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Define the connective $*$ for the Boolean variables $X$ and $Y$ as: $$X * Y = XY + X'Y'.$$ Let $Z = X * Y$. Consider the following expressions $P$, $Q$ and $R$.
$$P : X = Y * Z, \\ Q :Y = X * Z, \\ R : X *Y * Z = 1$$
Which of the following is TRUE?

  1. Only $P$ and $Q$ are valid.
  2. Only $Q$ and $R$ are valid.
  3. Only $P$ and $R$ are valid.
  4. All $P$, $Q$, $R$ are valid.
asked in Digital Logic by Veteran (59.6k points) | 1.9k views
0
Here, XNOR gate is given and XNOR gate does not obey associative law than how can someone put  X*Y instead of Z in 'R' ?

X*Y*Z = 0 always (because X*Y=Z means sum of 1's in X and Y is even and if I add Z in it than that will be even no. of 1's and that will make it odd term and is 0 always.)

If it is X*(Y*Z) than it is =1

otherwise for XNOR: x*Y*Z = 0

 

So only P & Q are valid.
0
Remarks : XNOR gate is commutative and associative.

3 Answers

+25 votes
Best answer
P:

Y * Z = Y * (X*Y)
= Y * (XY + X'Y')
= Y (XY + X'Y') + Y' (XY + X'Y')'
= XY + Y' ((X' + Y') (X+ Y))
= XY + Y' (X'Y + XY')
= XY + XY'
= X(Y + Y')
= X

So, P is valid.

Q:

X * Z = X * (X*Y)
= X * (XY + X'Y')
= X (XY + X'Y') + X' (XY + X'Y')'
= XY + X' ((X' + Y') (X+ Y))
= XY + X' (X'Y + XY')
= XY + X'Y
= Y(X + X')
= Y

So, Q is also valid.

R:

X * Y * Z = (X * Y) * (X * Y)

= (XY + X'Y') * (XY + X'Y')
= (XY + X'Y') (XY + X'Y') + (XY + X'Y')' (XY + X'Y')'
= (XY + X'Y') + (XY + X'Y')' (Since, AA = A)
 = 1 (Since A + A' = 1)

So, R is also valid.

Hence, D choice.
answered by Veteran (359k points)
selected by
0
Valid only means that it has to be a tautology p=x is contingency so jus satisfiable not valid.
+1
X = X is tautology.

X = X, Y=Y, 1 =1,
so, all are valid not just satisfiable.
0
X=x is a tautology u are right but P=x is not because p is a proposition and x is a propositional variable so value of p would depend on x and can be both 0 or 1 drpending on x being 0 or 1.
+1
where is P=x?

It is P: X = X after solving.

Similarly,
Q: Y = Y
R: 1 = 1
0
My bad.always
+1
you should have a big eye for GATE. On a PC screen, we often miss some characters.
+1
sir, why u have put x*y in place of z , how?
0
How x*y*z = (x*y)*(x*y) ?

Please explain
0
Let Z = X * Y. It is given in the question.

Now, X * Y * Z = X * Y * X * Y.

Moreover, X * Y = XY + X'Y' = X $\odot$ Y.

* is XNOR.

X*Y*Z = X $\odot$ Y $\odot$ X $\odot$ Y = (X $\odot$ X) $\odot$ (Y $\odot$ Y) = 1 $\odot$ 1 = 1
+13 votes

P: rhs=Y*Z =Y*(X*Y)=(Y*Y)*X=1*X=X=lhs  so valid

(we can use the fact that X*Y IS EQUIV TO X XNOR Y which is associative AND commutative)

Q:rhs=X*Z =X*(X*Y)=(X*X)*Y=1*Y=Y=lhs  so valid

R:X*Y*Z=X*Y*(X*Y) =(X*X)*(Y*Y)=1 so valid

answered by Loyal (8.1k points)
–1 vote
ans d)
answered by Loyal (5.2k points)
+1
They are all satisfiable but only R is valid no option matches.
Answer:

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