GATE CSE 2014 Set 2 | Question: 53

Question

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

• A. $((\,a\,\to\,b\,)\,\wedge\,(\,b\,\to\,c))\,\to\,(\,a\,\to\,c)$
• B. $(\,a\,\to\,c\,)\,\to\,(\,\sim b\,\to\,(a\,\wedge\,c))$
• C. $(\,a\,\wedge\,b\,\wedge\,c)\,\to\,(\,c\vee\,a)$
• D. $a\,\to\,(b\,\to\,a)$

Answer 1

Another way to solve it...

Implication $A\to B$ is not tautology if $B$ is false and $A$ is true.

For b option Let RHS ie. $b\to (a\wedge c)$ be false ie $b$ is false and $(a\wedge c)$ is false.

Now, $a \wedge c$ is false if either one of them is false.

Now, if $a$ and $c$ both are false then $a\to c$ is true. LHS is $\text{true}$ and RHS is $\text{false}.$

So option b is not tautology.

Comments

In option C if a and b are false and c is true then the proposition should be F --> T. Then, the proposition should be false and hence it must not be tautology. Am I correct ?

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No,you are not correct.Conditional statement P->q is false only when p is true and q is false and true otherwise.

You can also check in this way-

(a∧b∧c)→(c∨a)

≡~(a∧b∧c)∨ (c∨a)

≡(~a∨~b∨~c)∨ (c∨a)

≡(~a∨a)∨(~b∨b)∨ c

≡T∨T∨c

≡T∨c

≡T

Hence,(a∧b∧c)→(c∨a) is tautology.

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i am not able to get it for option a)
a) can be reduced like this:
((a`+b)(b`+c))`+a`+c
:= ab` + bc` + a` + c

now how this is a tautology?

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now for quick check put any combination of 0 and 1 you will get 1 always.

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thanks :)

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welcome :)

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@ Aayushi ,

:= ab` + bc` + a` + c  continuing ...

 = a'+ab' + c+c'b

 = a'+b' + c+b            for details see this http://cs.stackexchange.com/questions/24587/which-law-is-this-expression-x-x-y-xy

 = a'+c + b+b'

= a'+c + 1

= 1 (Tautology)

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LeenSharma  putting 0/1 can be time consuming because we may have to check all the combination and that can be frustrating in exam.

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Solving by boolean algebra rules is quite easy :)

Answer 2

$A.\ \ ((a \rightarrow b) \wedge (b \rightarrow c)) \rightarrow (a \rightarrow c)$

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

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

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

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

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

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

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

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

$\equiv T$

 

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

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

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

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

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

 $\equiv a \vee b $
 

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

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

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

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

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

$\equiv T$

 

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

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

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

$\equiv T \vee \sim b$

$\equiv T$

 

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

Comments

All doubts are cleared.Nice explanation.

Answer 3

option B reduces to A+B rest all reduces to 1

Hence, B is not a tautology.

Answer 4

here, option a,c,d are only give T value by evaluating each parts... And option b does not give any Truth value. hence, b is the answer.