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12 votes
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Which one of the following Boolean expressions is NOT a tautology?

  1. $((a \rightarrow b) \wedge (b \rightarrow c)) \rightarrow  (a \rightarrow c)$
  2. $(a \leftrightarrow c) \rightarrow (\sim b\rightarrow (a\wedge c))$
  3. $(a\wedge b \wedge c)\rightarrow (c \vee a)$
  4. $a\rightarrow (b\rightarrow a)$
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9 Answers

2 votes
2 votes
Option B

Solution:

(a <-> c) -> (~b -> (a˄c))

For the expression to be Tautology, T -> F should not arise. So taking a = False, c= False and ~b = True,

(~b -> (a˄c)) = False

and (a <-> c)  = True

So, True -> False, scenario is arising. So option B is the correct answer.

For all other expressions, True-> False scenario will not arise.
1 votes
1 votes
Answer B??
1 votes
1 votes

Option B is right choice it can be calculated from several methods:-like table method ,T->F method(The only condition which is false in implies so we focus to get T->F in this method),and may be others

0 votes
0 votes
(A)    (a → b) ^ (b → c)) →  (a → c)

         (a'+b)(b'+c) →(a'+c)

        ((a'+b)(b'+c))'+(a'+c)

            ab'+bc'+a'+c

         (ab'+a')+(bc'+c)

          (a'+b')+(c+b)

          1+a'+c      (becoz (b+b')=1

          1(true)......    so tautology

(C)        (a^b^c)→(c v a)

            abc→(c+a)

           (abc)'+c+a

             a'+b'+c'+c+a
 
            1+b'           (becoz (a+a')=1 and c+c' =1

             1(true)          so tautology

(D)         a→(b→a)

           a→(b'+a)
 
           a'+b+a

            1+b

              1(true)           so tautology

(B)              (a ↔ c) →(~b→(a^c))

                  (ac+a'c')→(b'→ac)

                    (ac+a'c')→(b+ac)

                    (ac+a'c')'+(b+ac)

                  (ac)'(a'c')'+b+ac

                 (a'+c')(a+c)+b+ac

                 a'c+ac'+b+ac

                  (a'c+ac)+(ac'+ac)+b           (becoz a+a+a+a+....+a =a)

                  c(a+a')+a(c+c')+b

                  a+b+c       so not a tautology
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