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.