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Unformatted text preview: Chapter 2
Fundamentals of Logic Section 2.1 —p. 54 1. The sentences in parts (a), (c), (d), and (f) are statements. The other two sentences are not.
3.3)0 b)0 c)1d)0
5. a) If triangle ABC is equilateral, then it is isosceles.
b) If triangle ABC is not isosceles, then it is not equilateral.
d) Triangle ABC is isosceles, but it is not equilateral.
7. a) If Darci practices her serve daily then she will have a good chance of winning the tennis
tournament.
b) If you do not ﬁx my air conditioner, then I shall not pay the rent.
c) If Mary is to be allowed on Larry’s motorcycle, then she must wear her helmet.
9. Statements (a), (e), (f), and (h) are tautologies.
11. a) 25 = 32 b) 2” 13. 11:0; r: 0; s10
15. a) m=3,n=6 b) m=3,n=9 c) m=18,n=9 d) m=4,n=9
e) m = 4, n = 9
2;, 17. Dawn 54 Solutions Section2.2 p.66 1. a) (1) p q qAr paw/W) p911 pqr (paw/Wqu
0 0 0 O 1 1 l 1
0 0 1 0 1 1 l 1
O l O 0 1 1 1 l
0 l l l l 1 1 l
l 0 0 0 0 0 0 O
1 O l O 0 0 1 0
l 1 0 0 0 1 0 0
1 1 l 1 l 1 1 1
(iii)
[7 qTr qu p—>(qu) p—HI r>(p>q)
0 0 0 O 1 l l
O O 1 1 1 1 1
0 l 0 1 1 1 l
0 1 1 1 l 1 1
1 0 0 0 0 0 0
l 0 l 1 1 O 1
1 1 O l 1 1 1
l 1 1 l 1 1 1
b) [P + (61 V r)] <=¢ [er —> (P —> (1)] From Part (iii) of Part (61)
42> [r —> (1p V q)] By the 2nd Substitution Rule, and (p —> q) ¢=> (pp Vq) <=> l“(_'P V q) —> 'r] By the lst Substitution Rule,
and (s —> t) <=> (t —> s) for any
primitive statements s, t {22> [(I—1p /\ ﬁq) —> r] By DeMorgan’s Law, Double Negation,
and the 2nd Substitution Rule
<::> [(p /\ q) —> r] By Double Negation and the 2nd Substitution Rule 3. a) For any primitive statement 3, s v us 42> To. Replace each occurrence of s by p v (q /\ r),
and the result follows by the lst Substitution Rule.
b) For any primitive statements 5, t, we have (s —> t) <=> (Ir ——> s). Replace each
occurrence of s by p v q, and each occurrence of t by r, and the result is a consequence of the
lst Substitution Rule. 5. a) Kelsey placed her studies before her interest in cheerleading, but she (still) did not get a
good education.
b) Norma is not doing her mathematics homework or Karen is not practicing her piano lesson.
c) Harold did pass his C++ course and he did ﬁnish his data structures project, but he did not
graduate at the end of the semester. 7.a
) p q (—pvq)A(pA(p/\q)) 17M
0 0 0 0
o 1 0 0
1 0 0 o
1 1 1 1 b) (ﬁpAq) thV(pvq)) 4% pvq 9. a) lfOlO = 0, then 1 +1 = 1. (FALSE)
Corztraposizive: If 1 + 1 7E 1, then 0 + 0 7é 0. (FALSE)
Converse: If1 + 1 = 1, then 0 + 0 = 0. (TRUE)
Inverse: IfO + 0 75 0, then 1 + 1 7E 1. (TRUE) Section 2.3 — p. 84 13. 15. 17. 19. Solutions b) If—l < 3 and 3 + 7 = 10, then sin (37") = —1. (TRUE) Converse: If sin : — 2 1, then —1 < 3 and 3 + 7 = 10. (TRUE) Inverse: If —1 Z 3 or 3 + 7 75 10, then sin 75 —1. (TRUE) Contrapositive: If sin (
11. a) (q —> r) v «p 37! 2 )7e ~17then —1 2 3 or3 +7 7é to. (TRUE)
b) (“'qu)V'—vp ﬂq I [(p<>q)Mq9r)A(r9p)] [(p—>q)A(q—>r)A(r>p)l
0 0 0 1 1
0 O 1 0 0
0 1 0 0 0
0 1 1 O O
1 0 0 0 0
1 0 1 0 0
1 1 O O O
1 1 1 1 1
a) (P1P) b)(pr)T(qTq) c)(qu)T(qu) d)PT(qTq)
e) (r T s) T (r T s), where r stands for p T (q T q) and s for q T (p T p)
17 q (p l q) (~11 T “(1) (p T q) (p 1 ~11)
0 0 0 0 O O
0 1 1 1 0 0
1 O 1 1 0 0
1 1 1 1 l 1
a) P V [P /\ (P V 61)] Reasons
<=> p v p Absorption Law
<::> p Idempotent Law of V c) [(ﬁpvﬁq)—>(pAq/\r)l
<::>—'(—'pv—uq)v(p/\q/\r)
:>(‘i—'p/\—'—'q)v(p/\q/\r)
<==>(p/\q)V(p/\q/\r) <=>p/\q Reasons 5 —> t ¢=> —vs V t
DeMorgan’s Laws Law of Double Negation
Absorption Law r P“)‘I (qu> (qu)—>r t—HP‘t‘OOOO ’E
r—i—‘OOi—v—‘OO Q 0
1
0
l
O
1
0
1 1
l
1
1
0
0
1
1 0 0
1
l
1
1
1
1 1
1
O
1
O
1
0
1 55 The validity of the argument follows from the results in the last row. (The ﬁrst seven rows may be ignored.) S6 Solutions c) J »——>—oooo "a q rqurlpVUIVr) —q er 0 0 0 O 1 0 0 l 1 1 1 1 l O 1 1 0 0 l 1 1 1 0 1 ‘s'.
0 0 o 1 1 1 *
0 1 1 1 1 1 1 0 1 1 O 1 1 1 1 1 0 l The results in rows 2, 5, and 6 establish the validity of the given argument. (The results in the
other ﬁve rows of the table may be disregarded.) b) When . a) If p has the truth value 0, then so does p /\ q. p V g has the truth value 0, then the truth value of p (and that of q) is 0. c) Ifq has truth value 0, then the truth value of [(p V q) /\ 'p] is 0, regardless of the truth
value of p. d) The statement q V s has truth value 0 only when each of q, s has truth value 0. Then (p —> q) has truth value 1 when p has truth value 0; (r —> s) has truth value 1 when r has truth
value 0. But then (p V r) must have truth value 0, not 1. . a) Rule of Conjunctive Simpliﬁcation b) Invalid—attempt to argue by the converse
c) Modus Tollens d) Rule of Disjunctive Syllogism e) Invalid— attempt to argue by the inverse .a) . 1) and 2) Premise
3) Steps (1) and (2) and the Rule of Detachment
4) Premise
5) Step (4) and (r —> ﬁq) <=> (ﬁ—wq —> —Ir) <2? (q —> vr)
6) Steps (3) and (5) and the Rule of Detachment
7) Premise
8) Steps (6) and (7) and the Rule of Disjunctive Syllogism
9) Step (8) and the Rule of Disjunctive Ampliﬁcation
1) Premise (The Negation of the Conclusion)
2) Step (1) and —'('q —> s) <=> '(—1q v s) ¢=> —(q V s) <:> q /\ Is
3) Step (2) and the Rule of Conjunctive Simpliﬁcation
4) Premise
5) Steps (3) and (4) and the Rule of Disjunctive Syllogism
6) Premise
7) Step (2) and the Rule of Conjunctive Simpliﬁcation
8) Steps (6) and (7) and Modus Tollens
9) Premise
10) Steps (8) and (9) and the Rule of Disjunctive Syllogism
11) Steps (5) and (10) and the Rule of Conjunction
12) Step (11) and the Method of Proof by Contradiction
b) 1) p ——> q Premise
2) eq —> op Step (1) and (p > q) e: (eq —> pp)
3) p v r Premise
4) Ip—>r Step (3) and (er)<:>(—Ip—>r)
5) ﬁg —> r Steps (2) and (4) and the Law of the Syllogism
6) Ir V 5 Premise
7) r —> 5 Step (6) and (—r V s) <=> (r —> s)
8) q —> 5 Steps (5) and (7) and the Law of the Syllogism Solutions 57
11. a) 1311 q 0 r21 c) p,q,r:1 5:0
b) p:0 qO r:00r1 d) p,q,r:1 5:0
p20 q 1 r21 13. a)
Pi—q r qu er (qu)A(pvr)i_qui[(pvq)A('pvr)]—>(qu)
0 0 0 0 1 0 0 1
0 0 1 O 1 O 1 1
0 1 0 1 1 1 1 1
0 1 1 1 1 1 1 1
1 0 0 1 0 0 0 1
1 0 1 1 1 1 1 1
1 1 O 1 0 O 1 1 1 1 1 1 1 1 1 From the last column of the twth table it follows that [(p V q) /\ (—p V r)] —> (q V r) is a
tautology. b) (i) Steps Reasons
1) p V (q /\ r) Premise
2) (p V q) /\ (p V 1') Step (1) and the Distributive Law of V over /\
3) 17 V r Step (2) and the Rule of Conjunctive Simpliﬁcation
4) p —> s Premise
S) —vst Step(4),p—>s<=>—'st
6) r V 5 Steps (3), (5), the Rule of Conjunction, and Resolution
(iii) Steps Reasons
1) p V q PremiSe
2) p —> r Premise
3) ﬁpvr Step(2),p—>r<=>p\/r
4) [(17 V q) /\ (—p V r)] Steps (1), (3), and the Rule of Conjunction
5) q V r Step (4) and Resolution
6) r —> s Premise
7) ﬁer Step(6),r—>s¢=>ﬁer
8) [(r V q) /\ (—ur V 5)] Steps (5), (7), the Commutative Law of V, and the Rule of Conjunction 9) q V 5 Step (8) and Resolution
(iv) Steps Reasons
1) mp V q V r Premise
2) q V ('p V r) Step (1) and the Commutative and
Associative Laws of V
3) —q Premise
4) —'q V (p V r) Step (3) and the Rule of Disjunctive
Ampliﬁcation
5) [[q V (mp V r)] /\ [—«q V (pp V r)]] Steps (2), (4), and the Rule of Conjunction
6) (—p V r) Step (5), Resolution, and the Idempotent
Law of /\
7) 'r Premise
8) —r V —'p Step (7) and the Rule of Disjunctive
Ampliﬁcation
9) [(r V —'p) /\ (r V p)] Steps (6), (8), the Commutative Law of V,
and the Rule of Conjunction
10) Ip Step (9), Resolution, and the Idempotent Law of V 58 Solutions Section 2.4—p. 100 c) Consider the following assignments. p: Jonathan has his driver’s license.
q: Jonathan’s new car is out of gas.
r: Jonathan likes to drive his new car. Then the given argument can be written in symbolic form as —‘P V q
p v —r
q V 'r —Ir
Steps Reasons
1) —'p V q Premise
2) p V —r Premise
3) (p V 'r) /\ (mp V q) Steps (2), (l), and the Rule of Conjunction
4) —r V q Step (3) and Resolution
5) q V —r Step (4) and the Commutative Law of V
6) —q V —ur Premise
7) (q V 'r) /\ ('q V —Ir) Steps (5), (6), and the Rule of Conjunction
8) w V 'r Step (7) and Resolution
9) r Step (8) and the Idempotent Law of V
1. a) False b) False c) False (1) True e) False f) False
3. Statements (a), (c), and (e) are true, and statements (b), (d), and (f) are false.
5. a) 3x [m(x) /\ C(x) /\ j(x)] True
b) 3x [s(x) /\ C(x) /\ 'm(x)] True
c) Vx [C(x) —> (m(x) Y p(x))] False
d) Vx [(g(x) /\ c(x)) —> —p(x)], or True
VX [(1706) /\ 606)) —> 1:06)], of
VX [(g(x) A p(x)) —> e600]
e) Vx [(c(x) /\ s(x)) —> (p(x) M e(x))] True
7. a) (i) 3x q(x) (ii) 3x lp(x) /\ 6106)] (iii) Vx [q (x) —> ut(x)] (iv) Vx [6106) > rim] (V) 3x [900 /\ 106)] (vi) Vx [(q(x) /\ r(x)) —> s(x)]
b) Statements (i), (ii), (v), and (vi) are true. Statements (iii) and (iv) are false; x = l0 provides
a counterexample for either statement.
c) (i) Ifx is a perfect square, then x > 0. (ii) Ifx is divisible by 4, then x is even. (iii) Ifx is divisible by 4, then x is not divisible by 5. (iv) There exists an integer that is divisible by 4, but it is not a perfect square. d) (i) Letx = 0. (iii) Letx = 20.
9. a) (i) True (ii) False Considerx = 3.
(iii) True (iv) True
c) (i) True (ii) True
(iii) True (iv) False For x = 2 or 5, the truth value of p(x) is 1
while that of r (x) is 0.
11. a) In this case the variable x is free, while the variables y, z are bound.
b) Here the variables x, y are bound; the variable z is free.
13. a) p(2, 3) /\ p(3, 3) /\ p(5, 3) b) [17(2, 2) V 17(2, 3) V 17(2» 5)] V [19(3. 2) V 11(3, 3) V 19(3, 5)] V [17(5, 2) V 13(5, 3) V [’05)] Section 2.5—p. 116 15. 17. 19. 21.
23. 25. Solutions 59 a) The proposed negation is correct and is a true statement. b) The proposed negation is wrong. A correct version of the negation is: For all rational numbers x, y, the sum x + y is rational. This correct version of the negation is a true statement. (1) The proposed negation is wrong. A correct version of the negation is: For all integers x, y, if x, y are both odd, then xy is even. The (original) statement is tnle. a) There exists an integer n such that n is not divisible by 2 but n is even (that is, not odd). b) There exist integers k, m, n such that k — m and m — n are odd, and k — n is odd. d) There exists a real number x such that Ix — 3 < 7 and either x g —4 or x 3 10. a) Statement: For all positive integers m, n, if m > n, then 1112 > nz. (TRUE) Converse: For all positive integers m, n, if m2 > nz, then m > n. (TRUE)
Inverse: For all positive integers m, n, ifm 5 n, then m2 3 n2. (TRUE)
Contrapositive: For all positive integers m, n, if m2 5 nz, then m 5 n. (TRUE) b) Statement: For all integers a, b, ifa > b, then a2 > b2. (FALSE—let a = l and b = —2.)
Converse: For all integers a, b, if a2 > b2, then a > b. (FALSE—let a = —5 and b = 3.)
Inverse: For all integers a, b, ifa 5 b, then a2 5 b2. (FALSE—let a = —5 and b = 3.)
Contrapositive: For all integers a, b, if a2 5 b2, then a f b. (FALSE—let a = l and
b = —2.) c) Statement: For all integers m, n, and p, ifm divides n and n divides p, then m divides p.
(TRUE) Converse: For all integers m and p, if m divides p, then for each integer n it follows that m
divides n and n divides p. (FALSE—let m = l, n = 2, and p = 3.) Inverse: For all integers m, n, and p, if m does not divide n or it does not divide p, then 111
does not divide p. (FALSE—let m = l, n = 2, and p = 3.) Contrapositive: For all integers in and p, ifm does not divide p, then for each integer n it follows that m does not divide n or 11 does not divide p. (TRUE) e) Statement: Vx [(x2 + 4x — 21 > 0) —> [(x > 3) v (x < #7)]] (TRUE) Converse: Vx [[(x > 3) V (x < —7)] —> (x2 + 4x — 21 > 0)] (TRUE) Inverse: Vx [(x2 + 4x — 21: O) —> [(x E 3) /\ (x 2 —7)]], or Vx [(x2 + 4x — 21 f 0) —>
(—7 E x E 3)] (TRUE) Contrapositive: Vx [[(x 5 3) /\ (x 3 —7)] —> (x2 + 4x — 2] 5 0)], or Vx [(—7 E x 5 3) —>
(x2 + 4x — 21 5 0)] (TRUE) a) True b) False c) False d) True e) False a) Va3b[a+b=b+a=0] b) Eana[an=na=a] c) Va7EOEIb[ab=ba=l] d) The statement in part (b) remains true, but the statement in part (c) is no longer true for this new universe. 3) 3x3y[(x>y)A(x—ySO)] b) 3x3y[(x<y)AVz[x2zvz2y]l . Although we may write 28 = 25 + l + l + l = 16 + 4 + 4 + 4, there is no way to express 28 as the sum of at most three perfect squares. 30=25+4+l 40=36+4 50=25+25
32=l6——16 42=25+l6+1 52=36——16
34=25——9 44=36—4+4 54=25+25+4
36:36 46=36—9+l 56=36~—16+4
38=36+l+1 48=l6+16+16 58=49~9 . a) The real number 71' is not an integer. c) All administrative directors know how to delegate authority.
d) Quadrilateral MN PQ is not equiangular. . a) When the statement 3x [p(x) V q(x)] is true, there is at least one element 0 in the prescribed universe where p(c) V q(c) is true. Hence at least one of the statements p(c), q(c)
has the truth value 1, so at least one of the statements 3x p(x) and Elx q(x) is true. Therefore, it
follows that ﬁx p(x) V Elx q(x) is true, and 3x [p(x) V q(x)] 2 Elx p(x) V Elx q(x).
Conversely, if 3x p(x) V Elx q(x) is true, then at least one of p(a), q(b) has the truth value 1, 5") Solutions 9. for some a, b in the prescribed universe. Assume without loss of generality that it is p(a). Then
[7(a) v q(a) has truth value 1 so Elx [p(x) V q(x)] is a true statement, and
316 I706) V 3x q(x) => 3x [p(x) V q(x)]~
b) First consider when the statement Vx [p(x) /\ q(x)] is true. This occurs when [9(a)ch [1(a) is
true for each a in the prescribed universe. Then p(a) is true [as is q(a)] for all a in the universe,
so the statements Vx p(x) and Vx q(x) are true. Therefore, the statement Vx p(x) A V37 q(x) is
true and Vx [p(x) A q(x)] 2: Vx p(x) A Vx q(x). Conversely, suppose that Vx p(x) /\ Vx q(x)
is a true statement. Then Vx p(x), Vx q(x) are both true. So now let 6 be any element in the
prescribed universe. Then p(c), q (c), and p(c) /\ q(c) are all true. And since c was chosen
arbitrarily, it follows that the statement Vx [p(x) A q(x)] is true, and
Vx p(x) /\ Vx q(x) :> Vx [p(x) /\ q(x)]. 1) Premise 2) Premise 3) Step (I) and the Rule of Universal Speciﬁcation 4) Step (2) and the Rule of Universal Speciﬁcation 5) Step (4) and the Rule of Conjunctive Simpliﬁcation 6) Steps (5) and (3) and Modus Ponens 7) Step (6) and the Rule of Conjunctive Simpliﬁcation 8) Step (4) and the Rule of Conjunctive Simpliﬁcation 9) Steps (7) and (8) and the Rule of Conjunction
10) Step (9) and the Rule of Universal Generalization 11. Consider the open statements w(x): x works for the credit union
€(x): x writes loan applications
C(x): x knows COBOL q(.x): x knows Excel and let r represent Roxe and 1' represent Imogene.
In symbolic form the given argument is as follows: Vx [w(x) —> c(x)]
V16 [(1006) A 1306)) —> 6106)]
Mr) /\ we)
£10) /\ —'C(i) jar) A ﬁw(i) The steps (and reasons) needed to verify this argument can now be presented. Steps Reasons
1) Vx [w(x) —> c(x)] Premise
2) q(i) A —c(z') Premise 3) 1c(i) Step (2) and the Rule of Conjunctive Simpliﬁcation
4) 112(1) —> C(i) Step (1) and the Rule of Universal Speciﬁcation 5) —uw(i) Steps (3) and (4) and Modus Tollens 6) Vx [(w(x) A Z(x)) —> q(x)] Premise 7) w(r) A —'q (r) Premise 8) WW)
9) (w(r>/\£’(r)) —> 40)
10) «women» Step (7) and the Rule of Conjunctive Simpliﬁcation
Step (6) and the Rule of Universal Speciﬁcation
Steps (8) and (9) and Modus Tollens 11) w(r) Step (7) and the Rule of Conjunctive Simpliﬁcation
12) —uw(r) V —£(r) Step (10) and DeMorgan’s Law
13) —115(r) Steps (11) and (12) and the Rule of Disjunctive Syllogism 14) '£(r) A 1w(i) Steps (13) and (5) and the Rule of Conjunction Supplementary
Exercises—p. 120 13. 15. 17. 19. 21. 23. Solutions 5” a) Contrapositive: For all integers k and 2, if k, E are not both odd, then k2 is not odd—OR,
For all integers k and E, if at least one of k, If is even, then k6 is even. Proof: Let us assume (without loss of generality) that k is even. Then k = 2c for some integer c—because of Deﬁnition 2.8. Then kZ '= (2c)€ = 2(cﬁ), by the associative law of
multiplication for integers—and (:6 is an integer. Consequently, k6 is even—once again, by
Deﬁnition 2.8. (Note that this result does not require anyming about the integer 8.) Proof: Assume that for some integer n, n2 is odd while 11 is not odd. Then 11 is even and we may
write it = 2a, for some integer a ~by Deﬁnition 2.8. Consequently, 11.2 = (201)2 = (2a)(2a) =
(2  2)(a  a), by the commutative and associative laws of multiplication for integers. Hence, we
may Write n2 = 2(2a2), with 2112 an integer—and this means that n2 is even. Thus we have
arrived at a contradiction, since we now have 112 both odd (at the start) and even. This
contradiction came about from the false assumption that n is not odd. Therefore, for every
integer n, it follows that 112 odd => it odd. Proof: (1) Since it is odd, we have n = 2a + 1 for some integer a. Then 17 + 11 = (2n + 1) + 11 =
2a + 12 = 2(a + 6), where a + 6 is an integer. So by Deﬁnition 2.8 it follows that
11 +11 is even. (2) Ifn + 11 is not even, then it is odd and we have n + 11 = 2b + 1, for some integer b. So
11  (2b 1 I) 11 — 2b 10 — 2(b 5), where b — 5 is an integer, and it follows from
Deﬁnition 2.8 that n is even—that is, not odd. (3) In this case we stay with the hypothesis — that n is odd — and also assume that n + 11 is
not evenJhence, odd. So we may mate it + 11 = 2b + l, for some integer b. This then
implies that n = 2(b — 5), for the integer b — 5. So by Deﬁnition 2.8 it follows that rt is
even. But with n both even (as shown) and odd (as in the hypothesis), we have arrived at
a contradiction, So our assumption was wrong, and it now follows that n + 11 is even for
every odd integer n. This result is not true, in general. For example, m = 4 = 22 and n = 1 = 12 are two positive
integers that are perfect squares, but 171 + n = 22 + 12 = 5 is not a perfect square. Proof: We shall prove the given result by establishing the truth of its (logically equivalent)
contrapositive. Let us consider the negation of the conclusion~that is, x < 50 and y < 50. Then with
x < 50 and y < 50 it follows that x + y < 50 + 50 = 100, and we have the negation of the
hypothesis. The given result now follows by this indirect method of proof (by the
contrapositive). Proof: 1er is odd, then it = 2k + l for some (particular) integer k. Then 7n + 8 = 7(2k + 1) +
8 — 14k .' 7 + 8 — 14k I 15 — 14k 1 14 l 1‘ 2(7k l 7) + 1. It then follows from Deﬁnition
2.8 that 711 + 8 is odd. To establish the converse, suppose that n is not odd. Then it is even, so we can write it = 2:,
for some (particular) integer I. But then 771 + 8 = 7(2t) —— 8 = l4t + 8 = 2(7t + 4), so it
follows from Deﬁnition 2.8 that 711 + 8 is even—that is, 711 + 8 is not odd. Consequently, the
converse follows by contraposition. ’—
t
/——/‘——\
p q Ls qAr I(SVr) [(qAr)—>—(svr)] p<—>t
0 O 0 0 0 1 1 O
0 0 0 1 0 O 1 0
0 O 1 O O 0 1 0
0 0 1 1 0 0 1 0
0 1 0 0 0 1 1 0
0 l O 1 O O 1 0
O 1 1 O 1 0 O 1
O l 1 l 1 O 0 1
a 512 Solutions t
r—'A—\
p q I s qAr (sVr) [(qAr)—>—(svr)] p<—>t
_l l—
1 0 O 0 0 1 1 1
1 O 0 1 0 O 1 1
1 0 1 0 O 0 1 1
1 0 1 1 0 0 1 1
1 1 O 0 0 1 1 1
1 1 0 1 0 0 1 1
1 1 1 0 1 0 0 0
1 1 1 1 1 0 0 0
3. a) p q r q<>r p<—>(q+>r) (INw) (p<>q)++r 0 0 0 1 0 1 0 0 0 1 0 1 1 I O 1 0 0 1 0 1 0 1 1 1 0 0 O 1 0 0 1 1 0 1 1 O 1 0 O 0 0 1 1 O 0 O 1 0 l 1 1 1 l 1 1 11. 13.
15. It follows from the results in columns 5 and 7 that [p 6 (q <—> r)] 42> [(p <—> q) 6 r].
b) The truth value assignments p: 0; q: 0; r: 0 result in the truth value 1 for [p —> (4 —> r)] and
the truth value 0 for [(p —> q) —> r]. Consequently, these statements are not logically equivalent. . (1) If Kaylyn does not practice her piano lessons, then she cannot go to the movies. (2) If Kaylyn is to go to the movies, then she will have to practice her piano lessons. .21) (—‘PV‘VD/WFOVWAP b) (nPanMWovm/w?
<=>(npvnq)A(pAp)
<=>(ﬁpvnq)/\p
<=>panvncD
<=><pA*p)V(p/\nq)
<=>F0V(p/\nq)
tapAnq F0 V P 42> P Idempotent Law of /\
Commutative Law of /\
Distributive Law of /\ over V
P /\ —‘P <=> F0 F0 is the identity for V. . a) contrapositive b) inverse c) contrapositive d) inverse e) converse
a) p q r qu (qu)Vr qu pV(qu)
0 0 0 0 0 O 0
O O 1 0 1 1 1
0 1 0 1 1 1 1
0 1 1 1 O O 0
l 0 O l l 0 l
1 O 1 1 0 1 0
1 1 0 0 0 1 0
l 1 1 0 1 0 1 It follows from the results in columns 5 and 7 that [(p V q) E r] <=> [p M (q \_/ r)]. b) The given statements are not logically equivalent. The truth value assignments p: 1; (/10;
r: 0 provide a counterexample. a) True b) False c) True d) True e) False f) False g) False h) True
Suppose that the 62 squares in this 8 X 8 chessboard (with two opposite missing comers) can be
covered with 31 dominos. The chessboard contains 30 blue squares and 32 white ones. Each Solutions 513 domino covers one blue and one white square —for a total of 31 blue squares and 31 white ones.
This contradiction tells us that we cannot cover this 62—square chessboard with the 31 dominos. ...
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This note was uploaded on 12/27/2011 for the course CMPSC 40 taught by Professor Egiceoclu during the Fall '09 term at UCSB.
 Fall '09
 Egiceoclu

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