Ralph p grimaldi discrete and combinatorial mathematics ,5e (instructors solution manual) pearson addison wesley (2004)

a Here we are dealing with the permutations of 30 objects the runners taken 8 the first ways... Since there are 14 ways to subdivide the books so that each shelf has at least one book, t

INSTRUCTOR'S SOLUTIONS MANUAL

DISCRETE AND COMBINAtORIAL MATHEMATICS

FIFTH EDITION

Ralph P Gritnaldi

Rose-Hulman Institute o/Technology

Boston San Fra.'1cisco New York London Toronto Sydney Tokyo Madrid Mexico City Munich Paris Town

Reproduced by Pearson Addison- Wesley from electronic files supplied by the author

Copyright © 2004 Pearson Education, Inc

Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116

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in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the Printed in the United States of America

Dedicated to the memory of

CONTENTS PART 1

Properties of the Integel's: Mathematical Induction Relations and Functions

Languages: Finite State Machines Relations: The Second Time Around

FURTHER TOPICS IN ENUMERATION

4:

Enumeration

PART 1

FUNDAMENTALS

OF

DISCRETE MATHEMATICS

CHAPTER 1

(b) Sim~e there are eight Republica,ns and Democrats, the rule of product we have

8 X 5 ::;:;; 40 possible pairs of opposing candidates

(c) The rule of sum in part (a); the rule of product in part (b)

2 By the rule of product there are 5 x 5 x 5 x 5 x 5 x 5 = 56 license plates where the first two symbols are vowels and the last four are even digits

3 By the rule of product there are (a) 4 X 12 X 3 x 2 = 288 distinct Buicks that can be manufactured Of these, (b) 4 x 1 x 3 x 2 = 24 are blue

4 ( a) From the rule of product there are 10 X 9 X 8 X 7 = P( 10, 4) = 5040 possible slates (b) (i) There are 3 x 9 x 8 x 7 = 1512 slates where a physician is nominated for president (ii) The number of slates with exactly one physician appearing is 4 x [3 x 7 x 6 x 5] = 2520 (iii) There are 7 x 6 x 5 x 4 = 840 slates where no physician is nominated for any of the four offices Consequently, 5040 - 840 = 4200 slates include at least one physician

that there are 2 x 2 x 1 x 10 x 10 x 2 = 800 different license plates

6 (a) Here we are dealing with the permutations of 30 objects (the runners) taken 8 (the first

ways

these 6 ways, there are P(28,6) ways for the other 6 finishers (in the top 8) to finish the

two runners a.U:lon,~ the top

are (a.) 12!

re31~nCl"lOl,LI); (b) (4!)( 8!) ways so that the four ,." "'

and (c) (4!)(51)(3!) where the top , """,.,

t-are PT(:JCe:8SE:<1

3

9 (a) (14)(12) = 168

(b) (14)(12)(6)(18) = 18,144

two nonempty shelves) of the 15 books we get 15! ways to arrange the books on the two shelves Since there are 14 ways to subdivide the books so that each shelf has at least one book, the total number of ways in which Pamela can arrange her books in this manner is (14)(15!)

11 ( a) There are four roads from town A to town B and three roads from town B to town

C, so by the rule of product there are 4 x 3 = 12 roads from A to C that pass through B Since there are two roads from A to C directly, there are 12 + 2 = 14 ways in which Linda can make the trip from A to C

(b) Using the result from part (a), together with the rule of product, we find that there are 14 X 14 = 196 different round trips (from A to C and back to A)

(c) Here there are 14 x 13 = 182 round trips

15 Here we must place a,b,c,d in the positions denoted by x: e A e A e A e A e By the rule

of product there are 4! ways to do this

(b) Here we arrange the six symbols D,T,G,R,M, AAA in 6! = 720

21 (a) 121/(3!2!2!2!)

(b) [11!/(3!2!212!)] (for AG) + [11!/(31212!2!)] (for GA)

seven symbols can be arranged in (7!)/(2!21) ways Since O,O,O,I,I,A can be arranged

in (6!)/(3!2!) ways, the number of arrangements with all vowels adjacent is

[7!/(2!2!)}[61/(3!2!)]

(Case 2: The leading digit is 6) (6!)/(2!)2

(Case 3: leading digit is 7) (61)/(21)2

In total there are (6!)/(2!)][1 + (1/2) + (1/2)J = 6! = 720 such positive integers n

23 Here the solution is the number of ways we can arrange 12 objects - 4 the first type,

3 of the second, 2 of the third, and 3 of the fourth There are 12!/( 4!3t2!31) = 277,200

26 Any such path fro:m, (0,0) to (7,7) or from (2,7) to (9,14) is an arrangement of 7 R's and

7 U's There are (141)/(7!7!) such arrangements

In general I for m, n nonnegative integers, and any real numbers a, b, the number of such paths from (a, b) to (a + m, b n) is (m n)!/{m!n!)

these 10 letters and this is the number of paths

numbers and m~ n, and p are nonnegative integers,

times, while those for j and k are py';·rut.PI'I 10-5+ 1 = 6

we the instrudio:o.s in

k '1-'''', resPei:trv'elY,

5

29 (3) & (b) By rrue product print statement is executed x 6 x 8 = 576 times

9 X 105 such six-digit integers

(i) (a) (9 x 8 x 7 x 6 x 5 x 1) (for the integers euding in 0) (8 x 8 x 7 x 6 x 5 x 4) (for the integers ending in 2,4,6, or 8) = 68,800 (b) When the digits may be repeated there are 9 x 10 x 10 x 10 x 10 x 5 = 450,000 six-digit even integers

(ii) (a) (9 X 8 x 7 x 6 x 5 x 1) (for the integers ending in 0) + (8 x 8 x 7 X 6 x 5 x 1) (for the integers ending in 5) = 28,560 (b) 9 x 10 x 10 x 10 x 10 x 2 = 180,000

(iii) We use the fact that an integer is divisible by 4 if a.nd only if the integer formed by the last two digits is divisible by 4 (a) (8 x 7 x 6 x 5 x 6) (last two digits are 04, 08, 20,

40, 60, or 80) + (7 x 7 x 6 x 5 x 16) (last two digits are 12, 16, 24, 28, 32, 36, 48, 52, 56,

64, 68, 72, 76, 84, 92, or 96) = 33,600 (b) 9 x 10 x 10 x 10 x 25 = 225, O~~

32 (a) For positive integers n, k, where n = 3k, n!/(3!)k is the number of ways to arrange

the n objects Xl! Xt, Xb Xz , X2, X2,' • ,Xk, Xk, XI; This must be an integer

(b) If n,k are positive integers with n = mk, then n!/(m!)" is an integer

33 (a) With 2 choices per question there are 210 = 1024 wa.ys to answer the examination

(b) Now there are 3 choices per question and 310 ways

34 (41/2!) (No 7'8) (4!) (One 7 and one 3) + (2)(4!/21) (One 7 all.d two 3's) (4!/2!) (Two

7'13 and no 3's) (2)(41/20 (Two 7'8 and one 3) + (41/(2121» (Two 7'8 and two 3'15) The total gives us 1.02 such four-digit Intc~l!elrS

(a) 6!

6

38 The nine women can be situated around the table in 8! Each such arrangement

39

1

2

provides nine spaces (between women) where °a man can placed VVe can

of these places situate a man each of them (:)6! = {) 8 7· 6 5· 4 ways

2,438, 553, 600

procedu.re SumOfFad( i, sum: positive integers; j,k: nonnegative integers;

factorial: array [0 9] of positive integers) begin

sum := factorial [iJ + factorial [y] factorial [kl

if (100 :+: i + 10 * j + k) = sum then print (100 * i + 10 * j + k)

five letters l, a, f, and t

(b) There are C(5, 3) = S!j[3!{5 - 3)IJ = 5!/(312!) = combinations of size 3 for the five letters m, l', a, f, and t They are

a,f,m a,r ,t

a,f,r f,m,r

a,f,t f,m,t

a,m,r f,r,t

a,m,t m,r,t

six) = (15)(4) ways

set G) ways vOllSe<luelllUJI the 12

8

(;~) - an even locations for (~~) ways for 0 < i < 5 Then for the positions selected there are tvvo choices; for the 10 -

'C.u,~C:WJ,,,,.uJ'5 positions there are also two choices - 1,3

20 (a) We can select 3 vertices from A, B, C, D, E, F, G, H in (:) ways, so there are (:) = 56 distinct inscribed triangles

1

2

3

Section 1.4

Let Xi, 1 :s; i :5 5, denote the arnounts given to the five "'u.J'~'UJ.""·"'4

(5+i~-l) = G~) Here n = 5, r = 10

(b) Giving each child one dime results in the equation Xl + X2 X3 X4 X5 = 5, 0 <

Xi, 1 :5 i 5 There are e+:-1 = (:) ways distribute the remaining five dimes

X2 Xa X4 X5 = 10, 0 :5 Xi, 1 :s; i :5 4, 2 :5 X5 is the number of solutions to

Yl Yz + + Y4 + Ys = 8, O:S; Yi, 1 i :5 5, which is C'+:-1) = (~2)

Let Xi, 1 i:5 5, denote the number of candy bars for the five children with Xl

the number for the youngest (Xl = 1): X2 + Xs + X4 Xs = 14 Here there are

(4+~:-1) = (i:) distributions (Xl = 2): X2 + X3 + X4 + X5 = 13 Here the number of distributions is (4+!~-1) = G:) The answer is G~) G:) by the rule of sum

(~H20-1) _ (23)

20 - 20

+1122-1) = G;)

(c) There are 31 ways to have 12 cones with the same flavor So there are (:;) - 31 ways

to order the 12 cones and have at least two flavors

10 Here we want the number of integer for Xl X2 X3 + X4 X.r; X6 = 100,

Xi ?: 3, 1 :s; i :s; 6 (For 1 ::; i :s; 6, Xi counts the number of times the face with i

dots is rolled.) This is equal to the numher of nonnegative integer solutions there are to

'113 +'115+116 = > 1:S i Consequently the answer is e+:;-l) = (:;)

11 (a) ('0+ 55-1) = e}4) (h) ('+:-1) SC'+:-l) + 3(7+;-1) + e+~~l) =

en -t 3(~O) 3(:) (~), where the first summand accounts the case where none of

12 (a) The number of solutions for Xl + X2 + X5 < 40, Xi ?: 0, 1 < i 5, is the same as the number for Xl + X2 ••• + X5 < 39, Xi ?: 0, 1 :S i < 5, and this equals the number of solutions for Xl + X2 ••• + X5 + X6 = 39, Xi 2: 0, 1 :S i ::; 6 There are C,+:-l) = (::) such solutions

(b) Let 'IIi = Xi 3, 1 :S i :5 5, and consider the inequality '111 + '112 + + '115 :s; 54, 'IIi > O

) = G)

(b) e+~-l) (container 4 has one marble) e+!-l) (container 4 has three marbles)

e+~-l) (container 4 has five marbles) + (3+~-1) (container 4 has seven marbles)

= Lr:o (~=~~)

(b) The terms in the expansion have the forID vaw b xCyd ze where a, h, c, d, e are

Consider one sum dist:dbutioll - the one where

shelves Here there are 24! ways for this to happen

18

19

20

WI W:;t + Ws + W19 = n -19, where Wi 0 for alll

(:"::-1~)' The number of positive integer solutions equation

ative integer solutions for

Here there are r = 4 nested for loops, so 1 < m ::; k ::; j ::; i ::; 20 We are making selections, with repetition, of size r = 4 from a collection of size n = 20 Hence the print statement is executed e O +,t 1) = (~3) times

Here there are r = 3 nested for loops and 1::; i ::; j < k ::; 15 So we are making selections, with repetition, of size r = 3 from a collection of size n = 15 Therefore the statement

counter := COftnier + 1

is executed C5+aS-l) = en times, and the final value of the variable counteris 10+ e:) =

690

this segment the value of the variable Bum is i = (220)(221 )/2 = 24,310

for i := 0 to 10 do

end

for j := 0 to 10 - i do

10 - t - J) .\

Xl + X2 + X3 X4 = 4, where -2::; Xi for 1::; i :5 4, is the number of integer solutions

to !II + !l2 + '!J3 - where !Ii 2:: 0 1 ::; i < 4 use this observation the following

procedure Selectio'n$2(i,j,k: nonnegative integers)

begin

for i := 0 to 12 do

end

for j := 0 to 12 - i do for k := 0 to 12 - i - j do

print (i,j,k, 12 - i - j - k)

25 If the smmnands must all be even, then consider one such composition - say,

26

20 = 10 + 4 + 2 + 4 = 2(5 + 2 + 1 + 2)

Here we notice that 5 + 2 + 1 + 2 provides a composition 10 Further, each composition

of 10, when multiplied through by 2, provides a composition 20, where each summand is

even, equals the nUlnher of compositions of 10 ~ namely, 210-1 = 29

the run we number of solutions for Xl

Xz + X3 X4 = 12, whe,re Xl + xa = 5, XI, X3 > 0 and X2 X", = XZ, X4 > This number is e+~-l) e+:-1) = (!) (:) = 4·6 = \Vhen the first run consists of tails we get (:) (:) = 6 4 = 24 arrangements

all there are 2(24) = 48 arrangements with four runs

d) If the first run starts with an H, thel1 we need the number of integer solutions for

Xl + Xz + Xs + X4 + Xs = Xl + Xs + X5 = Xl) Xa, X5 > 0 and X2 + X4 = 7,

Xz, X4 > O This is e+~-l) C+:-l) = (:) (:) = 36 For the case where the first run starts with a T, the number of arrangements is e+:-1) e+;-l) = (!) = 60

hI total tha'e are 36 60 = 96 ways for these 12 t()sses to determine five runs

e) e+:- 1) e+~-l) = (:) G) = 90 - the number of arrangel'uents which result in six runs, if the first run starts with an H But this is also the number when the first run starts with

a T Consequently, six runs come about in 2·90 = 180 ways

f) 2C+:-l) e~:-1)+2e+~-1) e+:-1

) +2(3+;-1) e+:-1)+2(4+!-1) (4+~-1)+2(5+6-1) e+~-l) =

2 L,t:::o (i~i) (6~i) = 2fl 1 4 () () 15 + 4 20 + 1 15] = 420

28 (a) For n 4, cIOn sider the strings made up of n bits - that is, a total of nO's and 1's

In particular, c()nsider those strings where there are (exactly) two ()ccurrences ()f 01 For example, if n = 6 we want to include strings such a ~ 010010 and 100101, but not 101111

or 010101 How many such strings are there?

(b) For n ~ 6, how many strings ()f nO's and l's contain (exactly) three occurrences of 01?

( c) Provide a combinatorial proof for the foll()wing:

For n ~ 1, 2ft = (n ~ 1) + (n 3 1)

( a) type of :1::1 followed X2 followed X3 1'8 foll()wed by

x" O's followed by Xs 1 '8 followed by X6 O's, where,

Xs > O •

"n",,,,.,,, the of ~J""""""""lD

_ (n+l)

- 'i • ( C) There are strings n 1 strings where there are k followed by n ,- k

for k = 0, '/: n n 1 strings contain no occur:rence~ of 01, so there are

(n;l) strings that contain (exactly) one occurrence of 01, (nil) strings with (exactly) two occurrences, (n~l) strings with (exactly) three occurrences, ) and for

n odd, we can have at most 1'1.;1 occurrences of 01 number of strings with occurrences of 01 is the number of integer solutions for

This is the same as the number of integer solutions for

1116+1 = n - (n - 1) = 1, where !II, 112, ••• ,111'1.+1 0

This number is (n+1)+1-1 1) = (n+l) _ (1'1,+1) _ ( n+l ) 1 - n - 2( ~)+1

(ii) n even, we can have at most ~ occurrences of The number of strings n 2" occurrences of 01 is the number of integer solutions for

This is the same as the number of integer solutions for

YI + Y2 +,., + YnH = n - n = 0, where Y' ~ 0 for 1 :::; i :::; n + 2

This number is (1'1.+2)+0-1) () = (1'1,+1) _ () - n+1 - ("'+1) _ ( 2(~)+1 1'1,+1 ) •

{ ~;~;~: : odd

follows

blO = 16796 bs)j 14(=

(b) For n ;:: 0 there are b n ( = (~) such from (0,0) to (n,n)

(b) (i) «(ab(c(de H> «(ab)(c(de»)f)

Oi) «ab((cd(e H> «ab)«cd)(ef»)

(iii) (a«(bc(de H> (a«(bc)(de»f»

9 (i) When n = 4 there are 14(= b 4 ) such diagrams

(li) For any n 0, there are b n different drawings of n semicircles on and above a horizontal

line, with no two semicircles intersecting Consider, for instance, the diagram part (f)

of the figure Going from left to right, write 1 the first time you encounter a semicircle

Here condition is violated, for the first time, after the third U Transform the

Here the entries up to and including the first violation remain unchanged, while those

cOITesponcience sbows us that the number of paths that violate the given condition the

such paths

Consequently, the answer is

(10) _ (10) _ lQi _ JQl _ 10!(S) _ 1O!(3} = (2) 1O! _ (1±1-3) (10)

[Note that when m = n, this becomes (n!l)(~)' the formula for the nth Catalan number.]

11 Consider one oUlle C~!l) e~6) = n) e:) ways in which the $5 and $10 bills can be arranged

- say,

(*) $5, $5, $10, $5, $5, $10, $10, $10, $5, $5, $10, $10

Here we consider the six $5 bills as indistinguishable -likewise, for the six $10 bills ever, we consider the patrons as distinct Hence, there are 6! ways for tbe six patrons, each with a $5 bill, to occupy positions 1, 2~ 4, 5, 9, and 10, in the arrangement (*) Likewise, there are 6f ways to locate the othe:r six patrons (each with a $10 hill) Consequently, here the number of arrangements is

How-seen in

Consequently, the largest UW,UVPJl

4 (a) e:t

(b) 3 e15) :I (~) (four hymns from one book, one from each the other two) 6 e15) e:) e:)

last (25th) Hence there are (34!)/(91) possible arrangements

(c) There are 251 ways to arrange the flags For each arrange:tnent consider the 24 spaces, one between each pair of flags Selecting 9 of these spaces provides a distribution among the 10 flagpoles where every flagpole has at least one flag and order is relevant Hence

of the first head; (2) One position between the i-th head and the (i l)-st head, where

1 SiS 44; and, (3) One position to the right of the 45-th (last) head To answer the question posed we need to select 15 of the 46 positions This we can do in (::) ways

In an alternate way, let Xi denote the number of heads to the left of the i-th tail, for

1 siS 15 Let X16 denote the number heads to the right of the 15th taiL Then we want the number of integer solutions for

Xl + XzXa + + Xli; + XIS = 45, where Xl > 0, Xi6 2 0, and Xi > 0 for 2 sis 15 This is th.e number of integer solutions for

(i) S12~e: there are 1 x 2 = 2

(li) Material, color: pair yields 1 X 4 = 4 such blocks

(iii) Material, shape: For this pair we obtain 1 x 5 := 5 such blocks

(iv) Size, we get 2 X 4 = 8 of the ULV !\.O

(v) Size, shape: This pair gives us 2 x 5 = 10 such blocks

(vi) Color, shape: this pair we find 4 x 5 = 20 the blocks we need count

large blue plfUltic hexagonal block in exactly two ways

10 Sinc e 'R' is the 18th letter of

en = (17)(16)/2 = 136 ways

alphabet, the first and middle initials can be chosen in

Alternately, since 'R' is the 18th letter of the alphabet, consider what happens when the middle initial is any letter between 'B' and 'Q' For middle initial 'Q' there are 16 possible first initials For middle initial 'P' there are 15 possible choices Continuing back to 'B' where there is only one choice (namely' A') for the first initial, we that the total

11 The number of linear arrangements of the 11 horses is 11!/(513!31) Each circular ment represents 11Unear arrangements, so there are (1/11)[111/(5!313!)J ways to arrange the horses on the carousel

UGAU."O can in 4 x 3 = 12 ways, the

348,364,800

can be arranged as a decreasing :l:O'!.U'-(l1f!lt UM,"-'.",,"'oI

'To complete the solution we ZIlust account for the decreasing four-digit integers where the

Consequently there are 2 (!) (:) = 343 such four-digit integers

(b) For each ~ondecrea8ing four-digit integer we four nonzero digits, with repetitions allowed These four digits can be selected e+!-l) = en ways And these same four digits account for a nonincreasing four-digit integer, So at this point we have 2cn - 9 of the four-digit integers we waut to count (The reason we subtract 9 because we have counted the nine integers 2222,3333, " 9999 twice 2(~2).)

We have not accounted for those nonincreasing four-digit integers where the units digit is

O There are Cil+ 3S-1) - 1 = (;2) - 1 of these four-digit integers (Here we subtracted 1

since we do not want to include 0000.)

Therefore there are [2C42) - 9} + {(;2) 1] = [2e,n (~)] - 10 = 1200 such four-digit integers

16 (a) (~,~,2)(1/2)2( _3)2 = 135/2

(b) Each term is of the form :e nt

yn2 zrl.S where each ni, 1 < i :s; 3, is a nonnegative integer and nl + n2 + n3 = 5 Consequently, there are e+:-1) = G) terms

e c ) Replace x, y, and z by 1 Then the sum of all the coefficients in the expansion is

«1/2) + 1 - 3)5 = (-3/2)5

17 (a) First place person A at the table There are five distinguishable places available for

A (e.g., any of the positions occupied by A,B,C,D,E in Fig 1.11(a» Then position the other nine people relative to A This can be done in 9! ways, so there are (5)(91) seating arrangements

(b) There are three distinct ways to position A,B 80 that are seated on longer sides

different ways, so the total nw::-qber of arrangements is (3)(8!)

18 (a) For Xl 3''>;1 + X3 = 6 are e+:-1

) = (:)nonnegative ultel:€~'

solutions for X4 - The number of <>'-P ",."" '''' of

can win sets e) ways, soores can be """I"nr-d"l1'>F1

Since B may be the winner, the final answer is 2[(~)74 (~)75]

(~)74 ways (:) 75] ways

20 We can choose r objects from ,n in (:) ways Once the r objects are selected they can

of the n objects taken 1" at a

21 For every positive integer n, 0 == (1 - l)n = (~)(1)o - (~)(1)1 + (;)(1Y' """' (;)(1)3 +

fill her bookshelf

(b)' There are en ways in which Francesca can select nine othe~ books Then she can arrange those nine books and the three books on tennis on her bookshelf in 12! ways

include Francesca's three books on tennis

10+(12-1+1)(1"-1 1)(2) [3+4 +(s-3+1)](4)+(12-3+1)(6)+(t-7 1)(8) ==

10 + (12)(1')(2) [(1/2)(8 - 3 + 1)(8 - 3 + 2) - 2 - 1}(4) + (10)(6) + (t - 6)(8) =

22 24,., + 8t + 2(8 - 2)(8 -1) - 12 = 14 + 24r + 8t 2S(8 - 3)

For 2k + 1 1 '5, where 0:::; k 8, there are 2k + 2 locations to select, with repetitions allowed The selection size is the number of 2'£1, which is (1/2)[17 - (2k + 1)) = 8 - k The selection can be made in (2k+2+(S-k)-1) a-Ie = (9+k) wavg and so the answer is ",8_ (9+k) =

29 1

5- 1 = 4 horizontal moves

in 111/(4!71) ways

move diagonal moves is hetween 0

Hexe we want certain paths from (1,1) to (14,4) where the moves are of the

(m, ~ (m + 1, n + 1), ifthe (n + l)-st ballot is for Katalin

(m, -+ + 1,11 1)) iHhc (n + l)~st ballot is for Donna

paths are ollesthat never touch or cross the horizontal

", !I""".,,,,,, pair (Tn, n) here indicates that m ballots have been counted

hy n votes number of ways to count the ballots according to

answer IS

33 are (:) = wa.ys to choose the

FrY[' of these choices ofiour (lUarh'!I's, there are 12·11, 10·9 ways to """"",,'p.,u

in total, there axe (~) ·12 ·11 ' 10·9 = 178,200 ways for to

Consider the as one unit Then we are trying to arra:n.ge

family and the eight other people around the table This can be dOllEl in

the family unit can be arranged in four WdYS, total number

'U'~"'f'"""'U"T' conditions is 4(8!)

Section 2.1

2 FUNDAMENTALS OF LOGIC

1 The sentences in parts (a), (c), (d), and (f) are statements

2 The statements in parts (a), (c), and (f) are primitive statements

3 Since p -+ q is false the truth value for p is 1 and that of q is O Consequently, the truth values for the given compound statements are

If triangle ABC is equilateral, then it is isosceles

If triangle ABC is not isosceles, then it is not equilateral

Triangle ABC is equilateral if and only if it is equiangular

Triangle ABC is isosceles but it is not equilateral

If triangle ABC is equiangular, then it is isosceles

(d) 0 (c) (sAr)-+q

12 (a) [(p 1\ q) 1\ rJ -} (8 V t) is false (0) when (p 1\ q) 1\ r is true (1) and s V t is false (0)

Hence p, q, and r must be true (1) while s and t must be false (0)

27

statements

guilty statement (2» and Tyler guilty (from statement (3»

again statements (3) and (4) do not contradict each other, and here we learn from statement

From part (iii) of part (a)

By the 2nd Substitution Rule, and (p -l> {::=:;} (.p V

By the 1st Substitution Rule, and (8 -l> t) {::=::} ( .t -l> -'s), for primitive statements s, t

and the 2nd Substitution Rule

2nd Substitution Rule

3 a) For a primitive statement 3, 8 V -'8 {=:::} To Replace each occurrence of s by

p V (q A r) and the result follows by the 1st Substitution Rule

b) For primitive statements s, t we have (s -l> t) ¢=} (-,t -l> -'8) Replace each occurrence of .5 by p V q, and each occurrence of t by r, and the result is a consequence

of the 1st Substitution Rule

4 (1) [(p A '1) A rJ V [(p A q) A -,r] {=} (p A q) A (r V -,r) ¢=} (p A q) A To ¢=} P A q

(2) [(p A q) V """Iq] {=:::} (p V .q) A (q V .q) {=:::} (p V -,q) A To ¢=} p V -'q

Therefore, given statement simplifies to (p V .q) -l> S or (q -l> p) -l> S

(b) p -;.(qAr){:::::? V(qAr), so(p -;.(qAr)]d{:=} ĂqVr)

(c) p +-7c q {:::::? (p -; q)/\(q -7 p) {:::::=? (-.pVq)ẶqVp), so (p +-7c q)d {:::=::> (-,pAq)V( qAp)

(d) P)/Jl ¢=::} (p A -,q) V ("'p /\ q), 80 (p'ịq)d {:::=::> (p V q) A (,p V q)

9 (a) If 0 0 = 0, then 2 + 2 = l

Let p : 0 + 0 = 0, q : 1 1 : 1

(The implication: p + q) - If 0 + 0 = 0, then 1 + 1 = 1 - Falsẹ

(The Converse of p -; q: q -7 p) - If 1 + 1 = 1, then 0 0 = Ọ - True

(The Inverse of p -7 q: -7 -.q) - If 0 + 0 =1= 0, then 1 + 1 =1= 1 - True

(The Contrapositive of p -7 q: -.q -7 p) - If 1 + 1 =1= 1, then 0 + 0 =1= ọ - False

(b) If -1 < 3 and 3 + 7 = 10, then sine;) = -1 (TRUE)

Converse: If sine;) = -1, then < 3 and 3 + 7 = 10 (TRUE)

Inverse: If -1 :;::: 3 or 3 + 7 =1= 10, then sine;) =1= -1 (TRUE)

Contrapositive: If sin(~) =1= -1, then -1 :;::: 3 or 3 7 =1= 10

So the given statement is not a tautology If we try to apply the second substitution rule

to the result in part (a) we would replace the first occurrence of p by p V q But this

does not result in a tautology because it is not a valid application of this substitution rule

- for p is not logically equivalent to p V q

q /\ {:> Fo (Inverse Law)

1'V Fo {:> l' (Identity Law) Reasons

Absorption Law (and the

Reasons Absorption Law

Reasons DeMorgan's Laws

Law Law

20 1\ (-,t' V q V -,q)J V V t V -'1') 1\ -,q) ¢::;::::> [p A (-'1' V V [(To V t) 1\

(1' 1\ V 1\ ~ l' V -q

(b) V (p A V (1' /\ q /\ /\ [{p /\ r 1\ V t] <¢=> P /\ t by the Absorption

rows of table are for assessing argument are rows

stal~en:Jlent q V s value 0 only each of 8 truth Then

(p -j> q) has truth value 1 when p has truth value -j> oS) has truth value 1 when r

(e) For (-,p V the truth value is 0 when both p, r have truth value 1 This then forces '1, s to have truth 'value 1, for (p -} q), ('1' -} to have value 1

(c) This is a repeat-until loop (Modus Ponens)

( d) Tim watched television ill the evening (Modus Tollens)

5 ( a) Rule of Conjunctive Simplification

(b) Invalid - attempt to argue by the converse

( c) Modus ToUens

(d) Rule of Disjunctive Syllogism

(e) Invalid - attempt to argue by the inverse

Step (2) and the Rule of Disjunctive Amplification Consequently, (q 1\ r) -} (q V r) is a tautology, or q A r =? q V r

(b) Consider the truth value assignments p : 0, q : 1, and r : O For these assignments

[p A (q A r)} V -,[p V (q 1\ r)] has truth value 1, while [p A (q V r)] V -,[p V (q V r)J has truth value O Therefore, P -j> PI is not a tautology, or P '::fo Pl'

(4)

Steps (6), (7) and the Rule of Disjunctive (8)

of Detachment

(5) Rule of Conjunctive

Rule

(11)

(12) Step [(-,p V q) -4 <====} [-,." -4 V q)]

9 (a)

(2) Step (1) and -,(- ,q -4 8) {=} -{"""q V 8) {=} (q V 8) {:.=.} -'q A ,s

-4, ,rV 8

r-4,fJ

(2)