TEAM 14

EX:-2.1

(QNO:-14) At the start of a program the integer variable n is assigned the value 7. Determine the value of n after each of the following successive statements is encountered during the execution of this program.[Here the value of n following the execution of the statement in part(a) becomes the value of n for the statement in part(b),and so on,through  the statement in part(d).For positive integer a,b[a/b] returns the integer part of the quotient-for example,[6/2]=3,[2/5]=0,and[8/3]=2.]

a)      if n>5 then n:= n + 2

b)      if((n+2=8) or (n-3=6)) then n:= 2 * n + 1

c)      if((n-3=16) and ([n/6]= 1)) then n:=  n + 3

d)     if((n≠21) and (n-7= 15)) then n:=  n – 4

Solution:-

a)      n > 5 and n=7 (given)

so n = n + 2

             = 7 + 2

             = 9

b)      n + 2 = 9 + 2 =11 ≠ 8 (false)

      Or

      n – 3 = 9 – 3 = 6 = 6 (true)

      so n = 2 * n + 1

             = 2 * 9 + 1

             = 18 + 1

             = 19

c)      n-3=16  (given)

19-3=16 (true)

And

      [n/6] = 1  (given)

[19/6]= 3 ≠ 1 (false)

Hence n = 19

d)     (n≠21)  given

n =19 ≠ 21 (true)

And

n-7 = 15  (given)

19 – 7 =12 ≠ 15 (false)

Hence n = 19

EX:-2.2

(QNO.14) For primitive statement p,q          

a)      Verify that p→[q→(p۸q)] is a tautology.

b)      Verify that (p۷q)→[q→q] is a tautology by using the result from part (a) along with the substitution rules and the law of logic.

c)      Is (p۷q)→[q→(p۸q)] a tautology?

Solution:-

p	q	p۸q	q→(p۸q)	p→[q→(p۸q)]
1	1	1	1	1
1	0	0	1	1
0	1	0	0	1
0	0	0	1	1

             a)

                  Hence the given compound statement is a tautology.

             b) Taking the compound statement from part (a)

                 p→[q→(p۸q)]

              = q→[q→(q۸q)]    (by substitution rule)

              = q→[q→q]           (by idempotent law)

                Take the compound statement from part (b)

                And let this statement is a tautology then

               (p۷q)→[q→q]

             = (q۷q)→[q→q]       (by substitution rule)

             = q→[q→q]              (by idempotent law)

                Hence both are logically equivalence

                So assumption is true.

                Thus (p۷q)→[q→q] is a tautology.

            c)

p	q	(p۸q)	q→(p۸q)	(p۷q)	(p۷q)→[q→(p۸q)]
1	1	1	1	1	1
1	0	0	1	1	1
0	1	0	0	1	0
0	0	0	1	0	1

                     

                           Hence the given compound statement is not a tautology.

EX:-2.4

(QNO:-7) For the universe of all integer,let p(x),q(x),r(x),s(x),and t(x) be the following open statements.

P(x): x > 0

q(x): x is even

r(x): x is a perfect square

s(x): x is (exactly) divisible by 4

t(x): x is (exactly) divisible by 5                    

a) Write the following statements in symbolic form.

   (1) At least one integer is even.

   (2) There exists a positive integer that is even.

   (3) If x is even, then x is not divisible by 5.

   (4) No even integer is divisible by 5.

   (5) There exists an even integer divisible by 5.

   (6) If x is even and x is a perfect square, then x is divisible by 4.

b) Determine whether each of the six statements in part (a) is true or false. For each false statement, provide a counterexample.

c) Express each of the following symbolic representations in words.

    1) For All x [r(x)→p(x)]             (2) For All x [s(x)→q(x)]

    3) For All x [s(x)→ ¬t(x)]           (4) For All x [s(x)۸ ¬r(x)]

d) Provide a counterexample for each false statement in part (c)

Solution:-

  a) (1) For Some x q(x)

      (2) For Some x [p(x)۸q(x)]

      (3) For All x [q(x)→ ¬t(x)]

      (4) For All x [q(x)→ ¬t(x)]

      (5) For Some x [q(x)۸t(x)]

      (6) For All x [q(x)۸r(x)→s(x)].

  b) (1) True

      (2) True

      (3) False (ex:- 10,20,30)

      (4) False (ex:- 10,20,30)

      (5) True

      (6) True

            c) (1) If x is a perfect square, then x > 0.

                (2) If x is divisible by 4 then x is a even no.

.               (3) If x is divisible by 4 then x is not divisible by 5.

                (4) There exists an integer divisible by 4 but it is not a perfect square)

  d) (1) False (ex:-0)

      (2) True

      (3) False (ex:- 20,60)

      (4) True

.

EX:-3.1

(QNO:-14) Give an example of three sets w,x,y such that w belongs to x and x belongs to y but w belongs to y.

Solution:-

               If  w belongs to x

                means all the elements of w will be in x.

.              If  x belongs to y

                means all the elements of x will be in y.

                hence all the elements of w will be in y . .

.             Therefore x belongs to y

               So, there is no such example.

SUBMITTED BY:- GROUP 14