Boolean Algebra Questions and Answers

Answer: A

The degree of the function is 4. The function F(A,B,C,D) = A`B`C`D` is a single product term in which each of the four variables appears once, in complemented form, so the term has four literals and its degree is 4. The values 8, 0 and 5 do not match the number of variables in the term, so only 4 is correct. Counting each variable in the product A`B`C`D` once gives the number 4, which is why the other numbers fail.

Answer: A

\(x\#y\) is equivalent to \(x\bar{y} + \bar{x}y\), the exclusive-OR.

The given properties are exactly those of XOR: \(x \oplus 0 = x\), \(x \oplus 1 = \bar{x}\), \(x \oplus x = 0\) and \(x \oplus \bar{x} = 1\). Check the candidate at \(y = 0\): \(x \cdot 1 + \bar{x} \cdot 0 = x\); at \(y = 1\): \(0 + \bar{x} = \bar{x}\). Both match.

\(xy + \bar{x}\bar{y}\) is XNOR, which gives \(x\#0 = \bar{x}\). \(x\bar{y} + \bar{x}\bar{y} = \bar{y}\) and \(\bar{x}y + xy = y\) ignore \(x\) entirely and fail \(x\#0 = x\).

Answer: C

The statement that always holds is \(\bar{x}_{1} \oplus \bar{x}_{3} = \bar{x}_{2} \oplus \bar{x}_{4}\).

XOR-ing \(x_{2} \oplus x_{4}\) onto both sides of \(x_{1} \oplus x_{2} \oplus x_{3} \oplus x_{4} = 0\) gives \(x_{1} \oplus x_{3} = x_{2} \oplus x_{4}\). Complementing both inputs of an XOR does not change it, \(\bar{a} \oplus \bar{b} = a \oplus b\), so

\[ \begin{aligned} \bar{x}_{1} \oplus \bar{x}_{3} &= x_{1} \oplus x_{3} \\ &= x_{2} \oplus x_{4} \\ &= \bar{x}_{2} \oplus \bar{x}_{4} \end{aligned} \]

The others fail for the assignment \(x_{1} = x_{2} = x_{3} = x_{4} = 1\), which satisfies the condition (four ones XOR to 0): then \(x_{1}x_{2}x_{3}x_{4} = 1\), \(x_{1}x_{3} + x_{2} = 1\) and \(x_{1} + x_{2} + x_{3} + x_{4} = 1\) as an OR.

Answer: B

The statement that is not a property is \(a\,.\,a' = 1\).

A variable ANDed with its own complement is always 0, never 1: if a = 1 then a′ = 0, and if a = 0 then the AND is 0 anyway. This is the complement law a . a′ = 0.

The other three are genuine laws of Boolean algebra: a + b = b + a and a . b = b . a are the commutative laws for OR and AND, and a + a′ = 1 is the complement law for OR, since one of a and a′ is always 1.

Answer: D

The incorrect identity is \((P \oplus \bar{P}) \oplus Q = (P \odot \bar{P}) \odot \bar{Q}\).

A variable and its complement always differ, so \(P \oplus \bar{P} = 1\) and \(P \odot \bar{P} = 0\). The left side is \(1 \oplus Q = \bar{Q}\). The right side is \(0 \odot \bar{Q}\), and XNOR with 0 complements its input, giving \(Q\). Since \(\bar{Q} \ne Q\), the identity fails.

The other three hold. XNOR is by definition the complement of XOR, so \(\overline{P \oplus Q} = P \odot Q\). Complementing one input of an XOR complements the result, so \(\bar{P} \oplus Q = P \odot Q\). Complementing both inputs cancels out, so \(\bar{P} \oplus \bar{Q} = P \oplus Q\).

Answer: C

The expression simplifies to \(\overline{A + B}\).

First multiply the last two factors:

\[ \begin{aligned} (\bar{A} + B)(A + \bar{B}) &= \bar{A}\bar{B} + AB \\ \overline{AB}\,(\bar{A}\bar{B} + AB) &= (\bar{A} + \bar{B})\bar{A}\bar{B} + 0 \\ &= \bar{A}\bar{B} = \overline{A + B} \end{aligned} \]

The product \(\overline{AB}\cdot AB\) is zero, leaving only \(\bar{A}\bar{B}\), which by De Morgan's law is the NOR \(\overline{A + B}\). Checking with \(A = B = 1\): the expression is 0, which rules out \(A + B\) and \(AB\); with \(A = 0, B = 1\) it is 0, which rules out \(\bar{A}B\).

Answer: D

The expression simplifies to \(\bar{C}\).

Every term contains \(\bar{C}\), so it can be taken out as a common factor. What is left is the sum of all four combinations of \(A\) and \(B\), which is always 1:

\[ \begin{aligned} Y &= \bar{C}(\bar{A}\bar{B} + \bar{A}B + A\bar{B} + AB) \\ &= \bar{C}\,[\bar{A}(\bar{B} + B) + A(\bar{B} + B)] \\ &= \bar{C}(\bar{A} + A) = \bar{C} \end{aligned} \]

So \(Y\) is 1 whenever \(C = 0\), whatever \(A\) and \(B\) are. \(A\bar{B}\) and \(\bar{A}B\bar{C}\) cover only one or two of the four minterms, and \(\bar{B}C\) requires \(C = 1\), where every term of \(Y\) is 0.

Answer: D

Boolean algebra does not apply to 0 - 1 = 1.

Boolean algebra defines only OR (written +), AND (written as multiplication) and NOT. There is no subtraction operation, so a relation written with a minus sign is not a Boolean relation at all.

The other three are valid Boolean OR relations: 0 + 0 = 0, 0 + 1 = 1, and 1 + 1 = 1. The last looks wrong by ordinary arithmetic, but in Boolean algebra it means 1 OR 1 = 1, which is correct; it is the classic trap the question sets.

Answer: C

The commutative law holds for both OR and AND gates.

In Boolean algebra, addition is the OR operation and multiplication is the AND operation. The commutative laws state

\[ A + B = B + A, \qquad A \cdot B = B \cdot A \]

so an OR gate and an AND gate give the same output whichever way their inputs are connected. Saying the law holds for OR only or AND only leaves out one of the two operations. A NOT gate has a single input, so the commutative law, which is about the order of two inputs, does not apply to it.