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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. $a\rightarrow (b\rightarrow a)$
in Mathematical Logic 4.1k views

9 Answers

9 votes
 
Best answer
$A.\ \ ((a \rightarrow b) \wedge (b \rightarrow c)) \rightarrow (a \rightarrow c)$

$\equiv (( \sim a \vee b) \wedge (\sim b \vee c)) \rightarrow (\sim a \vee c)$

$\equiv \sim (( \sim a \vee b) \wedge (\sim b \vee c)) \vee (\sim a \vee c)$

$\equiv (( a \ \wedge \sim b) \vee ( b \wedge \sim c)) \vee (\sim a \vee c)$

$\equiv (\sim a \vee ( a \ \wedge \sim b) )\vee( ( b \wedge \sim c) \vee c)$

$\equiv( (\sim a \vee a )\wedge(\sim a \vee \sim b) )\vee( ( b \vee c) \wedge( \sim c\vee c))$

$\equiv(T\wedge(\sim a \vee \sim b) )\vee( ( b \vee c) \wedge T)$

$\equiv\sim a \vee (\sim b \vee b) \vee c$

$\equiv\sim a \vee T \vee c$

$\equiv T$

 

$B.\ \ (a \leftrightarrow c)\rightarrow (\sim b \rightarrow(a \wedge c))$

$\equiv ((a \rightarrow c)\wedge(c \rightarrow a))\rightarrow ((\sim b \rightarrow(a \wedge c))$

$\equiv \sim((\sim a \vee c)\wedge(\sim c \vee a))\vee (( b \vee (a \wedge c))$

$\equiv\sim (( a \wedge c)(\sim a \wedge \sim c) )\vee (( b \vee (a \wedge c))$

$\equiv (( a \wedge \sim c)\vee(\sim a \wedge c) )\vee (( b \vee (a \wedge c))$

$\equiv (( a \wedge \sim c)\vee c (\sim a \wedge a) )\vee b$

$\equiv (( a \wedge \sim c)\vee c\vee b$

$\equiv a \vee b \vee c $

 

$C. \ \ (a\wedge b \wedge c) \rightarrow(c \vee a)$

$\equiv \sim(a\wedge b \wedge c) \vee (c \vee a)$

$\equiv \sim a \sim b \sim c \vee c \vee a$

$\equiv (a\vee\sim a )\vee \sim b \vee(\sim c \vee c)$

$\equiv T\vee \sim b \vee T$

$\equiv T$

 

$D.\ \ a\rightarrow (b\rightarrow a)$

$\equiv\sim a \vee (\sim b \vee a)$

$\equiv(\sim a\vee a)\vee \sim b$

$\equiv T \vee \sim b$

$\equiv T$

 

Hence, Option(B) $(a \leftrightarrow c)\rightarrow (\sim b \rightarrow(a \wedge c))$ is the correct choice.

selected by
9 votes

for A → B condition to be invalid we have to show that B is false when A is true.

So we assume that A ( left side of  → ) is true,  then we note the value of B (right side of  )  if it is false then A → B is not a tautology . 

a)     ((a → b) ^ (b → c)) →  (a → c)

left side of   '→'   is ((a → b) ^ (b → c)) which we assume to be true . for this to be true     both (a → b)  and (b → c) must be true. now if 'a' is false then right side (a → c)  is true so there is need to check bcoz we have to show right side as false in case of fallacy

 let 'a' is true then 'b' and  'c' must be true for left side ((a → b) ^ (b → c)) to be true .now as 'a' and 'c' is true (a → c) is true.

 so it is a tautology.

c )    (a^b^c)→(c v a))

 for left side (a^b^c) to be true 'a' ,'b' and 'c' must be true so right side (c v a) is also true.so it is a tautology.

d)     a→(b→a)

for left side to be true 'a'  must be true then (b→a) is also true bcoz we assume 'a' to be true.  so it is a tautology.

 b)     (a ↔ c) →(~b→(a^c))

 for left side to be true both ' a' and ' c' either be true or  false.

In case when both are 0 then (a^c) is 0.so for  b=0, a=0 and c=0 right side is false even if left side is true.As there is a case  when 1→0  so it is not a tautology 


edited by
5 votes
Option B is not tautology.

$(a \Leftrightarrow c) \rightarrow (b{}'\rightarrow (a\wedge c))$

$((a{}'\vee c)(a\vee c{}')) \rightarrow (b{}'\rightarrow (a\wedge c))$

$(ac+a{}'c{}')) \rightarrow (b{}'\rightarrow (a\wedge c))$

$(ac+a{}'c{}')){}' + (b{}'\rightarrow (a\wedge c))$

$((a{}'+c{}')(a+c)) + (b{}'\rightarrow (a\wedge c))$

$(ac{}'+a{}'c) + (b{}'\rightarrow (a\wedge c))$

$(ac{}'+a{}'c) + (b+ (a\wedge c))$

$(ac{}'+a{}'c) + (b+ a c)$

$(ac{}'+(a{}'+ a )c)+b$

$ac{}'+c+b$

$(ac{}'+c)+b$

$(a+c)+b$

$a+b+c$
0

HI
Can you please explain me how you are deducing these below steps . specificly on LHS side.

(ac)→(b ′ →(ac))
((a ′ ∨c)(ac ′ ))→(b ′ →(ac)) 
(ac+a ′ c ′ ))→(b ′ →(ac)) 

0
$a \Leftrightarrow c=\left ( \left ( a\rightarrow c \right )\left ( c\rightarrow a \right ) \right )$
                              =$\left ( \left ( a{}'+ c \right )\left ( c{}' +a \right ) \right )$
                              =$\left ( a{}'c{}'+a c \right )$
3 votes

A:   ((a→b)∧(b→c))→(a→c)

approach - find an assignment for which above statement becomes false.

The above statement becomes false only when  ((a→b)∧(b→c)) = T and (a→c) = F

to make (a→c) false .  put c=F and a=T.

so T→F = F

(a→b) will be true only when b=T.

by putting b = T

(b→c) becomes false which makes (a→b)∧(b→c) false 

so F→F = T

so there is no such assignment which makes ((a→b)∧(b→c))→(a→c) false . So its a tautology

C. (a∧b∧c)→(c∨a)

for the assignment a=F and c=F

(c∨a) becomes False.

but (a∧b∧c) also becomes false.

F→F =

so  there is no such assignment which makes ((a→b)∧(b→c))→(a→c) false . So it is also tautology.

D.  a→(b→a)

= ∼av(∼bva)

= ∼a v a v∼b

=T v ∼b

= T

it is also tautology. 

B. (a↔c)→(∼b→(a∧c))

for the assignment a=F b=F c=F the above statement becomes false so it is not a tautology.

So B is the correct option.

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 vote
Answer B??
0
I also got B :)
0
Solution steps please
0
Option A is a Tautology by Hypothetical Syllogism.Just refer Rosen.

Option C can be proven by simplification. If a,b,c are true then defenitely c or a will be true.

In the case of Option D , just expand it you will get a ,b and negation a. with OR gate inbetween them. 'a' and 'negation a' will give True. Anything 'OR'ed with True will be true. Hence that will also become a Tautology.
1

Option B

1 vote

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
(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

edited by
Answer:

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