MATH 250 - INTRODUCTION TO
DISCRETE MATHEMATICS - Boolean
Algebra
Question Bank - Set 4
Liberty University
Question 1
Question
Simplify the following Boolean expression using algebraic manipulation:
F=A′B+AB′+ (A+B′)(A+B)
Solution
Step 1: Apply the distributive law to simplify the expression:
F=A′B+AB′+ (A+B′)(A+B)
F=A′B+AB′+A(A+B′) + B′(A+B)
Step 2: Use the distributive law to expand the terms further:
F=A′B+AB′+AA +AB′+B′A+B′B
F=A′B+AB′+A+AB′+AB′+ 0
F=A′B+AB′+A
Step 3: Apply the absorption law to simplify the expression:
F=A′B+AB′+A
F=A+A′B+B′A
Step 4: Use the commutative law to rearrange the terms:
F=A+AB′+AB′
Step 5: Apply the idempotent law to simplify the expression:
F=A+AB′
Therefore, the simplified form of the Boolean expression is F=A+AB′.
Question 2
Question
Simplify the following Boolean expression: (A+B)(A′C+BC).
Solution
To simplify the given Boolean expression, we will use the distributive law, com-
plement law, and absorption law of Boolean algebra.
Step 1: Apply the distributive law (A+B)(A′C+BC).
(A+B)(A′C+BC) = AA′C+ABC +BA′C+BBC
= 0 + ABC + 0 + 0
=ABC
Step 2: Apply the absorption law XY +XZ =X(Y+Z) to simplify ABC.
ABC =A(BC)
Therefore, the simplified form of (A+B)(A′C+BC) is A(BC).
Question 3
Question
Given the Boolean expression F= (A+B)·(A+B), simplify the expression
using Boolean algebra laws.
Solution
To simplify the given Boolean expression F= (A+B)·(A+B), we can apply
various Boolean algebra laws such as the distributive law, complement law,
idempotent law, and absorption law.
Step 1: Apply the distributive law: (A+B)·(A+B) = A+B·B
Step 2: Apply the complement law: B·B= 0
Step 3: Simplify the expression: A+B·B=A+ 0 = A
Therefore, the simplified form of the Boolean expression F= (A+B)·(A+B)
is F=A.
Question 4
Question
Let F=A′BC +AB′C+ABC′. Simplify Fusing Boolean algebra laws and
theorems.
2
Solution
To simplify the given Boolean expression F=A′BC +AB′C+ABC′, we will
apply various Boolean algebra laws and theorems.
1. Use the commutative law: AB =BA
Step 1: Rearrange terms.
F=A′BC +AB′C+ABC′
=AB′C+A′BC +ABC′(commutative law)
2. Use the distributive law: A(B+C) = AB +AC
Step 2: Apply the distributive law.
F=AB′C+A′BC +ABC′
=B′C(A+A′) + BC(A+A′) (distributive law)
=B′C+BC
3. Use the absorption law: A+AB =A
Step 3: Apply the absorption law.
F=B′C+BC
=B′C+B′C′(complement law)
=B′(absorption law)
Therefore, the simplified form of Fis F=B′.
Question 5
Question
Simplify the following Boolean expression using laws of Boolean Algebra:
(A+B+C)(A+B+C)(A+B+C)
Solution
To simplify the given expression, we will use the basic laws of Boolean Algebra:
commutative law, associative law, identity law, complement law, and distribu-
tive law.
Step 1: Apply the distributive law to expand the given expression:
(A+B+C)(A+B+C)(A+B+C)
= (A(A+B+C) + B(A+B+C) + C(A+B+C))(A+B+C)
= (AA +AB +AC +BA +BB +BC +CA +CB +CC)(A+B+C)
= (A+AB +AC +BA +0+BC +CA + 0 + 0)(A+B+C)
= (A+AB +AC +BA +BC +CA)(A+B+C)
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Step 2: Apply the distributive law again to further simplify the expression:
(A+AB +AC +BA +BC +CA)(A+B+C)
= (A(A+B+C) + AB +AC +BA +BC +CA)(A+B+C)
= (AA +AB +AC +AB +AC +BA +BC +CA)(A+B+C)
= (0 + AB +AC +AB +AC +BA +BC +CA)(A+B+C)
= (AB +AC +AB +AC +BA +BC +CA)(A+B+C)
Therefore, the simplified form of the given expression is (AB +AC +AB +
AC +BA +BC +CA)(A+B+C).
Question 6
Question
Simplify the following Boolean expression: (A+B)(A+C+D)(A+B+D+E).
Solution
To simplify the given Boolean expression (A+B)(A+C+D)(A+B+D+E), we
will use the properties of Boolean Algebra, including distribution, absorption,
complement, and idempotent laws.
Step 1: Use the distributive law to expand the expression.
(A+B)(A+C+D)(A+B+D+E)
=A(A+C+D)(A+B+D+E) + B(A+C+D)(A+B+D+E)
Step 2: Use the distributive law to further simplify each of the two terms.
=AA +AC +AD +AB +BA +BC +BD +BB
= 0 + AC +AD +AB +0+BC +BD +B
Step 3: Apply some simplifications using complementation laws and iden-
tity laws.
=AC +AD +AB +BC +B
=AC +AB +BC +B
Step 4: Apply absorption law to further simplify the expression.
=A(C+B) + B(C+ 1)
=A(C+B) + B
Step 5: Apply distributive law and another simplification.
=AC +AB +B
=AC +B
Therefore, the simplified form of the expression (A+B)(A+C+D)(A+
B+D+E) is AC +B.
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Question 7
Question
Let A,B, and Cbe Boolean variables. Show that (A+B)·(A+B·C) = A+B·C.
Solution
To show that (A+B)·(A+B·C) = A+B·C, we will use Boolean algebra
laws and properties.
Step 1: Use the distributive law X·(Y+Z) = X·Y+X·Zto expand the
left-hand side.
(A+B)·(A+B·C) = A·A+A·B·C+B·A+B·B·C
Step 2: Apply the idempotent law X·X=Xto simplify the terms A·A
and B·B.
A·A+A·B·C+B·A+B·B·C=A+A·B·C+B·A+ 0
Step 3: Use the complementary law X·X= 0 to simplify the term A·B.
A+A·B·C+B·A+ 0 = A+0+B·A=A+B·A=A+B
Step 4: Apply the absorption law X+X·Y=Xto simplify the term B·A.
A+B
Step 5: Thus, (A+B)·(A+B·C) = A+B, which is equivalent to A+B·C.
So, (A+B)·(A+B·C) = A+B·Cis true in Boolean algebra.
Question 8
Question
Simplify the following Boolean expression: (A+B+C)(A′+B′C′).
Solution
To simplify the given Boolean expression, we will use the distributive law and
complement law of Boolean algebra.
Step 1: Apply the distributive law:
(A+B+C)(A′+B′C′) = AA′+AB′C′+AC +B′A′+B′B′C′+B′C′C′
Step 2: Apply the complement law:
AA′= 0, B′B′= 0, C′C′= 0
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Step 3: Simplify the expression:
(A+B+C)(A′+B′C′) = 0 + AB′C′+AC +0+0+0
Step 4: Further simplify:
(A+B+C)(A′+B′C′) = AB′C′+AC
Therefore, the simplified form of the expression (A+B+C)(A′+B′C′) is
AB′C′+AC.
Question 9
Question
Given the Boolean expression (A+B)(A′+B)(A+B′), simplify the expression
using Boolean algebra rules.
Solution
To simplify the given Boolean expression (A+B)(A′+B)(A+B′), we will use
Boolean algebra rules such as distribution, complement, and absorption.
Step 1: Distribute the terms (A+B)(A′+B)
(A+B)(A′+B) = AA′+AB +BA′+BB =A′A+AB +BA′+ 0 = AB +BA′
Step 2: Use commutativity (AB =BA) to simplify AB +BA′
AB +BA′=AB +AB =A(B+B) = A
Step 3: Apply distribution to the result of step 2 and the term (A+B′)
A(A+B′) = AA +AB′=A+AB′
Step 4: Apply absorption to simplify A+AB′
A+AB′=A(1 + B′) = A×1 = A
Therefore, the simplified form of the Boolean expression (A+B)(A′+B)(A+
B′) is A.
Question 10
Question
Let A,B, and Cbe Boolean variables. Given the expression (A+B)(A+C)(B+
C), simplify the expression using Boolean algebra laws.
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Solution
To simplify the expression (A+B)(A+C)(B+C), we will use various Boolean
algebra laws such as distribution, idempotent, and absorption laws.
Step 1: Expand using distributive law
(A+B)(A+C)(B+C) = A(A+C)(B+C) + B(A+C)(B+C)
=AA +AC +BC +BC
=A+AC +BC
Step 2: Apply idempotent law
=A+ (A+AC) + BC
Step 3: Apply absorption law
=A+BC
Therefore, the simplified expression for (A+B)(A+C)(B+C) is A+BC.
Question 11
Question
Simplify the Boolean expression (A+B)·(A′+B′) + A·B.
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
algebra such as identity, domination, idempotent, and absorption laws.
Step 1: Apply the distributive law: P·(Q+R) = P·Q+P·R, where P, Q,
and Rare Boolean variables.
(A+B)·(A′+B′) + A·B= (A·A′+A·B′+B·A′+B·B′) + A·B
= (0 + A·B′+B·A′+ 0) + A·B
= (A·B′+B·A′) + A·B
Step 2: Apply the commutative law: P+Q=Q+P.
(A·B′+B·A′) + A·B= (A·B′+A·B) + B·A′
=A·(B′+B) + B·A′
=A·1 + B·A′
=A+B·A′
Step 3: Apply the absorption law: P+P·Q=P.
A+B·A′=A
Therefore, the simplified Boolean expression is A.
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Question 12
Question
Let F(A, B, C) = A(B+C) + BC. Simplify the Boolean expression F(A, B, C)
using Boolean algebra laws.
Solution
To simplify the given Boolean expression F(A, B, C) = A(B+C) + BC, we will
apply various Boolean algebra laws step by step.
Step 1: Apply the distributive law: X(Y+Z) = XY +XZ
A(B+C) + BC =AB +AC +BC
Step 2: Apply the absorption law: X+XZ =X
AB +AC +BC =AB +AC
Step 3: Apply the absorption law: X+XZ =X
AB +AC =AB
Therefore, the simplified Boolean expression is F(A, B, C) = AB.
Question 13
Question
Given the Boolean expression F= (A+B)(A+B), simplify the expression
using Boolean algebra laws and theorems.
Solution
Step 1: Apply the distributive law: (A+B)(A+B) = AA +AB +BA +BB
Step 2: Apply the complement law: AA = 0, BA =AB =A+B, and
BB = 0, so the expression becomes: 0 + AB +A+B+ 0
Step 3: Simplify further: F=AB +A+B
Therefore, the simplified form of the Boolean expression F= (A+B)(A+B)
is F=AB +A+B.
Question 14
Question
Simplify the following Boolean expression using Boolean Algebra laws:
F=A′B′+AB +A′B
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Solution
Step 1: Apply the associative law:
F=A′B′+ (AB +A′B)
Step 2: Apply the distributive law:
F=A′B′+AB +A′B
Step 3: Apply the absorption law:
F=A′B′+AB
Step 4: Apply the consensus theorem:
F=A
Therefore, the simplified Boolean expression is F=A.
Question 15
Question
Simplify the following Boolean expression: (A+B)(A+B+C).
Solution
Step 1: Use the distributive property to expand the expression. Step 2: Apply
De Morgan’s laws to simplify the expression further. Step 3: Use the comple-
ment law to simplify the expression.
Step 1: Expanding the expression using the distributive property:
(A+B)(A+B+C) = AA +AB +AC +BA +BB +BC
Step 2: Applying De Morgan’s laws: AB =A+Band A+A= 1
= 0 + AB +AC +BA +0+BC
Step 3: Further simplifying using the complement law: AB +AC =A(B+
C) and BA +BC =B(A+C)
=A(B+C) + B(A+C)
Therefore, the simplified form of the given Boolean expression is A(B+C)+
B(A+C).
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Question 16
Question
Simplify the following Boolean expression:
F= (A+B)(A+B)(A+B)
Solution
To simplify the given Boolean expression, we will use the Boolean algebra laws:
commutative law, associative law, distributive law, and De Morgan’s law.
Step 1: Apply the distributive law to expand the expression.
F=AAB +AAB +AB2+AB2
Step 2: Apply the idempotent law XX = 0 to simplify the terms AA and
B2.
F= 0 + AAB +AB + 0
Step 3: Simplify further by combining the terms with AAB and AB.
F=AB +AB
Step 4: Apply the consensus theorem (X+Y)(X+Y) = X+Zwhere
Z=Yto simplify the expression.
F=A+B
Therefore, the simplified form of the given Boolean expression is F=A+B.
Question 17
Question
Simplify the following Boolean expression: (A+BC)(A+BC +C)
Solution
To simplify the given Boolean expression, we will first expand the expression
using the distributive property, then simplify using Boolean algebra rules.
Step 1: Expand the expression using the distributive property:
(A+BC)(A+BC +C)
=AA +ABC +AC +BCA +BCBC +BCC
=0+0+AC +0+BCBC + 0
Step 2: Simplify the expression using Boolean algebra rules:
=AC + 0 + 0
=AC
Therefore, the simplified form of the given Boolean expression is AC.
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Question 18
Question
Simplify the following Boolean expression: (a+b′+c′)(a′b+ab′).
Solution
To simplify the given Boolean expression, we will use the distributive property,
as well as the fact that x+x′= 1 and xx′= 0 for any variable x.
Step 1: (a+b′+c′)(a′b+ab′)
=aa′b+ab +ab′+a′b′+bb′+b′c′(Expand using the distributive property)
= 0 + ab +ab′+0+0+b′c′(Since aa′= 0 and bb′= 0)
=ab +ab′+b′c′
=b(a+a′) + b′c′(Factor out common terms)
=b·1 + b′c′(Since a+a′= 1)
=b+b′c′
Therefore, the simplified form of the given Boolean expression is b+b′c′.
Question 19
Question
Simplify the following Boolean expression:
(A+B)(A+C) + AC
Solution
To simplify the given Boolean expression, we will use the distributive property,
complement property (A+A′= 1, A+ 1 = 1, A∗0 = 0), and the identity
property (A+ 0 = A,A∗1 = A) of Boolean algebra.
Step 1: Apply the distributive property to expand (A+B)(A+C):
(A+B)(A+C) = AA +AC +AB +BC
Step 2: Simplify the expansion using the idempotent law and complemen-
tary law:
AA =A, AB =A·B, AC = 0 (complementary law)
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BC = 0 (complementary law)
So, the expression simplifies to:
(A+B)(A+C) = A+A·B+ 0 + 0 = A+A·B
Step 3: Now, simplify the expression obtained in Step 2 by applying the
identity law:
A+A·B=A(1 + B) = A·1 = A
Step 4: Finally, simplify the original expression (A+B)(A+C) + AC using
the result from Step 3:
(A+B)(A+C) + AC =A+AC =A(1 + C) = A·1 = A
Therefore, the simplified form of the Boolean expression (A+B)(A+C)+AC
is A.
Question 20
Question
Simplify the following Boolean expression: (A+B)·(A·B).
Solution
We will simplify the given Boolean expression step by step using basic rules of
Boolean algebra.
Step 1: Apply De Morgan’s law: (A·B) = A+B.
Step 2: Apply the distributive property: (A+B)·A·B.
Step 3: Apply the idempotent law: A·A= 0.
Step 4: Use the absorption law: X·(X+Y) = X.
Step 5: Use the complement law: X·X= 0.
Step 6: The final simplified expression is 0.
Therefore, (A+B)·(A·B) simplifies to 0.
Question 21
Question
Simplify the following Boolean expression:
F=A′B′C+A′BC +AB′C+ABC
12
Solution
To simplify the given Boolean expression, we will use the laws of Boolean al-
gebra such as identity, domination, double negation, idempotent, commutative,
associative, distributive, and De Morgan’s laws.
Step 1: Apply the commutative law to rearrange the terms:
F=A′B′C+A′BC +AB′C+ABC
Step 2: Factor out common terms:
F=A′B′C+A′BC +AB′C+ABC =A′B′(C+C′)+AB′(C+C′)+A′B(C+C′)
Step 3: Apply the idempotent law (X+X=X) to simplify the terms
inside the parentheses:
F=A′B′(1) + AB′(1) + A′B(1)
Step 4: Apply the identity law (X·1 = X) to further simplify the expression:
F=A′B+AB′+A′B=A′B+AB′+AB
Step 5: Apply the idempotent law again:
F=A′B+AB′+AB =A′B+AB′
Step 6: Apply the distributive law to simplify the expression:
F=A′B+AB′=B(A′+A) = B(1) = B
Therefore, the simplified Boolean expression is F=B.
Question 22
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+C)
Solution
Let’s simplify the given Boolean expression step by step:
Step 1: Apply the distributive property to expand the expression.
(A+B+C)(A+B+C)(A+B+C)
= (AA +AB +AC +BA +BB +BC +CA +CB +CC)
Step 2: Simplify the terms with complements of each variable.
= (0 + AB +AC +BA +0+0+0+CB + 0)
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= (AB +AC +BA +CB)
Step 3: Apply the commutative property to rearrange the variables.
= (AB +BA +AC +CB)
Step 4: Apply the complementary property XY +XY =X+Y.
= (A+B)+(A+C)
Therefore, the simplified Boolean expression is (A+B)+(A+C).
Question 23
Question
Simplify the following Boolean expression:
F= (A+B)·(A+B+C)·(A+B)
Solution
Step 1: Distribute the terms inside the parentheses using the distributive prop-
erty. Step 2: Simplify the expression by combining like terms and applying
Boolean identities.
Step 1:
F= (A+B)·(A+B+C)·(A+B)
F=A·A·B+A·A·C+A·B+B·A·B+B·A·C+B·B
Step 2: Apply the following Boolean identities:
X·X= 0
X+X= 1
1·X=X
0 + X=X
To simplify the expression:
F=0+0+A·B+0+0+B
F=A·B+B
F=B·(A+ 1)
F=B
Thus, the simplified Boolean expression for F is B.
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Question 24
Question
Simplify the following Boolean expression: (A+B)(A+C)(A+C).
Solution
To simplify the given Boolean expression, we will use the basic laws of Boolean
algebra, such as the distributive law, the complement law, and the identity law.
Step 1: Apply the distributive law to expand the expression.
(A+B)(A+C)(A+C) = A(A+C)(A+C) + B(A+C)(A+C)
Step 2: Use the complement law AA = 0 to simplify A(A+C).
A(A+C) = 0 + AC =AC
Step 3: Use the distributive law again to simplify B(A+C)(A+C).
B(A+C)(A+C) = BAA +BC +BAC
Step 4: Apply the complement law AA = 0 to simplify BAA.
BAA = 0
Step 5: Combine the simplified terms to get the final expression.
(A+B)(A+C)(A+C) = AC +BC +BAC =AC +BC
Therefore, the simplified Boolean expression is AC +BC.
Question 25
Question
Simplify the following Boolean expression: (A+B+C)(A+B)(A+B+C).
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
algebra such as the distributive law, identity law, and complement law.
Step 1: Use the distributive law to expand the expression.
(A+B+C)(A+B)(A+B+C) = A(A+B)(A+B+C)+B(A+B)(A+B+C)+C(A+B)(A+B+C)
Step 2: Apply the complement law.
=A(0 + B)(A+B+C) + B(A+ 1)(A+B+C) + C(A+B)(A+B+C)
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Step 3: Simplify the expressions.
=AB(A+B+C) + B(A+A+B+C) + C(A+B)(A+B+C)
=AB +ABB +ABC +BA +BA +BB +BC +CA +BC
Step 4: Simplify further using the complement law.
=AB +0+ABC +BA +BA +0+0+CA +BC
=AB +ABC +BA +BA +CA +BC
Step 5: Combine like terms.
=AB(1 + C) + BA +CA +BC
=AB +BA +CA +BC
Thus, the simplified Boolean expression is AB +BA +CA +BC.
Question 26
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+
C)(A+B+C).
Solution
Step 1: Apply the distributive property to expand the expression. Step 2:
Eliminate duplicate terms. Step 3: Use the idempotent law to simplify further.
Step 4: Apply the complement law to simplify the expression.
Step 1:
(A+B+C)(A+B+C)(A+B+C)(A+B+C)
Expanding using the distributive property:
=AA+AB+AC+AB+ABC+AC+AC+ACB+ACC+AB+AB+AC+AB+AC+ABC
Step 2: Eliminating duplicate terms:
=A+AB +AC +AB +ABC +AC +ACB +AB +AC +ABC
Step 3: Using the idempotent law: XX =Xfor any variable X:
=A+AB +AC +AB +AC +AB +AC +ABC
Step 4: Applying the complement law: XX = 0 for any variable X:
=A+AB +AC +AB +AC +BC
Therefore, the simplified Boolean expression is A+AB+AC+AB+AC+BC.
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Question 27
Question
Simplify the following Boolean expression: (A+B+C)(A′+B+C)(A+B′+
C′)(A+B+C′).
Solution
Step 1: Use the distributive property to expand the expression.
(A+B+C)(A′+B+C)(A+B′+C′)(A+B+C′)=(AA′+AB +AC +BA′+BB +BC +CA +CB +CC)(A+B′+C′)(A+B+C′)
= (0+AB +AC +0+0+0+0+CB +0)(A+B′+C′)(A+B+C′)
= (AB +AC +CB)(A+B′+C′)(A+B+C′)
Step 2: Apply the distributive property again to simplify further.
(AB +AC +CB)(A+B′+C′)(A+B+C′) = AB(A+B′+C′)(A+B+C′) + AC(A+B′+C′)(A+B+C′) + CB(A+B′+C′)(A+B+C′)
=ABA +ABB′+ABC′+ACA +ACB′+ACC′+CBA +CBB′+CBC′
=0+0+0+ACA +0+0+0+0+0
=ACA
Therefore, the simplified expression is ACA.
Question 28
Question
Simplify the Boolean expression: (A+B)(A+B)(A+B)(A+B).
Solution
Let’s simplify the given Boolean expression step by step.
Step 1: Use the distributive property to expand the expression.
(A+B)(A+B)(A+B)(A+B) = (A+B)(A+B)(AA +AB +AB +AB)
Step 2: Simplify the terms AA and AA.
(A+B)(A+B)(AA +AB +AB +AB)=(A+B)(A+B)(0 + AB +AB +AB)
Step 3: Simplify the expression AA by noting that it is always equal to 0.
(A+B)(A+B)(0 + AB +AB +AB) = (A+B)(A+B)(AB +AB +AB)
Step 4: Apply the distributive property again to expand the expression.
(A+B)(A+B)(AB +AB +AB) = AAB +AAB +BAB +BAB +ABA +ABA
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Step 5: Simplify the terms AA,BB, and AA.
AAB +AAB +BAB +BAB +ABA +ABA =0+0+0+0+0+0=0
Step 6: Therefore, the simplified form of the Boolean expression is 0 .
Question 29
Question
Simplify the following Boolean expression: (A+B)(A+C)(B+C).
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
Algebra, including the distributive property, idempotent laws, and complemen-
tary laws.
Step 1: Apply the distributive property to expand the expression.
(A+B)(A+C)(B+C) = AAB+AAC+ABC+BCB+BCC+CAC =A+AB+AC+BC+C
Step 2: Apply the idempotent law (A+A=A) to simplify the expression.
A+AB +AC +BC +C=A+AC +BC +C
Step 3: Apply the complementary law (A+A= 1) to simplify the expression
further.
A+AC +BC +C=A+ (A+C)(B+C) = A+ 1(B+C) = A+B+C
Therefore, the simplified form of (A+B)(A+C)(B+C) is A+B+C.
Question 30
Question
Simplify the following Boolean expression using Boolean Algebra laws:
F= (A+B)(A′+C)(B′+C′)
Solution
Step 1: Apply the Distributive Law to expand the given expression:
F= (A+B)(A′+C)(B′+C′)
F=AA′+AC +BB′+BC +A′B′+A′C+B′C′
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Step 2: Apply the Null Law (A+A′= 1) to simplify the term AA′:
F= 0 + AC +BB′+BC +A′B′+A′C+B′C′
F=AC +BB′+BC +A′B′+A′C+B′C′
Step 3: Apply the Complement Law to simplify the terms BB′and A′B′:
F=AC +0+BC +0+A′C+B′C′
F=AC +BC +A′C+B′C′
Step 4: Apply the Absorption Law to simplify the term AC +BC:
F=C(A+B) + A′C+B′C′
Step 5: Apply the Consensus Theorem: XY +X′Z+Y Z =XY +X′Zto
simplify the expression:
F=C(A+B) + A′C+B′C′
F=C+A′C+B′C′
Step 6: Apply the Idempotent Law to simplify the expression further:
F=C+A′C+B′C′
F=C+C′A+C′B′
F=C+A+B′
Question 31
Question
Simplify the Boolean expression (A+B)·(A′+C) + A·B·Cusing Boolean
algebra properties.
Solution
To simplify the given Boolean expression, we will apply various Boolean alge-
bra properties such as the distributive property, complement property, and the
idempotent property.
Step 1: Apply the distributive property: (A+B)·(A′+C) + A·B·C
⇒(AA′+AC +BB′+BC) + ABC
Step 2: Use the complement property AA′= 0 and BB′= 0:
⇒(0 + AC +0+BC) + ABC
19
⇒AC +BC +ABC
Step 3: Factor out a common term C:
⇒C(A+B+AB)
Step 4: Use the absorption property A+AB =A:
⇒C(A)
⇒AC
Therefore, the simplified form of the Boolean expression is AC.
Question 32
Question
Given the Boolean expression F(A, B, C) = AB +AC +BC, simplify the ex-
pression using Boolean Algebra rules.
Solution
To simplify the given Boolean expression F(A, B, C) = AB +AC +BC, we will
use various Boolean Algebra rules such as the Identity Law, Domination Law,
Idempotent Law, Complement Law, etc.
Step 1: Apply the Absorption Law: AB +AC = (A+A)(A+C) =
1(A+C) = A+C.
Therefore, the expression simplifies to F(A, B, C) = A+C+BC.
Step 2: Apply the Absorption Law again: A+C+BC = (A+C+B)(A+
C+C).
This simplifies further to F(A, B, C) = A+C+B.
Step 3: Apply the Consensus Theorem: A+C+B= (A+C)(A+B).
The final simplified expression is F(A, B, C)=(A+C)(A+B).
Therefore, the simplified Boolean expression for F(A, B, C) = AB+AC+BC
is (A+C)(A+B).
Question 33
Question
Simplify the following Boolean expression using laws of Boolean algebra:
F=A′B+AB +AC′+BC′+A′B′C
20
Solution
To simplify the given Boolean expression, we will use the laws of Boolean alge-
bra.
Step 1: Apply the absorption law XY +XZ =XY .
F=A′B+AB+AC′+BC′+A′B′C=A(B′+B)+AC′+BC′+A′B′C=A+AC′+BC′+A′B′C
Step 2: Apply the consensus theorem (X+Y)(X+Z)(Y+Z)=(X+
Y)(Y+Z).
F=A+AC′+BC′+A′B′C=A+AC′+BC′
Step 3: Apply the absorption law X+XY =X.
F=A+AC′+BC′=A+C′(A+B)
Hence, the simplified form of the Boolean expression is F=A+C′(A+B).
Question 34
Question
Let F(A, B, C) = ABC +ABC +ABC +ABC. Simplify F(A, B, C) using
Boolean algebra identities.
Solution
To simplify the given expression F(A, B, C), we will use Boolean algebra iden-
tities and properties.
21
Step 1: Apply the absorption law: XY +XY =X.
ABC +ABC +ABC +ABC =ABC +ABC +ABC +ABC +ABCC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC(C+C)
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC +A
=ABC +ABC +ABC +ABC +ABC +A(B+B)
=ABC +ABC +ABC +ABC +ABC +AB +AB
=ABC +ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +ABC +AB(C+C)
=ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +AB
=AB(C+C) + AB
=AB +AB
=AB
Therefore, the simplified expression for F(A, B, C) is AB.
Question 35
Question
Simplify the following Boolean expression: (A+B)(A′C′+AC).
22
Question 2
Question
Simplify the following Boolean expression: (A+B)(A′C+BC).
Solution
To simplify the given Boolean expression, we will use the distributive law, com-
plement law, and absorption law of Boolean algebra.
Step 1: Apply the distributive law (A+B)(A′C+BC).
(A+B)(A′C+BC) = AA′C+ABC +BA′C+BBC
= 0 + ABC + 0 + 0
=ABC
Step 2: Apply the absorption law XY +XZ =X(Y+Z) to simplify ABC.
ABC =A(BC)
Therefore, the simplified form of (A+B)(A′C+BC) is A(BC).
Question 3
Question
Given the Boolean expression F= (A+B)·(A+B), simplify the expression
using Boolean algebra laws.
Solution
To simplify the given Boolean expression F= (A+B)·(A+B), we can apply
various Boolean algebra laws such as the distributive law, complement law,
idempotent law, and absorption law.
Step 1: Apply the distributive law: (A+B)·(A+B) = A+B·B
Step 2: Apply the complement law: B·B= 0
Step 3: Simplify the expression: A+B·B=A+ 0 = A
Therefore, the simplified form of the Boolean expression F= (A+B)·(A+B)
is F=A.
Question 4
Question
Let F=A′BC +AB′C+ABC′. Simplify Fusing Boolean algebra laws and
theorems.
2
Solution
To simplify the given Boolean expression F=A′BC +AB′C+ABC′, we will
apply various Boolean algebra laws and theorems.
1. Use the commutative law: AB =BA
Step 1: Rearrange terms.
F=A′BC +AB′C+ABC′
=AB′C+A′BC +ABC′(commutative law)
2. Use the distributive law: A(B+C) = AB +AC
Step 2: Apply the distributive law.
F=AB′C+A′BC +ABC′
=B′C(A+A′) + BC(A+A′) (distributive law)
=B′C+BC
3. Use the absorption law: A+AB =A
Step 3: Apply the absorption law.
F=B′C+BC
=B′C+B′C′(complement law)
=B′(absorption law)
Therefore, the simplified form of Fis F=B′.
Question 5
Question
Simplify the following Boolean expression using laws of Boolean Algebra:
(A+B+C)(A+B+C)(A+B+C)
Solution
To simplify the given expression, we will use the basic laws of Boolean Algebra:
commutative law, associative law, identity law, complement law, and distribu-
tive law.
Step 1: Apply the distributive law to expand the given expression:
(A+B+C)(A+B+C)(A+B+C)
= (A(A+B+C) + B(A+B+C) + C(A+B+C))(A+B+C)
= (AA +AB +AC +BA +BB +BC +CA +CB +CC)(A+B+C)
= (A+AB +AC +BA +0+BC +CA + 0 + 0)(A+B+C)
= (A+AB +AC +BA +BC +CA)(A+B+C)
3
Step 2: Apply the distributive law again to further simplify the expression:
(A+AB +AC +BA +BC +CA)(A+B+C)
= (A(A+B+C) + AB +AC +BA +BC +CA)(A+B+C)
= (AA +AB +AC +AB +AC +BA +BC +CA)(A+B+C)
= (0 + AB +AC +AB +AC +BA +BC +CA)(A+B+C)
= (AB +AC +AB +AC +BA +BC +CA)(A+B+C)
Therefore, the simplified form of the given expression is (AB +AC +AB +
AC +BA +BC +CA)(A+B+C).
Question 6
Question
Simplify the following Boolean expression: (A+B)(A+C+D)(A+B+D+E).
Solution
To simplify the given Boolean expression (A+B)(A+C+D)(A+B+D+E), we
will use the properties of Boolean Algebra, including distribution, absorption,
complement, and idempotent laws.
Step 1: Use the distributive law to expand the expression.
(A+B)(A+C+D)(A+B+D+E)
=A(A+C+D)(A+B+D+E) + B(A+C+D)(A+B+D+E)
Step 2: Use the distributive law to further simplify each of the two terms.
=AA +AC +AD +AB +BA +BC +BD +BB
= 0 + AC +AD +AB +0+BC +BD +B
Step 3: Apply some simplifications using complementation laws and iden-
tity laws.
=AC +AD +AB +BC +B
=AC +AB +BC +B
Step 4: Apply absorption law to further simplify the expression.
=A(C+B) + B(C+ 1)
=A(C+B) + B
Step 5: Apply distributive law and another simplification.
=AC +AB +B
=AC +B
Therefore, the simplified form of the expression (A+B)(A+C+D)(A+
B+D+E) is AC +B.
4
Question 7
Question
Let A,B, and Cbe Boolean variables. Show that (A+B)·(A+B·C) = A+B·C.
Solution
To show that (A+B)·(A+B·C) = A+B·C, we will use Boolean algebra
laws and properties.
Step 1: Use the distributive law X·(Y+Z) = X·Y+X·Zto expand the
left-hand side.
(A+B)·(A+B·C) = A·A+A·B·C+B·A+B·B·C
Step 2: Apply the idempotent law X·X=Xto simplify the terms A·A
and B·B.
A·A+A·B·C+B·A+B·B·C=A+A·B·C+B·A+ 0
Step 3: Use the complementary law X·X= 0 to simplify the term A·B.
A+A·B·C+B·A+ 0 = A+0+B·A=A+B·A=A+B
Step 4: Apply the absorption law X+X·Y=Xto simplify the term B·A.
A+B
Step 5: Thus, (A+B)·(A+B·C) = A+B, which is equivalent to A+B·C.
So, (A+B)·(A+B·C) = A+B·Cis true in Boolean algebra.
Question 8
Question
Simplify the following Boolean expression: (A+B+C)(A′+B′C′).
Solution
To simplify the given Boolean expression, we will use the distributive law and
complement law of Boolean algebra.
Step 1: Apply the distributive law:
(A+B+C)(A′+B′C′) = AA′+AB′C′+AC +B′A′+B′B′C′+B′C′C′
Step 2: Apply the complement law:
AA′= 0, B′B′= 0, C′C′= 0
5
Step 3: Simplify the expression:
(A+B+C)(A′+B′C′) = 0 + AB′C′+AC +0+0+0
Step 4: Further simplify:
(A+B+C)(A′+B′C′) = AB′C′+AC
Therefore, the simplified form of the expression (A+B+C)(A′+B′C′) is
AB′C′+AC.
Question 9
Question
Given the Boolean expression (A+B)(A′+B)(A+B′), simplify the expression
using Boolean algebra rules.
Solution
To simplify the given Boolean expression (A+B)(A′+B)(A+B′), we will use
Boolean algebra rules such as distribution, complement, and absorption.
Step 1: Distribute the terms (A+B)(A′+B)
(A+B)(A′+B) = AA′+AB +BA′+BB =A′A+AB +BA′+ 0 = AB +BA′
Step 2: Use commutativity (AB =BA) to simplify AB +BA′
AB +BA′=AB +AB =A(B+B) = A
Step 3: Apply distribution to the result of step 2 and the term (A+B′)
A(A+B′) = AA +AB′=A+AB′
Step 4: Apply absorption to simplify A+AB′
A+AB′=A(1 + B′) = A×1 = A
Therefore, the simplified form of the Boolean expression (A+B)(A′+B)(A+
B′) is A.
Question 10
Question
Let A,B, and Cbe Boolean variables. Given the expression (A+B)(A+C)(B+
C), simplify the expression using Boolean algebra laws.
6
Solution
To simplify the expression (A+B)(A+C)(B+C), we will use various Boolean
algebra laws such as distribution, idempotent, and absorption laws.
Step 1: Expand using distributive law
(A+B)(A+C)(B+C) = A(A+C)(B+C) + B(A+C)(B+C)
=AA +AC +BC +BC
=A+AC +BC
Step 2: Apply idempotent law
=A+ (A+AC) + BC
Step 3: Apply absorption law
=A+BC
Therefore, the simplified expression for (A+B)(A+C)(B+C) is A+BC.
Question 11
Question
Simplify the Boolean expression (A+B)·(A′+B′) + A·B.
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
algebra such as identity, domination, idempotent, and absorption laws.
Step 1: Apply the distributive law: P·(Q+R) = P·Q+P·R, where P, Q,
and Rare Boolean variables.
(A+B)·(A′+B′) + A·B= (A·A′+A·B′+B·A′+B·B′) + A·B
= (0 + A·B′+B·A′+ 0) + A·B
= (A·B′+B·A′) + A·B
Step 2: Apply the commutative law: P+Q=Q+P.
(A·B′+B·A′) + A·B= (A·B′+A·B) + B·A′
=A·(B′+B) + B·A′
=A·1 + B·A′
=A+B·A′
Step 3: Apply the absorption law: P+P·Q=P.
A+B·A′=A
Therefore, the simplified Boolean expression is A.
7
Question 12
Question
Let F(A, B, C) = A(B+C) + BC. Simplify the Boolean expression F(A, B, C)
using Boolean algebra laws.
Solution
To simplify the given Boolean expression F(A, B, C) = A(B+C) + BC, we will
apply various Boolean algebra laws step by step.
Step 1: Apply the distributive law: X(Y+Z) = XY +XZ
A(B+C) + BC =AB +AC +BC
Step 2: Apply the absorption law: X+XZ =X
AB +AC +BC =AB +AC
Step 3: Apply the absorption law: X+XZ =X
AB +AC =AB
Therefore, the simplified Boolean expression is F(A, B, C) = AB.
Question 13
Question
Given the Boolean expression F= (A+B)(A+B), simplify the expression
using Boolean algebra laws and theorems.
Solution
Step 1: Apply the distributive law: (A+B)(A+B) = AA +AB +BA +BB
Step 2: Apply the complement law: AA = 0, BA =AB =A+B, and
BB = 0, so the expression becomes: 0 + AB +A+B+ 0
Step 3: Simplify further: F=AB +A+B
Therefore, the simplified form of the Boolean expression F= (A+B)(A+B)
is F=AB +A+B.
Question 14
Question
Simplify the following Boolean expression using Boolean Algebra laws:
F=A′B′+AB +A′B
8
Solution
Step 1: Apply the associative law:
F=A′B′+ (AB +A′B)
Step 2: Apply the distributive law:
F=A′B′+AB +A′B
Step 3: Apply the absorption law:
F=A′B′+AB
Step 4: Apply the consensus theorem:
F=A
Therefore, the simplified Boolean expression is F=A.
Question 15
Question
Simplify the following Boolean expression: (A+B)(A+B+C).
Solution
Step 1: Use the distributive property to expand the expression. Step 2: Apply
De Morgan’s laws to simplify the expression further. Step 3: Use the comple-
ment law to simplify the expression.
Step 1: Expanding the expression using the distributive property:
(A+B)(A+B+C) = AA +AB +AC +BA +BB +BC
Step 2: Applying De Morgan’s laws: AB =A+Band A+A= 1
= 0 + AB +AC +BA +0+BC
Step 3: Further simplifying using the complement law: AB +AC =A(B+
C) and BA +BC =B(A+C)
=A(B+C) + B(A+C)
Therefore, the simplified form of the given Boolean expression is A(B+C)+
B(A+C).
9
Question 16
Question
Simplify the following Boolean expression:
F= (A+B)(A+B)(A+B)
Solution
To simplify the given Boolean expression, we will use the Boolean algebra laws:
commutative law, associative law, distributive law, and De Morgan’s law.
Step 1: Apply the distributive law to expand the expression.
F=AAB +AAB +AB2+AB2
Step 2: Apply the idempotent law XX = 0 to simplify the terms AA and
B2.
F= 0 + AAB +AB + 0
Step 3: Simplify further by combining the terms with AAB and AB.
F=AB +AB
Step 4: Apply the consensus theorem (X+Y)(X+Y) = X+Zwhere
Z=Yto simplify the expression.
F=A+B
Therefore, the simplified form of the given Boolean expression is F=A+B.
Question 17
Question
Simplify the following Boolean expression: (A+BC)(A+BC +C)
Solution
To simplify the given Boolean expression, we will first expand the expression
using the distributive property, then simplify using Boolean algebra rules.
Step 1: Expand the expression using the distributive property:
(A+BC)(A+BC +C)
=AA +ABC +AC +BCA +BCBC +BCC
=0+0+AC +0+BCBC + 0
Step 2: Simplify the expression using Boolean algebra rules:
=AC + 0 + 0
=AC
Therefore, the simplified form of the given Boolean expression is AC.
10
Question 18
Question
Simplify the following Boolean expression: (a+b′+c′)(a′b+ab′).
Solution
To simplify the given Boolean expression, we will use the distributive property,
as well as the fact that x+x′= 1 and xx′= 0 for any variable x.
Step 1: (a+b′+c′)(a′b+ab′)
=aa′b+ab +ab′+a′b′+bb′+b′c′(Expand using the distributive property)
= 0 + ab +ab′+0+0+b′c′(Since aa′= 0 and bb′= 0)
=ab +ab′+b′c′
=b(a+a′) + b′c′(Factor out common terms)
=b·1 + b′c′(Since a+a′= 1)
=b+b′c′
Therefore, the simplified form of the given Boolean expression is b+b′c′.
Question 19
Question
Simplify the following Boolean expression:
(A+B)(A+C) + AC
Solution
To simplify the given Boolean expression, we will use the distributive property,
complement property (A+A′= 1, A+ 1 = 1, A∗0 = 0), and the identity
property (A+ 0 = A,A∗1 = A) of Boolean algebra.
Step 1: Apply the distributive property to expand (A+B)(A+C):
(A+B)(A+C) = AA +AC +AB +BC
Step 2: Simplify the expansion using the idempotent law and complemen-
tary law:
AA =A, AB =A·B, AC = 0 (complementary law)
11
BC = 0 (complementary law)
So, the expression simplifies to:
(A+B)(A+C) = A+A·B+ 0 + 0 = A+A·B
Step 3: Now, simplify the expression obtained in Step 2 by applying the
identity law:
A+A·B=A(1 + B) = A·1 = A
Step 4: Finally, simplify the original expression (A+B)(A+C) + AC using
the result from Step 3:
(A+B)(A+C) + AC =A+AC =A(1 + C) = A·1 = A
Therefore, the simplified form of the Boolean expression (A+B)(A+C)+AC
is A.
Question 20
Question
Simplify the following Boolean expression: (A+B)·(A·B).
Solution
We will simplify the given Boolean expression step by step using basic rules of
Boolean algebra.
Step 1: Apply De Morgan’s law: (A·B) = A+B.
Step 2: Apply the distributive property: (A+B)·A·B.
Step 3: Apply the idempotent law: A·A= 0.
Step 4: Use the absorption law: X·(X+Y) = X.
Step 5: Use the complement law: X·X= 0.
Step 6: The final simplified expression is 0.
Therefore, (A+B)·(A·B) simplifies to 0.
Question 21
Question
Simplify the following Boolean expression:
F=A′B′C+A′BC +AB′C+ABC
12
Solution
To simplify the given Boolean expression, we will use the laws of Boolean al-
gebra such as identity, domination, double negation, idempotent, commutative,
associative, distributive, and De Morgan’s laws.
Step 1: Apply the commutative law to rearrange the terms:
F=A′B′C+A′BC +AB′C+ABC
Step 2: Factor out common terms:
F=A′B′C+A′BC +AB′C+ABC =A′B′(C+C′)+AB′(C+C′)+A′B(C+C′)
Step 3: Apply the idempotent law (X+X=X) to simplify the terms
inside the parentheses:
F=A′B′(1) + AB′(1) + A′B(1)
Step 4: Apply the identity law (X·1 = X) to further simplify the expression:
F=A′B+AB′+A′B=A′B+AB′+AB
Step 5: Apply the idempotent law again:
F=A′B+AB′+AB =A′B+AB′
Step 6: Apply the distributive law to simplify the expression:
F=A′B+AB′=B(A′+A) = B(1) = B
Therefore, the simplified Boolean expression is F=B.
Question 22
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+C)
Solution
Let’s simplify the given Boolean expression step by step:
Step 1: Apply the distributive property to expand the expression.
(A+B+C)(A+B+C)(A+B+C)
= (AA +AB +AC +BA +BB +BC +CA +CB +CC)
Step 2: Simplify the terms with complements of each variable.
= (0 + AB +AC +BA +0+0+0+CB + 0)
13
= (AB +AC +BA +CB)
Step 3: Apply the commutative property to rearrange the variables.
= (AB +BA +AC +CB)
Step 4: Apply the complementary property XY +XY =X+Y.
= (A+B)+(A+C)
Therefore, the simplified Boolean expression is (A+B)+(A+C).
Question 23
Question
Simplify the following Boolean expression:
F= (A+B)·(A+B+C)·(A+B)
Solution
Step 1: Distribute the terms inside the parentheses using the distributive prop-
erty. Step 2: Simplify the expression by combining like terms and applying
Boolean identities.
Step 1:
F= (A+B)·(A+B+C)·(A+B)
F=A·A·B+A·A·C+A·B+B·A·B+B·A·C+B·B
Step 2: Apply the following Boolean identities:
X·X= 0
X+X= 1
1·X=X
0 + X=X
To simplify the expression:
F=0+0+A·B+0+0+B
F=A·B+B
F=B·(A+ 1)
F=B
Thus, the simplified Boolean expression for F is B.
14
Question 24
Question
Simplify the following Boolean expression: (A+B)(A+C)(A+C).
Solution
To simplify the given Boolean expression, we will use the basic laws of Boolean
algebra, such as the distributive law, the complement law, and the identity law.
Step 1: Apply the distributive law to expand the expression.
(A+B)(A+C)(A+C) = A(A+C)(A+C) + B(A+C)(A+C)
Step 2: Use the complement law AA = 0 to simplify A(A+C).
A(A+C) = 0 + AC =AC
Step 3: Use the distributive law again to simplify B(A+C)(A+C).
B(A+C)(A+C) = BAA +BC +BAC
Step 4: Apply the complement law AA = 0 to simplify BAA.
BAA = 0
Step 5: Combine the simplified terms to get the final expression.
(A+B)(A+C)(A+C) = AC +BC +BAC =AC +BC
Therefore, the simplified Boolean expression is AC +BC.
Question 25
Question
Simplify the following Boolean expression: (A+B+C)(A+B)(A+B+C).
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
algebra such as the distributive law, identity law, and complement law.
Step 1: Use the distributive law to expand the expression.
(A+B+C)(A+B)(A+B+C) = A(A+B)(A+B+C)+B(A+B)(A+B+C)+C(A+B)(A+B+C)
Step 2: Apply the complement law.
=A(0 + B)(A+B+C) + B(A+ 1)(A+B+C) + C(A+B)(A+B+C)
15
Step 3: Simplify the expressions.
=AB(A+B+C) + B(A+A+B+C) + C(A+B)(A+B+C)
=AB +ABB +ABC +BA +BA +BB +BC +CA +BC
Step 4: Simplify further using the complement law.
=AB +0+ABC +BA +BA +0+0+CA +BC
=AB +ABC +BA +BA +CA +BC
Step 5: Combine like terms.
=AB(1 + C) + BA +CA +BC
=AB +BA +CA +BC
Thus, the simplified Boolean expression is AB +BA +CA +BC.
Question 26
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+
C)(A+B+C).
Solution
Step 1: Apply the distributive property to expand the expression. Step 2:
Eliminate duplicate terms. Step 3: Use the idempotent law to simplify further.
Step 4: Apply the complement law to simplify the expression.
Step 1:
(A+B+C)(A+B+C)(A+B+C)(A+B+C)
Expanding using the distributive property:
=AA+AB+AC+AB+ABC+AC+AC+ACB+ACC+AB+AB+AC+AB+AC+ABC
Step 2: Eliminating duplicate terms:
=A+AB +AC +AB +ABC +AC +ACB +AB +AC +ABC
Step 3: Using the idempotent law: XX =Xfor any variable X:
=A+AB +AC +AB +AC +AB +AC +ABC
Step 4: Applying the complement law: XX = 0 for any variable X:
=A+AB +AC +AB +AC +BC
Therefore, the simplified Boolean expression is A+AB+AC+AB+AC+BC.
16
Question 27
Question
Simplify the following Boolean expression: (A+B+C)(A′+B+C)(A+B′+
C′)(A+B+C′).
Solution
Step 1: Use the distributive property to expand the expression.
(A+B+C)(A′+B+C)(A+B′+C′)(A+B+C′)=(AA′+AB +AC +BA′+BB +BC +CA +CB +CC)(A+B′+C′)(A+B+C′)
= (0+AB +AC +0+0+0+0+CB +0)(A+B′+C′)(A+B+C′)
= (AB +AC +CB)(A+B′+C′)(A+B+C′)
Step 2: Apply the distributive property again to simplify further.
(AB +AC +CB)(A+B′+C′)(A+B+C′) = AB(A+B′+C′)(A+B+C′) + AC(A+B′+C′)(A+B+C′) + CB(A+B′+C′)(A+B+C′)
=ABA +ABB′+ABC′+ACA +ACB′+ACC′+CBA +CBB′+CBC′
=0+0+0+ACA +0+0+0+0+0
=ACA
Therefore, the simplified expression is ACA.
Question 28
Question
Simplify the Boolean expression: (A+B)(A+B)(A+B)(A+B).
Solution
Let’s simplify the given Boolean expression step by step.
Step 1: Use the distributive property to expand the expression.
(A+B)(A+B)(A+B)(A+B) = (A+B)(A+B)(AA +AB +AB +AB)
Step 2: Simplify the terms AA and AA.
(A+B)(A+B)(AA +AB +AB +AB)=(A+B)(A+B)(0 + AB +AB +AB)
Step 3: Simplify the expression AA by noting that it is always equal to 0.
(A+B)(A+B)(0 + AB +AB +AB) = (A+B)(A+B)(AB +AB +AB)
Step 4: Apply the distributive property again to expand the expression.
(A+B)(A+B)(AB +AB +AB) = AAB +AAB +BAB +BAB +ABA +ABA
17
Step 5: Simplify the terms AA,BB, and AA.
AAB +AAB +BAB +BAB +ABA +ABA =0+0+0+0+0+0=0
Step 6: Therefore, the simplified form of the Boolean expression is 0 .
Question 29
Question
Simplify the following Boolean expression: (A+B)(A+C)(B+C).
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
Algebra, including the distributive property, idempotent laws, and complemen-
tary laws.
Step 1: Apply the distributive property to expand the expression.
(A+B)(A+C)(B+C) = AAB+AAC+ABC+BCB+BCC+CAC =A+AB+AC+BC+C
Step 2: Apply the idempotent law (A+A=A) to simplify the expression.
A+AB +AC +BC +C=A+AC +BC +C
Step 3: Apply the complementary law (A+A= 1) to simplify the expression
further.
A+AC +BC +C=A+ (A+C)(B+C) = A+ 1(B+C) = A+B+C
Therefore, the simplified form of (A+B)(A+C)(B+C) is A+B+C.
Question 30
Question
Simplify the following Boolean expression using Boolean Algebra laws:
F= (A+B)(A′+C)(B′+C′)
Solution
Step 1: Apply the Distributive Law to expand the given expression:
F= (A+B)(A′+C)(B′+C′)
F=AA′+AC +BB′+BC +A′B′+A′C+B′C′
18
Step 2: Apply the Null Law (A+A′= 1) to simplify the term AA′:
F= 0 + AC +BB′+BC +A′B′+A′C+B′C′
F=AC +BB′+BC +A′B′+A′C+B′C′
Step 3: Apply the Complement Law to simplify the terms BB′and A′B′:
F=AC +0+BC +0+A′C+B′C′
F=AC +BC +A′C+B′C′
Step 4: Apply the Absorption Law to simplify the term AC +BC:
F=C(A+B) + A′C+B′C′
Step 5: Apply the Consensus Theorem: XY +X′Z+Y Z =XY +X′Zto
simplify the expression:
F=C(A+B) + A′C+B′C′
F=C+A′C+B′C′
Step 6: Apply the Idempotent Law to simplify the expression further:
F=C+A′C+B′C′
F=C+C′A+C′B′
F=C+A+B′
Question 31
Question
Simplify the Boolean expression (A+B)·(A′+C) + A·B·Cusing Boolean
algebra properties.
Solution
To simplify the given Boolean expression, we will apply various Boolean alge-
bra properties such as the distributive property, complement property, and the
idempotent property.
Step 1: Apply the distributive property: (A+B)·(A′+C) + A·B·C
⇒(AA′+AC +BB′+BC) + ABC
Step 2: Use the complement property AA′= 0 and BB′= 0:
⇒(0 + AC +0+BC) + ABC
19
⇒AC +BC +ABC
Step 3: Factor out a common term C:
⇒C(A+B+AB)
Step 4: Use the absorption property A+AB =A:
⇒C(A)
⇒AC
Therefore, the simplified form of the Boolean expression is AC.
Question 32
Question
Given the Boolean expression F(A, B, C) = AB +AC +BC, simplify the ex-
pression using Boolean Algebra rules.
Solution
To simplify the given Boolean expression F(A, B, C) = AB +AC +BC, we will
use various Boolean Algebra rules such as the Identity Law, Domination Law,
Idempotent Law, Complement Law, etc.
Step 1: Apply the Absorption Law: AB +AC = (A+A)(A+C) =
1(A+C) = A+C.
Therefore, the expression simplifies to F(A, B, C) = A+C+BC.
Step 2: Apply the Absorption Law again: A+C+BC = (A+C+B)(A+
C+C).
This simplifies further to F(A, B, C) = A+C+B.
Step 3: Apply the Consensus Theorem: A+C+B= (A+C)(A+B).
The final simplified expression is F(A, B, C)=(A+C)(A+B).
Therefore, the simplified Boolean expression for F(A, B, C) = AB+AC+BC
is (A+C)(A+B).
Question 33
Question
Simplify the following Boolean expression using laws of Boolean algebra:
F=A′B+AB +AC′+BC′+A′B′C
20
Solution
To simplify the given Boolean expression, we will use the laws of Boolean alge-
bra.
Step 1: Apply the absorption law XY +XZ =XY .
F=A′B+AB+AC′+BC′+A′B′C=A(B′+B)+AC′+BC′+A′B′C=A+AC′+BC′+A′B′C
Step 2: Apply the consensus theorem (X+Y)(X+Z)(Y+Z)=(X+
Y)(Y+Z).
F=A+AC′+BC′+A′B′C=A+AC′+BC′
Step 3: Apply the absorption law X+XY =X.
F=A+AC′+BC′=A+C′(A+B)
Hence, the simplified form of the Boolean expression is F=A+C′(A+B).
Question 34
Question
Let F(A, B, C) = ABC +ABC +ABC +ABC. Simplify F(A, B, C) using
Boolean algebra identities.
Solution
To simplify the given expression F(A, B, C), we will use Boolean algebra iden-
tities and properties.
21
Step 1: Apply the absorption law: XY +XY =X.
ABC +ABC +ABC +ABC =ABC +ABC +ABC +ABC +ABCC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC(C+C)
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC +A
=ABC +ABC +ABC +ABC +ABC +A(B+B)
=ABC +ABC +ABC +ABC +ABC +AB +AB
=ABC +ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +ABC +AB(C+C)
=ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +AB
=AB(C+C) + AB
=AB +AB
=AB
Therefore, the simplified expression for F(A, B, C) is AB.
Question 35
Question
Simplify the following Boolean expression: (A+B)(A′C′+AC).
22
Question 2
Question
Simplify the following Boolean expression: (A+B)(A′C+BC).
Solution
To simplify the given Boolean expression, we will use the distributive law, com-
plement law, and absorption law of Boolean algebra.
Step 1: Apply the distributive law (A+B)(A′C+BC).
(A+B)(A′C+BC) = AA′C+ABC +BA′C+BBC
= 0 + ABC + 0 + 0
=ABC
Step 2: Apply the absorption law XY +XZ =X(Y+Z) to simplify ABC.
ABC =A(BC)
Therefore, the simplified form of (A+B)(A′C+BC) is A(BC).
Question 3
Question
Given the Boolean expression F= (A+B)·(A+B), simplify the expression
using Boolean algebra laws.
Solution
To simplify the given Boolean expression F= (A+B)·(A+B), we can apply
various Boolean algebra laws such as the distributive law, complement law,
idempotent law, and absorption law.
Step 1: Apply the distributive law: (A+B)·(A+B) = A+B·B
Step 2: Apply the complement law: B·B= 0
Step 3: Simplify the expression: A+B·B=A+ 0 = A
Therefore, the simplified form of the Boolean expression F= (A+B)·(A+B)
is F=A.
Question 4
Question
Let F=A′BC +AB′C+ABC′. Simplify Fusing Boolean algebra laws and
theorems.
2
Solution
To simplify the given Boolean expression F=A′BC +AB′C+ABC′, we will
apply various Boolean algebra laws and theorems.
1. Use the commutative law: AB =BA
Step 1: Rearrange terms.
F=A′BC +AB′C+ABC′
=AB′C+A′BC +ABC′(commutative law)
2. Use the distributive law: A(B+C) = AB +AC
Step 2: Apply the distributive law.
F=AB′C+A′BC +ABC′
=B′C(A+A′) + BC(A+A′) (distributive law)
=B′C+BC
3. Use the absorption law: A+AB =A
Step 3: Apply the absorption law.
F=B′C+BC
=B′C+B′C′(complement law)
=B′(absorption law)
Therefore, the simplified form of Fis F=B′.
Question 5
Question
Simplify the following Boolean expression using laws of Boolean Algebra:
(A+B+C)(A+B+C)(A+B+C)
Solution
To simplify the given expression, we will use the basic laws of Boolean Algebra:
commutative law, associative law, identity law, complement law, and distribu-
tive law.
Step 1: Apply the distributive law to expand the given expression:
(A+B+C)(A+B+C)(A+B+C)
= (A(A+B+C) + B(A+B+C) + C(A+B+C))(A+B+C)
= (AA +AB +AC +BA +BB +BC +CA +CB +CC)(A+B+C)
= (A+AB +AC +BA +0+BC +CA + 0 + 0)(A+B+C)
= (A+AB +AC +BA +BC +CA)(A+B+C)
3
Step 2: Apply the distributive law again to further simplify the expression:
(A+AB +AC +BA +BC +CA)(A+B+C)
= (A(A+B+C) + AB +AC +BA +BC +CA)(A+B+C)
= (AA +AB +AC +AB +AC +BA +BC +CA)(A+B+C)
= (0 + AB +AC +AB +AC +BA +BC +CA)(A+B+C)
= (AB +AC +AB +AC +BA +BC +CA)(A+B+C)
Therefore, the simplified form of the given expression is (AB +AC +AB +
AC +BA +BC +CA)(A+B+C).
Question 6
Question
Simplify the following Boolean expression: (A+B)(A+C+D)(A+B+D+E).
Solution
To simplify the given Boolean expression (A+B)(A+C+D)(A+B+D+E), we
will use the properties of Boolean Algebra, including distribution, absorption,
complement, and idempotent laws.
Step 1: Use the distributive law to expand the expression.
(A+B)(A+C+D)(A+B+D+E)
=A(A+C+D)(A+B+D+E) + B(A+C+D)(A+B+D+E)
Step 2: Use the distributive law to further simplify each of the two terms.
=AA +AC +AD +AB +BA +BC +BD +BB
= 0 + AC +AD +AB +0+BC +BD +B
Step 3: Apply some simplifications using complementation laws and iden-
tity laws.
=AC +AD +AB +BC +B
=AC +AB +BC +B
Step 4: Apply absorption law to further simplify the expression.
=A(C+B) + B(C+ 1)
=A(C+B) + B
Step 5: Apply distributive law and another simplification.
=AC +AB +B
=AC +B
Therefore, the simplified form of the expression (A+B)(A+C+D)(A+
B+D+E) is AC +B.
4
Question 7
Question
Let A,B, and Cbe Boolean variables. Show that (A+B)·(A+B·C) = A+B·C.
Solution
To show that (A+B)·(A+B·C) = A+B·C, we will use Boolean algebra
laws and properties.
Step 1: Use the distributive law X·(Y+Z) = X·Y+X·Zto expand the
left-hand side.
(A+B)·(A+B·C) = A·A+A·B·C+B·A+B·B·C
Step 2: Apply the idempotent law X·X=Xto simplify the terms A·A
and B·B.
A·A+A·B·C+B·A+B·B·C=A+A·B·C+B·A+ 0
Step 3: Use the complementary law X·X= 0 to simplify the term A·B.
A+A·B·C+B·A+ 0 = A+0+B·A=A+B·A=A+B
Step 4: Apply the absorption law X+X·Y=Xto simplify the term B·A.
A+B
Step 5: Thus, (A+B)·(A+B·C) = A+B, which is equivalent to A+B·C.
So, (A+B)·(A+B·C) = A+B·Cis true in Boolean algebra.
Question 8
Question
Simplify the following Boolean expression: (A+B+C)(A′+B′C′).
Solution
To simplify the given Boolean expression, we will use the distributive law and
complement law of Boolean algebra.
Step 1: Apply the distributive law:
(A+B+C)(A′+B′C′) = AA′+AB′C′+AC +B′A′+B′B′C′+B′C′C′
Step 2: Apply the complement law:
AA′= 0, B′B′= 0, C′C′= 0
5
Step 3: Simplify the expression:
(A+B+C)(A′+B′C′) = 0 + AB′C′+AC +0+0+0
Step 4: Further simplify:
(A+B+C)(A′+B′C′) = AB′C′+AC
Therefore, the simplified form of the expression (A+B+C)(A′+B′C′) is
AB′C′+AC.
Question 9
Question
Given the Boolean expression (A+B)(A′+B)(A+B′), simplify the expression
using Boolean algebra rules.
Solution
To simplify the given Boolean expression (A+B)(A′+B)(A+B′), we will use
Boolean algebra rules such as distribution, complement, and absorption.
Step 1: Distribute the terms (A+B)(A′+B)
(A+B)(A′+B) = AA′+AB +BA′+BB =A′A+AB +BA′+ 0 = AB +BA′
Step 2: Use commutativity (AB =BA) to simplify AB +BA′
AB +BA′=AB +AB =A(B+B) = A
Step 3: Apply distribution to the result of step 2 and the term (A+B′)
A(A+B′) = AA +AB′=A+AB′
Step 4: Apply absorption to simplify A+AB′
A+AB′=A(1 + B′) = A×1 = A
Therefore, the simplified form of the Boolean expression (A+B)(A′+B)(A+
B′) is A.
Question 10
Question
Let A,B, and Cbe Boolean variables. Given the expression (A+B)(A+C)(B+
C), simplify the expression using Boolean algebra laws.
6
Solution
To simplify the expression (A+B)(A+C)(B+C), we will use various Boolean
algebra laws such as distribution, idempotent, and absorption laws.
Step 1: Expand using distributive law
(A+B)(A+C)(B+C) = A(A+C)(B+C) + B(A+C)(B+C)
=AA +AC +BC +BC
=A+AC +BC
Step 2: Apply idempotent law
=A+ (A+AC) + BC
Step 3: Apply absorption law
=A+BC
Therefore, the simplified expression for (A+B)(A+C)(B+C) is A+BC.
Question 11
Question
Simplify the Boolean expression (A+B)·(A′+B′) + A·B.
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
algebra such as identity, domination, idempotent, and absorption laws.
Step 1: Apply the distributive law: P·(Q+R) = P·Q+P·R, where P, Q,
and Rare Boolean variables.
(A+B)·(A′+B′) + A·B= (A·A′+A·B′+B·A′+B·B′) + A·B
= (0 + A·B′+B·A′+ 0) + A·B
= (A·B′+B·A′) + A·B
Step 2: Apply the commutative law: P+Q=Q+P.
(A·B′+B·A′) + A·B= (A·B′+A·B) + B·A′
=A·(B′+B) + B·A′
=A·1 + B·A′
=A+B·A′
Step 3: Apply the absorption law: P+P·Q=P.
A+B·A′=A
Therefore, the simplified Boolean expression is A.
7
Question 12
Question
Let F(A, B, C) = A(B+C) + BC. Simplify the Boolean expression F(A, B, C)
using Boolean algebra laws.
Solution
To simplify the given Boolean expression F(A, B, C) = A(B+C) + BC, we will
apply various Boolean algebra laws step by step.
Step 1: Apply the distributive law: X(Y+Z) = XY +XZ
A(B+C) + BC =AB +AC +BC
Step 2: Apply the absorption law: X+XZ =X
AB +AC +BC =AB +AC
Step 3: Apply the absorption law: X+XZ =X
AB +AC =AB
Therefore, the simplified Boolean expression is F(A, B, C) = AB.
Question 13
Question
Given the Boolean expression F= (A+B)(A+B), simplify the expression
using Boolean algebra laws and theorems.
Solution
Step 1: Apply the distributive law: (A+B)(A+B) = AA +AB +BA +BB
Step 2: Apply the complement law: AA = 0, BA =AB =A+B, and
BB = 0, so the expression becomes: 0 + AB +A+B+ 0
Step 3: Simplify further: F=AB +A+B
Therefore, the simplified form of the Boolean expression F= (A+B)(A+B)
is F=AB +A+B.
Question 14
Question
Simplify the following Boolean expression using Boolean Algebra laws:
F=A′B′+AB +A′B
8
Solution
Step 1: Apply the associative law:
F=A′B′+ (AB +A′B)
Step 2: Apply the distributive law:
F=A′B′+AB +A′B
Step 3: Apply the absorption law:
F=A′B′+AB
Step 4: Apply the consensus theorem:
F=A
Therefore, the simplified Boolean expression is F=A.
Question 15
Question
Simplify the following Boolean expression: (A+B)(A+B+C).
Solution
Step 1: Use the distributive property to expand the expression. Step 2: Apply
De Morgan’s laws to simplify the expression further. Step 3: Use the comple-
ment law to simplify the expression.
Step 1: Expanding the expression using the distributive property:
(A+B)(A+B+C) = AA +AB +AC +BA +BB +BC
Step 2: Applying De Morgan’s laws: AB =A+Band A+A= 1
= 0 + AB +AC +BA +0+BC
Step 3: Further simplifying using the complement law: AB +AC =A(B+
C) and BA +BC =B(A+C)
=A(B+C) + B(A+C)
Therefore, the simplified form of the given Boolean expression is A(B+C)+
B(A+C).
9
Question 16
Question
Simplify the following Boolean expression:
F= (A+B)(A+B)(A+B)
Solution
To simplify the given Boolean expression, we will use the Boolean algebra laws:
commutative law, associative law, distributive law, and De Morgan’s law.
Step 1: Apply the distributive law to expand the expression.
F=AAB +AAB +AB2+AB2
Step 2: Apply the idempotent law XX = 0 to simplify the terms AA and
B2.
F= 0 + AAB +AB + 0
Step 3: Simplify further by combining the terms with AAB and AB.
F=AB +AB
Step 4: Apply the consensus theorem (X+Y)(X+Y) = X+Zwhere
Z=Yto simplify the expression.
F=A+B
Therefore, the simplified form of the given Boolean expression is F=A+B.
Question 17
Question
Simplify the following Boolean expression: (A+BC)(A+BC +C)
Solution
To simplify the given Boolean expression, we will first expand the expression
using the distributive property, then simplify using Boolean algebra rules.
Step 1: Expand the expression using the distributive property:
(A+BC)(A+BC +C)
=AA +ABC +AC +BCA +BCBC +BCC
=0+0+AC +0+BCBC + 0
Step 2: Simplify the expression using Boolean algebra rules:
=AC + 0 + 0
=AC
Therefore, the simplified form of the given Boolean expression is AC.
10
Question 18
Question
Simplify the following Boolean expression: (a+b′+c′)(a′b+ab′).
Solution
To simplify the given Boolean expression, we will use the distributive property,
as well as the fact that x+x′= 1 and xx′= 0 for any variable x.
Step 1: (a+b′+c′)(a′b+ab′)
=aa′b+ab +ab′+a′b′+bb′+b′c′(Expand using the distributive property)
= 0 + ab +ab′+0+0+b′c′(Since aa′= 0 and bb′= 0)
=ab +ab′+b′c′
=b(a+a′) + b′c′(Factor out common terms)
=b·1 + b′c′(Since a+a′= 1)
=b+b′c′
Therefore, the simplified form of the given Boolean expression is b+b′c′.
Question 19
Question
Simplify the following Boolean expression:
(A+B)(A+C) + AC
Solution
To simplify the given Boolean expression, we will use the distributive property,
complement property (A+A′= 1, A+ 1 = 1, A∗0 = 0), and the identity
property (A+ 0 = A,A∗1 = A) of Boolean algebra.
Step 1: Apply the distributive property to expand (A+B)(A+C):
(A+B)(A+C) = AA +AC +AB +BC
Step 2: Simplify the expansion using the idempotent law and complemen-
tary law:
AA =A, AB =A·B, AC = 0 (complementary law)
11
BC = 0 (complementary law)
So, the expression simplifies to:
(A+B)(A+C) = A+A·B+ 0 + 0 = A+A·B
Step 3: Now, simplify the expression obtained in Step 2 by applying the
identity law:
A+A·B=A(1 + B) = A·1 = A
Step 4: Finally, simplify the original expression (A+B)(A+C) + AC using
the result from Step 3:
(A+B)(A+C) + AC =A+AC =A(1 + C) = A·1 = A
Therefore, the simplified form of the Boolean expression (A+B)(A+C)+AC
is A.
Question 20
Question
Simplify the following Boolean expression: (A+B)·(A·B).
Solution
We will simplify the given Boolean expression step by step using basic rules of
Boolean algebra.
Step 1: Apply De Morgan’s law: (A·B) = A+B.
Step 2: Apply the distributive property: (A+B)·A·B.
Step 3: Apply the idempotent law: A·A= 0.
Step 4: Use the absorption law: X·(X+Y) = X.
Step 5: Use the complement law: X·X= 0.
Step 6: The final simplified expression is 0.
Therefore, (A+B)·(A·B) simplifies to 0.
Question 21
Question
Simplify the following Boolean expression:
F=A′B′C+A′BC +AB′C+ABC
12
Solution
To simplify the given Boolean expression, we will use the laws of Boolean al-
gebra such as identity, domination, double negation, idempotent, commutative,
associative, distributive, and De Morgan’s laws.
Step 1: Apply the commutative law to rearrange the terms:
F=A′B′C+A′BC +AB′C+ABC
Step 2: Factor out common terms:
F=A′B′C+A′BC +AB′C+ABC =A′B′(C+C′)+AB′(C+C′)+A′B(C+C′)
Step 3: Apply the idempotent law (X+X=X) to simplify the terms
inside the parentheses:
F=A′B′(1) + AB′(1) + A′B(1)
Step 4: Apply the identity law (X·1 = X) to further simplify the expression:
F=A′B+AB′+A′B=A′B+AB′+AB
Step 5: Apply the idempotent law again:
F=A′B+AB′+AB =A′B+AB′
Step 6: Apply the distributive law to simplify the expression:
F=A′B+AB′=B(A′+A) = B(1) = B
Therefore, the simplified Boolean expression is F=B.
Question 22
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+C)
Solution
Let’s simplify the given Boolean expression step by step:
Step 1: Apply the distributive property to expand the expression.
(A+B+C)(A+B+C)(A+B+C)
= (AA +AB +AC +BA +BB +BC +CA +CB +CC)
Step 2: Simplify the terms with complements of each variable.
= (0 + AB +AC +BA +0+0+0+CB + 0)
13
= (AB +AC +BA +CB)
Step 3: Apply the commutative property to rearrange the variables.
= (AB +BA +AC +CB)
Step 4: Apply the complementary property XY +XY =X+Y.
= (A+B)+(A+C)
Therefore, the simplified Boolean expression is (A+B)+(A+C).
Question 23
Question
Simplify the following Boolean expression:
F= (A+B)·(A+B+C)·(A+B)
Solution
Step 1: Distribute the terms inside the parentheses using the distributive prop-
erty. Step 2: Simplify the expression by combining like terms and applying
Boolean identities.
Step 1:
F= (A+B)·(A+B+C)·(A+B)
F=A·A·B+A·A·C+A·B+B·A·B+B·A·C+B·B
Step 2: Apply the following Boolean identities:
X·X= 0
X+X= 1
1·X=X
0 + X=X
To simplify the expression:
F=0+0+A·B+0+0+B
F=A·B+B
F=B·(A+ 1)
F=B
Thus, the simplified Boolean expression for F is B.
14
Question 24
Question
Simplify the following Boolean expression: (A+B)(A+C)(A+C).
Solution
To simplify the given Boolean expression, we will use the basic laws of Boolean
algebra, such as the distributive law, the complement law, and the identity law.
Step 1: Apply the distributive law to expand the expression.
(A+B)(A+C)(A+C) = A(A+C)(A+C) + B(A+C)(A+C)
Step 2: Use the complement law AA = 0 to simplify A(A+C).
A(A+C) = 0 + AC =AC
Step 3: Use the distributive law again to simplify B(A+C)(A+C).
B(A+C)(A+C) = BAA +BC +BAC
Step 4: Apply the complement law AA = 0 to simplify BAA.
BAA = 0
Step 5: Combine the simplified terms to get the final expression.
(A+B)(A+C)(A+C) = AC +BC +BAC =AC +BC
Therefore, the simplified Boolean expression is AC +BC.
Question 25
Question
Simplify the following Boolean expression: (A+B+C)(A+B)(A+B+C).
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
algebra such as the distributive law, identity law, and complement law.
Step 1: Use the distributive law to expand the expression.
(A+B+C)(A+B)(A+B+C) = A(A+B)(A+B+C)+B(A+B)(A+B+C)+C(A+B)(A+B+C)
Step 2: Apply the complement law.
=A(0 + B)(A+B+C) + B(A+ 1)(A+B+C) + C(A+B)(A+B+C)
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Step 3: Simplify the expressions.
=AB(A+B+C) + B(A+A+B+C) + C(A+B)(A+B+C)
=AB +ABB +ABC +BA +BA +BB +BC +CA +BC
Step 4: Simplify further using the complement law.
=AB +0+ABC +BA +BA +0+0+CA +BC
=AB +ABC +BA +BA +CA +BC
Step 5: Combine like terms.
=AB(1 + C) + BA +CA +BC
=AB +BA +CA +BC
Thus, the simplified Boolean expression is AB +BA +CA +BC.
Question 26
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+
C)(A+B+C).
Solution
Step 1: Apply the distributive property to expand the expression. Step 2:
Eliminate duplicate terms. Step 3: Use the idempotent law to simplify further.
Step 4: Apply the complement law to simplify the expression.
Step 1:
(A+B+C)(A+B+C)(A+B+C)(A+B+C)
Expanding using the distributive property:
=AA+AB+AC+AB+ABC+AC+AC+ACB+ACC+AB+AB+AC+AB+AC+ABC
Step 2: Eliminating duplicate terms:
=A+AB +AC +AB +ABC +AC +ACB +AB +AC +ABC
Step 3: Using the idempotent law: XX =Xfor any variable X:
=A+AB +AC +AB +AC +AB +AC +ABC
Step 4: Applying the complement law: XX = 0 for any variable X:
=A+AB +AC +AB +AC +BC
Therefore, the simplified Boolean expression is A+AB+AC+AB+AC+BC.
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Question 27
Question
Simplify the following Boolean expression: (A+B+C)(A′+B+C)(A+B′+
C′)(A+B+C′).
Solution
Step 1: Use the distributive property to expand the expression.
(A+B+C)(A′+B+C)(A+B′+C′)(A+B+C′)=(AA′+AB +AC +BA′+BB +BC +CA +CB +CC)(A+B′+C′)(A+B+C′)
= (0+AB +AC +0+0+0+0+CB +0)(A+B′+C′)(A+B+C′)
= (AB +AC +CB)(A+B′+C′)(A+B+C′)
Step 2: Apply the distributive property again to simplify further.
(AB +AC +CB)(A+B′+C′)(A+B+C′) = AB(A+B′+C′)(A+B+C′) + AC(A+B′+C′)(A+B+C′) + CB(A+B′+C′)(A+B+C′)
=ABA +ABB′+ABC′+ACA +ACB′+ACC′+CBA +CBB′+CBC′
=0+0+0+ACA +0+0+0+0+0
=ACA
Therefore, the simplified expression is ACA.
Question 28
Question
Simplify the Boolean expression: (A+B)(A+B)(A+B)(A+B).
Solution
Let’s simplify the given Boolean expression step by step.
Step 1: Use the distributive property to expand the expression.
(A+B)(A+B)(A+B)(A+B) = (A+B)(A+B)(AA +AB +AB +AB)
Step 2: Simplify the terms AA and AA.
(A+B)(A+B)(AA +AB +AB +AB)=(A+B)(A+B)(0 + AB +AB +AB)
Step 3: Simplify the expression AA by noting that it is always equal to 0.
(A+B)(A+B)(0 + AB +AB +AB) = (A+B)(A+B)(AB +AB +AB)
Step 4: Apply the distributive property again to expand the expression.
(A+B)(A+B)(AB +AB +AB) = AAB +AAB +BAB +BAB +ABA +ABA
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Step 5: Simplify the terms AA,BB, and AA.
AAB +AAB +BAB +BAB +ABA +ABA =0+0+0+0+0+0=0
Step 6: Therefore, the simplified form of the Boolean expression is 0 .
Question 29
Question
Simplify the following Boolean expression: (A+B)(A+C)(B+C).
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
Algebra, including the distributive property, idempotent laws, and complemen-
tary laws.
Step 1: Apply the distributive property to expand the expression.
(A+B)(A+C)(B+C) = AAB+AAC+ABC+BCB+BCC+CAC =A+AB+AC+BC+C
Step 2: Apply the idempotent law (A+A=A) to simplify the expression.
A+AB +AC +BC +C=A+AC +BC +C
Step 3: Apply the complementary law (A+A= 1) to simplify the expression
further.
A+AC +BC +C=A+ (A+C)(B+C) = A+ 1(B+C) = A+B+C
Therefore, the simplified form of (A+B)(A+C)(B+C) is A+B+C.
Question 30
Question
Simplify the following Boolean expression using Boolean Algebra laws:
F= (A+B)(A′+C)(B′+C′)
Solution
Step 1: Apply the Distributive Law to expand the given expression:
F= (A+B)(A′+C)(B′+C′)
F=AA′+AC +BB′+BC +A′B′+A′C+B′C′
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Step 2: Apply the Null Law (A+A′= 1) to simplify the term AA′:
F= 0 + AC +BB′+BC +A′B′+A′C+B′C′
F=AC +BB′+BC +A′B′+A′C+B′C′
Step 3: Apply the Complement Law to simplify the terms BB′and A′B′:
F=AC +0+BC +0+A′C+B′C′
F=AC +BC +A′C+B′C′
Step 4: Apply the Absorption Law to simplify the term AC +BC:
F=C(A+B) + A′C+B′C′
Step 5: Apply the Consensus Theorem: XY +X′Z+Y Z =XY +X′Zto
simplify the expression:
F=C(A+B) + A′C+B′C′
F=C+A′C+B′C′
Step 6: Apply the Idempotent Law to simplify the expression further:
F=C+A′C+B′C′
F=C+C′A+C′B′
F=C+A+B′
Question 31
Question
Simplify the Boolean expression (A+B)·(A′+C) + A·B·Cusing Boolean
algebra properties.
Solution
To simplify the given Boolean expression, we will apply various Boolean alge-
bra properties such as the distributive property, complement property, and the
idempotent property.
Step 1: Apply the distributive property: (A+B)·(A′+C) + A·B·C
⇒(AA′+AC +BB′+BC) + ABC
Step 2: Use the complement property AA′= 0 and BB′= 0:
⇒(0 + AC +0+BC) + ABC
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⇒AC +BC +ABC
Step 3: Factor out a common term C:
⇒C(A+B+AB)
Step 4: Use the absorption property A+AB =A:
⇒C(A)
⇒AC
Therefore, the simplified form of the Boolean expression is AC.
Question 32
Question
Given the Boolean expression F(A, B, C) = AB +AC +BC, simplify the ex-
pression using Boolean Algebra rules.
Solution
To simplify the given Boolean expression F(A, B, C) = AB +AC +BC, we will
use various Boolean Algebra rules such as the Identity Law, Domination Law,
Idempotent Law, Complement Law, etc.
Step 1: Apply the Absorption Law: AB +AC = (A+A)(A+C) =
1(A+C) = A+C.
Therefore, the expression simplifies to F(A, B, C) = A+C+BC.
Step 2: Apply the Absorption Law again: A+C+BC = (A+C+B)(A+
C+C).
This simplifies further to F(A, B, C) = A+C+B.
Step 3: Apply the Consensus Theorem: A+C+B= (A+C)(A+B).
The final simplified expression is F(A, B, C)=(A+C)(A+B).
Therefore, the simplified Boolean expression for F(A, B, C) = AB+AC+BC
is (A+C)(A+B).
Question 33
Question
Simplify the following Boolean expression using laws of Boolean algebra:
F=A′B+AB +AC′+BC′+A′B′C
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Solution
To simplify the given Boolean expression, we will use the laws of Boolean alge-
bra.
Step 1: Apply the absorption law XY +XZ =XY .
F=A′B+AB+AC′+BC′+A′B′C=A(B′+B)+AC′+BC′+A′B′C=A+AC′+BC′+A′B′C
Step 2: Apply the consensus theorem (X+Y)(X+Z)(Y+Z)=(X+
Y)(Y+Z).
F=A+AC′+BC′+A′B′C=A+AC′+BC′
Step 3: Apply the absorption law X+XY =X.
F=A+AC′+BC′=A+C′(A+B)
Hence, the simplified form of the Boolean expression is F=A+C′(A+B).
Question 34
Question
Let F(A, B, C) = ABC +ABC +ABC +ABC. Simplify F(A, B, C) using
Boolean algebra identities.
Solution
To simplify the given expression F(A, B, C), we will use Boolean algebra iden-
tities and properties.
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Step 1: Apply the absorption law: XY +XY =X.
ABC +ABC +ABC +ABC =ABC +ABC +ABC +ABC +ABCC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC(C+C)
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC
=ABC +ABC +ABC +ABC +ABC +A
=ABC +ABC +ABC +ABC +ABC +A(B+B)
=ABC +ABC +ABC +ABC +ABC +AB +AB
=ABC +ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +ABC +AB(C+C)
=ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +ABC +AB
=ABC +ABC +ABC +AB
=AB(C+C) + AB
=AB +AB
=AB
Therefore, the simplified expression for F(A, B, C) is AB.
Question 35
Question
Simplify the following Boolean expression: (A+B)(A′C′+AC).
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Solution
To simplify the given Boolean expression (A+B)(A′C′+AC), we will use the
distributive property and other Boolean algebra rules.
(A+B)(A′C′+AC) = AA′C′+AAC +BA′C′+BAC
=A′C′+AC +BA′C′+BAC (Since AA =A)
=A′C′+AC +AB +BAC (Since A′C′+BA′C′=A′C′+AB)
= (A′C′+AC)+(AB +BAC)
=A′C′+AC +AB +C
=A′C′+AB +AC +C(Since AC +C=C)
=C(A′+A) + AB (Since X+XY =X)
=C+AB (Since A′+A= 1)
Therefore, the simplified form of (A+B)(A′C′+AC) is C+AB.
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