, but A could be a knight or a knave.
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Solution 5.24. Let Q be the question. Then, Q = A = A B, i.e. Q = -*B. In words, ask A whether B is a knave. D Solution 6.5. Suppose that I and m are given numbers such that for all numbers n, Instantiating n to I (which is allowed because n and I are assumed to have the same type), we get
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l^l = l^m { < is reflexive } true = l^m { true is the unit of equivalence } Thus, we conclude that I ^ m. Symmetrically, instantiating n to m (which is allowed because n and m are assumed to have the same type), we get and, hence, m ^ I. The conjunction of I ^ m and m^l, together with the fact that the at-most relation is antisymmetric establishes that I = m, as required. n
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274 Solution 6.6.
Solutions to Exercises
(a) We have, for all n, [x+raj { definition of floor } { arithmetic } n-m^x { definition of floor [x\ { arithmetic } The result follows by indirect equality. (b) We have, for all n,
n^ [x/m\ { definition of floor } n^x/m { arithmetic, m is positive }
definition of floor }
{ arithmetic, m is positive } [x\/m { definition of floor } The result follows by indirect equality. D
Solution 6.7. The second 'definition of floor' step is invalid since m/n is not an integer. Taking m, n and x to be 1, 2 and f, we have
[|J=0 }
2 ^ L3j
< Ul
This suggests that 2x [f J * I 2 x | j , which indeed is the case as 0 * 1.
Solutions to Exercises
Solution 6.11.
I (-i) + (-i)-i I
which equals 3 and is clearly different from f }l, which is 1.
Solution 6.12. The second step (with the hint 'inequalities') is invalid. The rule m<k+l = ra < fc is only valid for integers m and k, and not for real numbers. The mistake made here is an easy one to make because of the overloading of the symbols < and ^ for ordering both real and integer numbers. In this case, the mistake is easily spotted, but in other circumstances it may not be so easy to spot. The moral is: beware of overloaded operators! D Solution 6.13. We have, for all n,
negation } definition of floor } negation } definition of ceiling } D
r-xi .
Thus, by indirect equality, the function / is the ceiling function. Solution 6.14. The defining equation is
k ^ m+n = kxn ^ m .
Indirect equality is used to show that Im I m+n = \ in] We have, for all integers k,
k ^ m+n
= = { above definition } arithmetic }
kxn ^ m
{ ra "^ n
276 = { definition of the floor function }