Solutions to Exercises in .NET

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Solutions to Exercises
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"If]In the case that m+n is the smallest integer k such that kxn^m, the definition becomes
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k ^ m+n = kxn ^ m .
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Solution 7.4. For continued equivalences, pairs of repeated terms cancel each other out. So for continued equivalences, an even number of occurrences of the same term reduces to none, and an odd number of repeated terms reduces to one. D Solution 7.7. p v true { p y ( p = p) { py p =pv { true . Solution 7.10.
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reflexivity of equivalence (5.4) } disjunction distributes over equivalence (7.5) } p reflexivity of equivalence (5.4) }
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{ golden rule, p,q := p,p } p = p= pyp { disjunction is idempotent } p =p = p { reflexivity of equivalence (5.4) }
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Solution 7.12. = p/\(pyq) { golden rule: p,q := p,pvq }
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Solutions to Exercises
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associativity and idempotence of disjunction }
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reflexivity of equivalence (5.4) }
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p = pyq = pyq P
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Solution 7.13.
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(pyq) A(pyr)
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pyq = pyr = ( p v q ) v ( p y r )
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associativity, symmetry, idempotence of disjunction } disjunction distributes over equivalence } golden rule }
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py (q = r = qvr)
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Solution 7.14. Modus ponens:
golden rule, p,q := p,p = q }
p = p = q = py (p = q)
{ disjunction distributes over equivalence } p = p = q = pyp = pyq { simplification of continued equivalence, disjunction is idempotent } p = q = pvq
pAq .
golden rule }
Solutions to Exercises De Morgan. The more complicated side is the right side (because it contains two negations rather than one).
->p v ->q
= = { { definition of negation } disjunction distributes over equivalence (applied twice) } p y q = false v q = p v false = false v false = { { false is unit of disjunction } rearranging terms (using symmetry and associativity of equivalence) } p = q = py q = false { = { golden rule } definition of negation } p /\q = false py q = q = p = false (p = false) v (q = false)
De Morgan. Again we begin with the right side.
-~>p A ~>q
= { { golden rule } contrapositive applied to ->p = ->q rule just proved: ->p v ->q = -^(p A q) applied to third term } p = q = ^(p^q) = = { { { definition of negation } golden rule } definition of negation } p = q = pAq = false p v q = false
Solutions to Exercises
DistributMty of conjunction over equivalence.
p / \ ( q = r)
{ = golden rule, p,q := p,q = r }
p = q = r = p v (q = r) { distributivity of disjunction over equivalence } p = q = r = pvq = pvr
{ rearrange terms (using symmetry and associativity of equivalence) and add p twice in order to head for the golden rule. }
p = q = pvq
= p = p = r =
golden rule (twice once with p,q := p,q , once with p,q := p,r) }
p Aq = p = p AT .
We thus conclude that
p A ( q = r) = pAq = p = p f \ r .
Rearranging terms we get the required result. D Solution 7.15. In this solution, simplification of continued equivalences and disjunctions using the basic laws (symmetry, associativity, idempotence and constants) is not spelt out. We begin by deriving the equality between (a) and (b).
( p y q ) A (qvr) A (r v p)
= { { golden rule, p,q :- p vq , (q vr) A (r v p) = pvqv((qvr) A ( r v p } } distributivity of disjunction over conjunction and simplification } pvq = (qvr) A ( r v p ) = pvqvr = { golden rule, p,q := qvr ,rv p, simplification of continued disjunction } pvq = qvr = rvp = pv qvr = pv qvr = { simplification of continued equivalences } } pyq = (qvr) A (rvp)
pvq = qvr = rvp .
Solutions to Exercises Now the equality between (d) and (c) is obtained by replacing conjunction everywhere by disjunction and vice versa.
(p Aq) v (qAr) v (r Ap)
golden rule, p,q := p Aq , (qAr) v (r A p) } distributwity of conjunction over disjunction and simplification }
p Aq ~ (qAr) v (r Ap) = pAqA((qAr) v (r A p ) )