# GATE2014-2-53

3.5k views

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

1. $((\,a\,\to\,b\,)\,\wedge\,(\,b\,\to\,c))\,\to\,(\,a\,\to\,c)$
2. $(\,a\,\to\,c\,)\,\to\,(\,\sim b\,\to\,(a\,\wedge\,c))$
3. $(\,a\,\wedge\,b\,\wedge\,c)\,\to\,(\,c\vee\,a)$
4. $a\,\to\,(b\,\to\,a)$

edited

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, $a$ AND $c$ is false if both $a$ and $c$ are false or one of them is true and other is false.

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

So option b is not tautology..

edited
0
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 ?
0

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.

0
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?
1
now for quick check put any combination of 0 and 1 you will get 1 always.
1
thanks :)
0
welcome :)
3

@ 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)

0

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

1
Solving by boolean algebra rules is quite easy :)
$A.\ \ ((a \rightarrow b) \wedge (b \rightarrow c)) \rightarrow (a \rightarrow c)$

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

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

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

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

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

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

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

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

$\equiv T$

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

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

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

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

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

$\equiv a \vee b$

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

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

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

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

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

$\equiv T$

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

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

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

$\equiv T \vee \sim b$

$\equiv T$

Hence, Option(B) $(a \rightarrow c)\rightarrow (\sim b \rightarrow(a \wedge c))$ is the correct choice.
0
All doubts are cleared.Nice explanation.

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

Hence, B is not a tautology.

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.
we can also solve it by satisfying the T(try to make LHS true) and F(try to make RHS False) condition of ->(implication), if it satisfied then it is NOT tautology. if it is not then it is Tautology. Answer is B

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