PAq = (qAr) v (r A p ) = p A qAr in .NET

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PAq = (qAr) v (r A p ) = p A qAr
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{ golden rule, p,q := qAr ,r Ap, simplification of continued conjunction }
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pAq = qAr = r Ap = p A qAr = p A qAr
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= { simplification of continued equivalences }
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p Aq = qAr = r Ap .
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Now we prove that (b) and (c) are equal.
p Aq = qAr = r Ap
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{ { golden rule, applied to each conjunct } simplification of continued equivalences }
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pvq = qvr = rvp .
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Solution 7.22.
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p Aq => p . q => pvq .
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Solution 7.23. false <= true { = { definition } true is zero of disjunction } false = false v true false s true
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Solutions to Exercises { false . true is unit of equivalence }
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Solution 7.24.
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= = { definition } false is unit of disjunction } disjunction distributes through equivalence } definition of negation }
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p = p vq
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{ { { p v false = p v q p v (false = q)
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p v ->q .
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Solution 7.25. { { { true .
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(7.24) } symmetry and associativity of disjunction } excluded middle (twice) }
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p v ~>q v -<p v q p v ~>p v q v ~>q
Solution 7.26. Contrapositive: { { definition } De Morgan }
ip = -1/7 A ->q p s -.(p vq)
282 = {
p = pvq
Solutions to Exercises contrapositive } definition }
{ p<=q . Contradiction: p=> false = = = Distributivity: (p = q)<=r = { { = Distributivity: = { { { { { { p = p/\ false { p = false {
definition } false is zero of conjunction } definition }
definition } distributivity of conjunction over equivalence } simplification of continued equivalences }
r = (p = q) ^r
r = p AT = qf\r = r p /\r = qAr .
definition } distributivity of disjunction over equivalence } rearranging terms } definition }
p = q = (p = q)vr p = q = pvr = qvr
p = p\j r = q = qy r
Solutions to Exercises
Shunting: { { { { { definition (applied twice) } distributivity of disjunction over equivalence } distributivity of disjunction over equivalence } golden rule } definition }
p = pv q = (p = pv q)vr
p = p v q = pvr = pvqvr
p = py (q = r = qvr)
p = pv(q/\r)
p <= q AT .
Solution 7.29. Mutual implication: definition }
(p =
^(q = pyq)
substitution of equals for equals (7.27), first occurrence of p v q replaced by q }
(p = q) { (p =
= pyq)
substitution of equals for equals (7.27), second occurrence of p replaced by q }
= qyq)
disjunction is idempotent, properties of true }
p =q .
Distributivity:
p <=q vr {
definition of <= }
p = p v q vr
Solutions to Exercises
substitution of equals for equals (7.28):
(p = pvqvr) A (pvq = pvqvr) { substitution of equals for equals (7.27): specifically, p v q for first occurrence of p v q v r } (p = pvq) A ( p y q = pvqvr) { substitution of equals for equals (7.27): specifically, p for last two occurrences of p v q }
(p = pvq) A (p = pvr)
{ Distributive ty: { = {
definition of <= }
(p<=q) /\(p<=r) .
definition } substitution of equals for equals (7.27) specifically, p A r for last two occurrences of r }
(r = p A T ) A (r = q Ar)
(r = p Ar) A ( p A r = p A < | A r ) { substitution of equals for equals (7.27), specifically, p A q A r for first occurrence of p A r } (r = p /\qs\r) A (p /\r = p A g Ar) { substitution of equals for equals (7.28), with g.x = (p AX) } r = p A,qAr = {
p Aq <=r .
definition }
Solution 7.30. (b) Simplifying A = ->A => B using (7.19), we get A = ^A = ->A A B, which equals A v - . So, either A is a knight or B is a knave.
Solutions to Exercises
(c) Simplifying A ~ A => ->B using (7.19), we get A A ~^B. So A is a knight and B is a knave. (d) Simplifying A == ->A=> ->JB using (7.19), we get A = ->A = ->A A -ifi, which equals Av B, So, at least one of A or B is a knight. (f) Simplifying A = ~^B=>A using (7.18), we get A = A = -iJ5 v A, which equals -iB v A. So, either A is a knight or B is a knave. (g) Simplifying A = B => ->A using (7.18), we get A = -iA = ->A v B, which equals A A -i . So A is a knight and B is a knave. (h) Simplifying A = ->B=> ->A using (7.18), we get A = -iA = ~rA v -ifi, which equals A/\B. So, both A and B are knights.
Solution 7.31. Let the natives be A and B. Following the analysis given in Section 5.5, to determine whether both are knights, the question to be posed is A = A A B. This is the same as A=*B. So, in words, the question is 'is it the case that, if you are a knight, then your colleague is also a knight '. To determine whether at least one is a knight, the question to be posed is A = A v B, This is the same as A<=#. So, in words, the question is 'is it the case that you are a knight if your colleague is a knight '. n Solution 7.32. We are required to simplify A = -iA v B. A = ->Av = = { { { definition of negation } disjunction distributes through equivalence } associativity of equivalence, false is unit of disjunction }