Boolean Algebra Practice Questions

An essential prime implicant (EPI) is a prime implicant that covers at least one minterm that cannot be covered by any other prime implicant. To find the EPIs, we can use a Karnaugh map (K-map) for the given function $F(P,Q,R,S)=\sum m(0,2,5,7,9,11)+d(3,8,10,12,14)$.

By grouping the minterms and don't-cares on the K-map, we identify the prime implicants. Then, we check which of these are essential:

1. The group covering minterms (0, 2, 8, 10) is an EPI because minterm 0 is covered only by this group.
2. The group covering minterms (3, 7, 11) is an EPI because minterm 11 is covered only by this group.
3. The group covering minterms (5, 7, 9, 11) is an EPI because minterm 5 is covered only by this group.

Since there are three such groups, the number of essential prime implicants is 3.

Let's analyze the expression in option D: $(P \bigoplus \overline{P})\bigoplus Q=(P\bigodot\overline{P})\bigodot \overline{Q}$.

First, evaluate the Left Hand Side (LHS):
The expression $P \bigoplus \overline{P}$ is always true, evaluating to 1.
So, the LHS becomes $1 \bigoplus Q$. By the definition of XOR, $1 \bigoplus Q = (1 \cdot \overline{Q}) + (\overline{1} \cdot Q) = \overline{Q}$.

Next, evaluate the Right Hand Side (RHS):
The expression $P \bigodot \overline{P}$ is always false, evaluating to 0.
So, the RHS becomes $0 \bigodot \overline{Q}$. By the definition of XNOR, $0 \bigodot \overline{Q} = (0 \cdot \overline{Q}) + (\overline{0} \cdot \overline{\overline{Q}}) = 1 \cdot Q = Q$.

The statement thus simplifies to $\overline{Q} = Q$, which is not a valid Boolean identity. Therefore, this statement is not correct.

To find the minimum Product-of-Sums (POS) form, we group the 0s in the Karnaugh map.
The function is given by minterms $f(w,x,y,z) = \sum_{m}(0,1,2,3,7,8,10)$ and don't cares $\sum_{d}(5,6,11,15)$.
The cells corresponding to 0s are those not in the minterm or don't-care lists: {4, 9, 12, 13, 14}.

We create a K-map for these 0s and group them to find the simplified POS expression.
The K-map for the 0s (with don't cares) is:

   yz
wx 00 01 11 10
00  .  .  .  .
01  0  X  .  X
11  0  0  X  0
10  .  0  X  .

We can form two maximal groups of 0s (including don't cares):

1. A group of four covering cells (4, 6, 12, 14). For this group, $x=1$ and $z=0$ is not true for all cells. The common variable is $z=0$ and $x$ changes. This is not a valid group. Let's re-examine. The group for $(\bar{x}+z)$ corresponds to cells where $x=1$ and $z=0$. These are cells 4, 6, 12, 14. Values: 4(0), 6(X), 12(0), 14(0). This is a valid group. The SOP term for $\bar{f}$ is $x\bar{z}$, which gives the POS term $(\bar{x}+z)$ for $f$.
2. A group of four covering cells (9, 11, 13, 15). Values: 9(0), 11(X), 13(0), 15(X). This is a valid group. The SOP term for $\bar{f}$ is $wz$, which gives the POS term $(\bar{w}+\bar{z})$ for $f$.

Combining these two terms gives the minimal POS form: $f = (\bar{w}+\bar{z})(\bar{x}+z)$.

The OR function is expressed as $Y = A + B$. By applying De Morgan's Law, we can rewrite this as $Y = \overline{\overline{A + B}} = \overline{\overline{A} \cdot \overline{B}}$.
To implement this using NAND gates, we first convert two NAND gates into NOT gates by tying their inputs together: $G_1 = \overline{A \cdot A} = \overline{A}$ and $G_2 = \overline{B \cdot B} = \overline{B}$.
We then feed these inverted signals into a third NAND gate: $Y = \overline{G_1 \cdot G_2} = \overline{\overline{A} \cdot \overline{B}} = A + B$.
This configuration uses a total of three NAND gates to perform the logic of an OR gate.

The 2-input XOR gate performs the Exclusive OR operation, represented by the logical expression $Y = A \oplus B$.
The truth table for this operation is as follows: $(0,0) \to 0$, $(0,1) \to 1$, $(1,0) \to 1$, and $(1,1) \to 0$.
As observed, the output $Y$ is high ($1$) only for the input combinations where $A \neq B$.
These combinations correspond to the scenarios where one input is $0$ and the other is $1$.
Thus, the XOR gate produces a high output exclusively when the input values are different.

The question asks to identify the incorrect Boolean identity. Let's analyze option C.
The expression is $(w\bar{x}(y+x\bar{z})+\bar{w}\bar{x})y=x\bar{y}$.

Let's simplify the Left Hand Side (LHS):
LHS = $(w\bar{x}y + w\bar{x}x\bar{z} + \bar{w}\bar{x})y$ (Distributive law)
Since $x \cdot \bar{x} = 0$, the term $w\bar{x}x\bar{z}$ becomes 0.
LHS = $(w\bar{x}y + \bar{w}\bar{x})y$
LHS = $w\bar{x}y \cdot y + \bar{w}\bar{x}y$ (Distributive law)
Using the idempotent law ($y \cdot y = y$):
LHS = $w\bar{x}y + \bar{w}\bar{x}y$
Factoring out $\bar{x}y$:
LHS = $(w+\bar{w})\bar{x}y$
Using the complement law ($w+\bar{w}=1$):
LHS = $1 \cdot \bar{x}y = \bar{x}y$

The simplified LHS is $\bar{x}y$, which is not equal to the Right Hand Side (RHS) $x\bar{y}$. Thus, the statement is incorrect.

The Karnaugh map simplification is the first step. The PDF solution simplifies the function to $F(a,b,c,d) = \bar{a}c$. The question requires implementing this function using only 2-input NOR gates, and the complements of the inputs are available.

The expression $\bar{a}c$ can be rewritten for a NOR-gate implementation using De Morgan's laws:
$F = \bar{a}c = \overline{\overline{(\bar{a}c)}} = \overline{\overline{\bar{a}} + \bar{c}} = \overline{a + \bar{c}}$
Since the inputs $a$ and $\bar{c}$ are directly available, this expression can be implemented with a single 2-input NOR gate.

Given the Boolean expression $E = (P + \bar{Q} + \bar{R}) \cdot (P + Q + R) \cdot (P + Q + \bar{R})$.
Apply the identity $(A + B)(A + C) = A + BC$ to the last two factors:
$(P + Q + R) \cdot (P + Q + \bar{R}) = (P + Q) + (R \cdot \bar{R}) = P + Q$.
Now, substitute this back into the original expression:
$E = (P + \bar{Q} + \bar{R}) \cdot (P + Q)$.
Distributing the terms gives: $P + (\bar{Q} + \bar{R}) \cdot Q = P + (\bar{Q} \cdot Q) + (\bar{R} \cdot Q) = P + 0 + Q \cdot \bar{R}$.
The simplified expression is $E = P + Q \cdot \bar{R}$.

The circuit consists of three AND gates receiving inputs $(A, B)$, $(A, C)$, and $(B, C)$ respectively, which generate the product terms $AB$, $AC$, and $BC$. These three outputs are subsequently connected to the inputs of a single OR gate. The OR gate combines these terms, resulting in the final Boolean expression $F = AB + AC + BC$. This logic represents the majority function of three variables.

The properties of the operator can be used to construct its truth table.
For $y=0$, the output is $x\#0 = x$.
For $y=1$, the output is $x\#1 = \bar{x}$.

This gives the following results:

• $0\#0 = 0$
• $0\#1 = \bar{0} = 1$
• $1\#0 = 1$
• $1\#1 = \bar{1} = 0$

The output column (0, 1, 1, 0) matches the truth table for the XOR operation. The Boolean expression for XOR is $x\bar{y}+\bar{x}y$.

To find the minimum number of $\text{NAND}$ gates, simplify the Boolean expression $F = A + A\bar{B} + A\bar{B}C$ using Boolean algebra rules.
Factor out $A$ from the expression: $F = A(1 + \bar{B} + \bar{B}C)$.
According to the redundancy law and the property of ORing with 1, $1 + \text{anything} = 1$.
Thus, the expression simplifies to $F = A(1) = A$.
Since the output function $F$ is identical to the input variable $A$, no logic gates are required to implement it (i.e., a direct connection suffices).
Therefore, the number of $\text{NAND}$ gates required is $0$.

The output of the given logic circuit is $F = w' + (w'z) + (z'xy)$, where the inputs to the OR gate (Gate 4) are $w'$, $w'z$, and $z'xy$. Using the Boolean absorption law, the first two terms simplify as $w' + w'z = w'(1 + z) = w'$. Consequently, the expression becomes $F = w' + z'xy$. Since the term $w'z$ produced by Gate 2 is completely covered by the $w'$ signal directly fed into Gate 4, the output of Gate 2 is redundant and does not affect the final result.

The given equation is $x_{1} \oplus x_{2} \oplus x_{3} \oplus x_{4}=0$. The XOR operation is associative and commutative. We can manipulate the equation using the property that $a \oplus a = 0$. By XORing both sides of the original equation with $(x_{2} \oplus x_{4})$, we get:
$(x_{1} \oplus x_{2} \oplus x_{3} \oplus x_{4}) \oplus (x_{2} \oplus x_{4}) = 0 \oplus (x_{2} \oplus x_{4})$
Rearranging the terms:
$x_{1} \oplus x_{3} \oplus (x_{2} \oplus x_{2}) \oplus (x_{4} \oplus x_{4}) = x_{2} \oplus x_{4}$
This simplifies to $x_{1} \oplus x_{3} = x_{2} \oplus x_{4}$.

The given Boolean expression is ${[{D}'+A{B}'+{A}'C+A{C}'D+{A}'{C}'{D}]}'$. We first simplify the expression inside the brackets.

1. Factor ${C}'D$ from the last two terms: ${D}'+A{B}'+{A}'C+{C}'D(A+{A}')$.
2. Since $A+{A}'=1$, this becomes ${D}'+A{B}'+{A}'C+{C}'D$.
3. Using the adjacency rule ($X+X'Y = X+Y$), we simplify ${D}'+{C}'D$ to ${D}'+{C}'$. The expression is now ${D}'+{C}'+A{B}'+{A}'C$.
4. Similarly, ${C}'+{A}'C$ simplifies to ${C}'+{A}'$. The expression is ${D}'+{C}'+{A}'+A{B}'$.
5. Again, ${A}'+A{B}'$ simplifies to ${A}'+{B}'$. The expression is ${D}'+{C}'+{A}'+{B}'$.
   Applying the final complement (De Morgan's Law) to $({D}'+{C}'+{A}'+{B}')$ gives $D \cdot C \cdot A \cdot B$. This simplified form, $ABCD$, is a single product term, which represents exactly one min-term.

A prime implicant is a product term that cannot be further simplified by combining with other terms in a Karnaugh map. For the function $f(w,x,y,z)= \Sigma (0,2,4,5,6,10)$, we can identify the prime implicants by grouping the 1s.

1. A group of four 1s for minterms (0, 2, 4, 6), which corresponds to the term $w'z'$.
2. A group of two 1s for minterms (4, 5), which corresponds to the term $w'xy'$.
3. A group of two 1s for minterms (2, 10), which corresponds to the term $x'yz'$.
   These three groups cover all the 1s and cannot be expanded further. Thus, there are 3 prime implicants.

The binary operator is defined as $X\#Y = X' + Y'$.
Statement S2 claims commutativity: $Q\#R = R\#Q$.
$Q\#R = Q' + R'$.
$R\#Q = R' + Q'$.
Since the Boolean OR operation is commutative, $Q' + R' = R' + Q'$, so S2 is true.

Statement S1 claims associativity: $(P\#Q)\#R = P\#(Q\#R)$.
Let's use a counterexample: $P=1, Q=1, R=0$.
LHS: $(1\#1)\#0 = (1'+1')\#0 = (0+0)\#0 = 0\#0 = 0'+0' = 1+1 = 1$.
RHS: $1\#(1\#0) = 1\#(1'+0') = 1\#(0+1) = 1\#1 = 1'+1' = 0+0 = 0$.
Since LHS $\neq$ RHS, S1 is false.

The Boolean function is F = P' + QR. To find its canonical forms, we can create a truth table or use a Karnaugh map. The minterms are the input combinations for which F=1.
F is true if P=0, or if Q=1 and R=1.
The minterms are:
PQR=000 (m0), 001 (m1), 010 (m2), 011 (m3), and 111 (m7).
So, F = $\Sigma m(0, 1, 2, 3, 7)$. This makes statement (S2) true and (S1) false.
The maxterms are the remaining combinations: {4, 5, 6}.
So, F = $\Pi M(4, 5, 6)$. This makes statement (S3) true and (S4) false.

To find the complement of the expression $E = AB(\bar{B}C + AC)$, we apply De Morgan's Theorem, which states that $\overline{XY} = \bar{X} + \bar{Y}$ and $\overline{X+Y} = \bar{X} \cdot \bar{Y}$.
The complement is $\overline{E} = \overline{AB(\bar{B}C + AC)}$.
By treating $AB$ and $(\bar{B}C + AC)$ as two terms, we have $\overline{E} = \overline{AB} + \overline{(\bar{B}C + AC)}$.
Expanding $\overline{AB}$ gives $(\bar{A} + \bar{B})$.
Applying the second part of De Morgan's Theorem to the term $\overline{\bar{B}C + AC}$ yields $(\overline{\bar{B}C}) \cdot (\overline{AC})$.
Further expanding using $\overline{XY} = \bar{X} + \bar{Y}$ results in $(B + \bar{C}) \cdot (\bar{A} + \bar{C})$.
Combining these, we get $\overline{E} = (\bar{A} + \bar{B}) + (B + \bar{C}) \cdot (\bar{A} + \bar{C})$.

Given the expression $F = AB + AB' + A'C + AC$.
Factoring $A$ from the first two terms gives $A(B + B') + A'C + AC$, which simplifies to $A + A'C + AC$ since $B + B' = 1$.
Further grouping $C$ gives $A + C(A' + A)$, and since $A' + A = 1$, this simplifies to $A + C$.
The final form $A + C$ does not contain the variable $B$.
Therefore, the expression is independent of $B$.