MATH 250 - INTRODUCTION TO
DISCRETE MATHEMATICS - Boolean
Algebra
Question Bank - Set 5
Liberty University
Question 1
Question
Let F=AB +A′C+BC. Show that F=AB +A′Cusing Boolean algebra
laws and properties.
Solution
Step 1: Apply the absorption law A+AB =A.
Step 2: Simplify the expression AB +A′C+BC using the absorption law.
F=AB +A′C+BC =AB + (A′C+BC) = AB +A′C
Therefore, F=AB +A′C.
Question 2
Question
Let f(x, y, z) be a Boolean function defined as f(x, y, z) = xyz +xyz+xyz +xyz.
Simplify the function fusing Boolean Algebra laws.
Solution
To simplify the given Boolean function f(x, y, z), we will use various Boolean
Algebra laws such as identity laws, complement laws, etc.
Step 1: Apply the distributive law: xy +xy =x(y+y) = x
Step 2: Simplify each term in the given function:
Starting with xyz, we can see that this term will be the result because it
does not have any terms to combine with.
Step 3: Simplify xyz:
xyz =xyz(y+y) = xyzy +xyzy = 0 + xyzy =xyz
Step 4: Simplify xyz:
xyz =xyz +xyz +xyzy =xy(z+z) + 0 = xy
Step 5: Simplify xyz:
xyz =xyz +xyz(y+y) = xyz + 0 = xyz
Step 6: Combine all simplified terms together:
f(x, y, z) = xyz +xyz +xyz +xyz =xyz +xyz +xy +xyz =xy +xyz
Therefore, the simplified form of the Boolean function f(x, y, z) is xy +xyz.
Question 3
Question
Simplify the following Boolean expression: (A+B·C)·(A+B+C)
Solution
To simplify the given Boolean expression, we will use the basic rules of Boolean
algebra: X+X=X,X·X=X, and De Morgan’s laws: A·B=A+B,
A+B=A·B.
Step 1: Apply distributive law.
(A+B·C)·(A+B+C)
=A·A+A·B+A·C+B·C·A+B·C·B+B·C·C
Step 2: Simplify using X·X= 0 and X+X= 1.
(A·A)+(A·B)+(A·C)+(B·C·A)+(B·C·B)+(B·C·C)
= 0 + A·B+A·C+0+0+0
=A·B+A·C
Step 3: Final simplified form of the given Boolean expression is A·B+A·C.
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Question 4
Question
Let A,B, and Cbe Boolean variables. Simplify the following Boolean expres-
sion:
(A+B·C)·(A·B+C)
Solution
To simplify the given Boolean expression, we will use basic Boolean algebra
rules and laws.
Step 1: Distribute the terms
(A+B·C)·(A·B+C) = A·A·B+A·C+B·C·A·B+B·C·C
Step 2: Apply the complement law
A·A= 0
Step 3: Simplify the expression
0·B+A·C+0+B·C·0 = A·C+B·0 = A·C
Step 4: Final answer Therefore, the simplified form of (A+B·C)·(A·B+C)
is A·C.
Question 5
Question
Simplify the following Boolean expression using Boolean algebra laws:
F=A′B′C+AB′C′+ABC +AB′C
Solution
To simplify the given Boolean expression F=A′B′C+AB′C′+ABC +AB′C,
we will use the laws of Boolean algebra.
Step 1: Apply absorption law
AB′C′+AB′C=AB′C′
Now we have:
F=A′B′C+AB′C′+ABC +AB′C′
Step 2: Apply absorption law again
ABC +AB′C′=AB(C+C′) = AB
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Now we have:
F=A′B′C+AB′C′+AB
Step 3: Apply consensus theorem (or theorem of consensus) law
A′B′C+AB =A′C+AB
Now we have:
F=A′C+AB′C′
Step 4: No further simplification is possible, hence the simplified
expression is
F=A′C+AB′C′
Question 6
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: Use the distributive law to expand the expression.
(A+B+C)(A+B+C)(A+B+C) = A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)
Step 2: Further simplify each term by distributing.
=AA +AB +AC +BA +AB +BB +BC +CA +BC +CC
Since XX =Xfor any Boolean variable X, we simplify AB,AB,BC, and
CC terms:
=A+A+AC +AB +B+0+BC +CA +0+C
Step 3: Combine like terms.
=A+AC +AB +B+BC +CA +C
Step 4: Use the consensus theorem, XY +XZ +Y Z =XY +XZ, to
simplify AC +AB +BC.
=A(C+B) + BC
Step 5: Apply the distributive law once again to obtain the final simplified
expression.
=A+BC
Therefore, (A+B+C)(A+B+C)(A+B+C) simplifies to A+BC.
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Question 7
Question
Simplify the following Boolean expression using algebraic manipulation:
F=A′B′C′+A′BC′+AB′C+ABC
Solution
Step 1: Apply the distributive law to factor out C′:
F=A′B′C′(1 + A) + AB′C(1 + A)
Step 2: Apply the distributive law again to factor out A:
F=A′C′(A′+B′) + AB′C(A+ 1)
Step 3: Use the complement laws (XX′= 0 and X+X′= 1) to simplify
the terms:
F=A′C′+AB′C
Step 4: Apply the distributive law to factor out C:
F=C(A′+AB′)
Step 5: Apply the absorption law (X+XY =X) to simplify the expression:
F=C(A′+B′)
So, the simplified Boolean expression is F=C(A′+B′).
Question 8
Question
Simplify the Boolean expression: (A+B+C)(A+D)(B+C+D)(A+B+C+D).
Solution
Step 1: Use the distributive law to expand the given expression.
(A+B+C)(A+D)(B+C+D)(A+B+C+D)
= (A2+AD +AB +BD +AC +CD +BC +BD)(A+B+C+D)
Step 2: Simplify the expression using the complement law (AA′= 0 and
A+A′= 1).
(A2+AD +AB +BD +AC +CD +BC +BD)(A+B+C+D)
= (AD +AB +BD +AC +CD +BC)(A+B+C+D)
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Step 3: Expand the expression further.
(AD +AB +BD +AC +CD +BC)(A+B+C+D)
=ADA +ADB +ADC +ADB +ACD +ACC +BCD +BCB
Step 4: Simplify using the idempotent law (AA =A).
ADA +ADB +ADC +ADB +ACD +ACC +BCD +BCB
= 0 + ADB +ADC +ADB +ACD +0+BCD + 0
Step 5: Simplify further by combining terms.
0 + ADB +ADC +ADB +ACD +0+BCD + 0
=ADB +ADB +ACD +BCD
=AD(B+B+C) + CD(A+B)
=AD +CD
Therefore, the simplified form of the Boolean expression is AD +CD.
Question 9
Question
Simplify the following Boolean algebra expression: (A+B+C)(A′+B+C)(A+
B′+C)(A+B+C′).
Solution
To simplify the given Boolean algebra expression, we will use the properties of
Boolean algebra including the distributive law, complement law, identity law,
and simplification rules.
Step 1: Apply the distributive law: (A+B+C)(A′+B+C)(A+B′+
C)(A+B+C′)
(A+B+C)(A′+B+C)(A+B′+C)(A+B+C′)=(A+B)(A+B′+C)(A+B+C′)
Step 2: Apply the distributive law again: (A+B)(A+B′+C)(A+B+C′)
(A+B)(A+B′+C)(A+B+C′) = AA +AB′+AC +BBA +BBB′+BBC′
(A+B)(A+B′+C)(A+B+C′) = A+AB′+AC +0+B′+ 0
Step 3: Simplify the expression: A+AB′+AC +B′
A+AB′+AC +B′=A(1 + B′) + B′(1 + A)
A+AB′+AC +B′=A+B′
Therefore, the simplified Boolean algebra expression is A+B′.
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Question 10
Question
Simplify the following Boolean expression using laws of Boolean Algebra: (A+
B)·(A′+B)·(A+B′).
Solution
Step 1: Apply the distributive law X·(Y+Z) = X·Y+X·Z.
= (A+B)·(A+B′)+(A+B)·(A′+B)
Step 2: Apply the distributive law again.
=AA +AB′+BA +BB′+AA′+AB
Step 3: Simplify by applying the idempotent law XX =Xand the identity
law X+X′= 1.
=A+AB′+AB +B+A′+AB
Step 4: Apply the idempotent law and simplify further.
=A+B+A′
Step 5: Apply the idempotent law one more time to get the final simplified
expression.
=A+B+A
Step 6: Apply the idempotent law for the last time to obtain the simplest
form of the expression.
=A+B
Therefore, (A+B)·(A′+B)·(A+B′) simplifies to A+B.
Question 11
Question
Simplify the Boolean expression F=A′B+AC +BC′.
Solution
To simplify the Boolean expression F=A′B+AC +BC′, we will use Boolean
algebra laws and rules to simplify the expression step by step.
Step 1: Apply the distributive law to factor out A.
F=A′B+AC +BC′=A(B′+C) + BC′
Step 2: Apply the distributive law again to factor out B.
F=A(B′+C) + BC′=AB′+AC +BC′
Step 3: Apply the consensus theorem, AB +A′C+BC =AB +A′C, where
AB is the consensus term.
F=AB′+AC +BC′=AB′+A′C
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Therefore, the simplified form of the Boolean expression F=A′B+AC+BC′
is F=AB′+A′C.
Question 12
Question
Simplify the Boolean expression (A+B)·(A+B·C) using Boolean algebra
laws.
Solution
Step 1: Apply the Distributive Law: X·(Y+Z) = X·Y+X·Z
(A+B)·(A+B·C) = A·A+A·B·C+B·A+B·B·C
=A+AB ·C+A·B+ 0
=A+AB ·C+A·B
Step 2: Apply the Idempotent Law: X+X=X
A+AB ·C+A·B=A+AB ·C
Therefore, the simplified form of the Boolean expression (A+B)·(A+B·C)
is A+AB ·C.
Question 13
Question
Simplify the following Boolean expression: (A+B)(A+B+C) using Boolean
algebra laws.
Solution
To simplify the expression (A+B)(A+B+C), we will use the distributive law
and De Morgan’s law.
Step 1: Apply distributive law:
(A+B)(A+B+C) = A(A+B+C) + B(A+B+C)
Step 2: Apply distributive law again:
=AA +AB +AC +BA +BB +BC
Step 3: Apply the complement law (AA = 0 and BB = 0):
= 0 + AB +AC +BA +0+BC
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Step 4: Rearrange the terms:
=AC +BC +AB +BA
Step 5: Apply the absorption law (AC +BC =C):
=C+AB+BA
Step 6: Apply the commutative law:
=C+BA +AB
Therefore, the simplified form of the given Boolean expression is C+BA +
AB.
Question 14
Question
Simplify the Boolean expression (A+B+C)(A′+C′)(A′+B).
Solution
We will simplify the given Boolean expression step by step using the laws of
Boolean Algebra.
Step 1: Apply the Distributive Law: X(Y+Z) = XY +XZ.
(A+B+C)(A′+C′)(A′+B)=(A+B+C)(A′A′+A′C′+AB +BC′+AC′+BC)
= (A+B+C)(0 + A′C′+AB +BC′+AC′+BC)
=A′C′(A+B+C) + AB(A+B+C) + BC′(A+B+C)
=A′C′A+A′C′B+A′C′C+ABA +ABB +ABC +BCA +BCB +BCC
=0+0+0+AB +0+0+0+0+0
=AB
Step 2: Apply the Idempotent Law: X+X=X.
AB =AB
Therefore, the simplified Boolean expression is AB.
Question 15
Question
Simplify the Boolean expression F=A′B′C+A′BC +AB′C′+ABC′using
Boolean algebra laws and theorems.
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Solution
To simplify the given Boolean expression F=A′B′C+A′BC +AB′C′+ABC′,
we will use the laws and theorems of Boolean algebra.
Step 1: Apply the absorption law: XY +XZ =X(Y+Z).
F=A′B′C+A′BC +AB′C′+ABC′
=A′B′C+A′BC +A′B(C′+C′) + AB′C′(Applying absorption law)
=A′B′C+A′BC +A′B+AB′C′
=A′B′C+A′BC +A′B+A′B′C′(Reordering terms)
Step 2: Apply the consensus theorem: XY +X′Z+Y Z =XY +X′Z.
F=A′B′C+A′BC +A′B+A′B′C′
=A′B′C+A′BC +A′B+A′B′C(Applying consensus theorem)
Therefore, the simplified Boolean expression is F=A′B′C+A′BC +A′B+
A′B′C.
Question 16
Question
Simplify the following expression using Boolean Algebra:
F=A′B′C+AB′C+ABC′+ABC
Solution
To simplify the expression F=A′B′C+AB′C+ABC′+ABC, we will use the
properties of Boolean Algebra.
Step 1: Apply the absorption law XY +XY ′=X.
F=A′B′C+AB′C+ABC′+ABC
=A′B′C+AB′C+A(B+BC′)
Step 2: Apply the absorption law XY +X′Z=XY +X′Z+Y Z twice.
F=A′B′C+AB′C+A(B+BC′)
=A′B′C+AB′C+AB +ABC′+ABC
Step 3: Apply the consensus theorem (X+Y)(X+Z)(Y+Z)=(X+
Y)(Y+Z).
F=A′B′C+AB′C+AB +ABC′+ABC
=A′B′C+AB′C+AB +ABC
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Step 4: Apply the absorption law XY +X′Z=XY +X′Z+Y Z.
F=A′B′C+AB′C+AB +ABC
=A+AB +AC
Step 5: Apply the consensus theorem (X+Y)(X+Z)(Y+Z)=(X+
Y)(Y+Z).
F=A+AB +AC
=A(1 + B) + AC
Step 6: Apply the identity law X+X′Y=X+Y.
F=A+AC
=A(1 + C)
Thus, the simplified expression for F=A′B′C+AB′C+ABC′+ABC is
F=A(1 + C).
Question 17
Question
Simplify the following Boolean expression using Boolean algebra rules:
F=A′B+AB′+ABC′+BC
Solution
To simplify the Boolean expression F=A′B+AB′+ABC′+BC, we will use
the rules of Boolean algebra to manipulate the expression.
Step 1: Apply the absorption law (X+XY =X) to A′B+AB′.
A′B+AB′=A′B+AB′+ABC′+BC
Step 2: Apply the consensus theorem (X+X′Y=X+Y) to A′B+ABC′.
A′B+AB′+ABC′=A′B+BC
Step 3: Apply the consensus theorem to A′B+BC.
A′B+BC =A′B+A′C+BC
Step 4: Apply the absorption law to A′B+A′C.
A′B+A′C+BC =A′B+A′C+BC +BC
Step 5: Apply the idempotent law (X+X=X) to BC +BC.
A′B+A′C+BC +BC =A′B+A′C+BC
Step 6: The simplified expression is A′B+A′C+BC.
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Question 18
Question
Given the Boolean expression (A∧B)∨(¬A∧C)∨(B∧ ¬C), simplify the
expression using Boolean algebra laws.
Solution
To simplify the given Boolean expression, we will use the laws of Boolean alge-
bra, including the commutative law, associative law, distributive law, identity
law, complement law, etc.
Step 1: Apply the distributive law to the given expression.
(A∧B)∨(¬A∧C)∨(B∧ ¬C)=(A∨ ¬A)∧(A∨C)∧(B∨C)
Step 2: Apply the complement law to simplify (A∨ ¬A).
(A∨ ¬A)∧(A∨C)∧(B∨C) = 1∧(A∨C)∧(B∨C)
Step 3: Apply the identity law to simplify 1∧(A∨C).
1∧(A∨C)∧(B∨C) = A∨C∧(B∨C)
Step 4: Apply the absorption law to simplify (A∨C)∧(B∨C).
A∨C∧(B∨C) = A∨C
Therefore, the simplified Boolean expression is A∨C.
Question 19
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+C)
Solution
To simplify the given Boolean expression, we will use the distributive property
and Boolean algebra laws.
Step 1: Apply the distributive property to expand the expression.
(A+B+C)(A+B+C)(A+B+C)
= (A+B+C)(A+B+C)(A+B+C)
=A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)
Step 2: Apply the Absorption Law XY +XY =X.
=A(A+B+C) + B(A+B+C) + C(A+B+C)
=A+B+C
Therefore, the simplified form of the given Boolean expression is A+B+C.
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Question 20
Question
Simplify the following Boolean expression: (A+B)(A+B)(A+B).
Solution
To simplify the given Boolean expression, we will use the Boolean algebra laws,
such as the distributive law, the complement law, and the idempotent law.
Step 1: Apply the distributive law to expand the expression.
(A+B)(A+B)(A+B)=(AA +AB +BA +BB)(A+B)
Step 2: Simplify the expression by applying the idempotent law (X+X=
X) and the complement law (XX = 0).
(0 + AB +BA +B)(A+B) = (AB +BA +B)(A+B)
Step 3: Apply the distributive law to expand the expression further.
(AB +BA +B)(A+B) = ABA +ABB +BA +BB
Step 4: Simplify the expression by applying the idempotent law and the
complement law.
ABA +0+BA + 0 = AB +BA
Step 5: Rearrange the terms using the commutative law (XY =Y X).
AB +BA =AB +AB
Step 6: Apply the idempotent law to simplify the expression.
AB +AB =AB
Therefore, (A+B)(A+B)(A+B) simplifies to AB.
Question 21
Question
Simplify the following Boolean expression using algebraic manipulation:
F=A′B+AB′+AB
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Solution
Step 1: Apply the commutative property of OR (+) to rearrange the terms.
F=A′B+AB′+AB
=A′B+AB +AB′
=A′B+AB +BA (Step 1)
Step 2: Apply the idempotent law to simplify AB +BA.
F=A′B+AB +BA
=A′B+A(B+B)
=A′B+A(1)
=A′B+A(Step 2)
Step 3: Apply the absorption law to simplify A′B+A.
F=A′B+A
=A+A′B
=A(1 + B)
=A(Step 3)
Therefore, the simplified expression is F=A.
Question 22
Question
Simplify the following Boolean expression using Boolean algebra laws: (A+B+
C)(A′+B′C) + A′B′.
Solution
To simplify the given Boolean expression, we will use the laws of Boolean algebra
to manipulate the expression step by step.
Step 1: Apply the distributive law: (A+B+C)(A′+B′C) + A′B′.
Step 2: Use the distributive property: AA′+AB′C+BA′+BB′C+CA′+
CB′C+A′B′.
Step 3: Simplify the terms involving complement pairs: 0 + AB′C+0+
0+0+0+A′B′.
Step 4: Further simplify the expression: AB′C+A′B′.
Step 5: Apply the complement law: AB′C+A′B′=AB′C+A′B′(C+C′).
Step 6: Use the distributive property: AB′C+A′B′C+A′B′C′.
Step 7: Combine terms with Cusing the consensus theorem: AB′C+
A′B′C+A′B′C′=AB′C+A′B′C′.
Therefore, the simplified Boolean expression is AB′C+A′B′C′.
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Question 23
Question
Simplify the following Boolean expression: (A+B)(A+B)(A+B).
Solution
To simplify the given Boolean expression, we will use the distributive property
and De Morgan’s laws.
Step 1: Apply the distributive property to expand the expression.
(A+B)(A+B)(A+B)
= (A+B)((A·B)+(A·B)+(B·A)+(B·B)) (Expanding)
= (A+B)(AB +AB +AB + 0) (Complement law: B·B= 0)
= (A+B)(AB +AB)
Step 2: Apply the distributive property again to simplify further.
(A+B)(AB +AB)
=A(AB +AB) + B(AB +AB) (Distributive property)
=AAB +AAB +BAB +BAB (Expanding)
=AB + 0B+AB + 0 (Complement law: A·A= 0 and B·A= 0)
=AB +AB
=AB
Therefore, the simplified form of the given Boolean expression is AB.
Question 24
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+C).
Solution
To simplify the given Boolean expression (A+B+C)(A+B+C)(A+B+C),
we can expand the expression using the distributive property and then simplify
it using the rules of Boolean Algebra.
Step 1: Expand the expression by distributing:
(A+B+C)(A+B+C)(A+B+C)
=A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)
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Step 2: Apply the distributive property to each term:
=AAA +AAB +AAC +ABA +ABB +ABC +ACA +ACB +ACC+
+BAA +BAB +BAC +BBA +BBB +BBC +BCA +BCB +BCC+
+CAA +CAB +CAC +CBA +CBB +CBC +CCA +CCB +CCC
Step 3: Simplify the expression using the rules of Boolean Algebra. Note
that XX =Xand XX = 0:
= 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0
= 0
Therefore, the simplified Boolean expression is 0.
Question 25
Question
Simplify the Boolean expression (A+BC)(AB′C+A′B).
Solution
To simplify the given Boolean expression, we will use the basic laws of Boolean
algebra: commutative law, associative law, distributive law, identity law, com-
plement law, and absorption law.
Step 1: Apply the distributive law (A+BC)(AB′C+A′B).
=AAB′C+AA′B+ABCB′C+ABCB
=A′B+A′B+ABC +ABC
=A′B+ABC
Therefore, (A+BC)(AB′C+A′B) simplifies to A′B+ABC.
Question 26
Question
Simplify the Boolean expression F=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 X+XY =Xto simplify AB +AC.
AB +AC =A(B+C)
Step 2: Apply the absorption law again to simplify A(B+C).
A(B+C) = A
Step 3: Using the identity X′X= 0, simplify A′B′C′.
A′B′C′= 0
Step 4: Rewrite the simplified expressions and combine them to get the
final simplified expression for F.
F=AB +AC +BC +A′B′C′=A+BC + 0 = A+BC
Therefore, the simplified Boolean expression is F=A+BC.
Question 27
Question
Simplify the following Boolean expression: (A+B)(A+C)+(A+B).
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
algebra.
Step 1: Apply the distributive law:
(A+B)(A+C)+(A+B) = A(A+C) + B(A+C) + A+B.
Step 2: Apply the distributive law again:
A(A+C) + B(A+C) + A+B=AA +AC +BA +BC +A+B.
Step 3: Use the idempotent law (AA =A) and zero property (XA = 0 if
Xis a variable and not complemented):
AA +AC +BA +BC +A+B=A+AC +BA +BC +A+B.
Step 4: Apply the idempotent law once more:
A+AC +BA +BC +A+B=A+BA +BC +B.
Step 5: Apply the absorption law (X+XY =X):
A+BA +BC +B=A+B(1 + C) = A+B.
Therefore, (A+B)(A+C)+(A+B) simplifies to A+B.
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Question 28
Question
Simplify the Boolean expression (A+B)(A+B+C)(A+B+C).
Solution
Step 1: Apply the distributive property to expand the expression.
(A+B)(A+B+C)(A+B+C) = AA+BA+BA+BB+BC+AB+BC+BB+AB+ABC
Step 2: Simplify the terms involving complement pairs.
0+0+0+0+0+AB +BC +0+AB +ABC
Step 3: Simplify the remaining terms.
AB +BC +AB +ABC =AB(1 + C) + AB(1 + C)
Step 4: Use the distributive property to factor out a common term.
AB(1 + C) + AB(1 + C) = AB +AB
Step 5: Apply the absorption law X+X′Y=X+Y.
AB +AB =A+B
Therefore, the simplified Boolean expression is A+B.
Question 29
Question
Simplify the Boolean expression: AB +A′B+A′C+BC +B′C
Solution
To simplify the Boolean expression AB +A′B+A′C+BC +B′C, we can use
Boolean algebra laws and properties to simplify the expression step by step.
Step 1: Apply the absorption law A+AB =A.
AB +A′B+A′C+BC +B′C=AB +A′B+A′C+BC +B′C+ 0
=AB +A′B+A′C+ABC +B′C
=AB(1 + C) + A′C(1 + B)
=AB +A′C
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Step 2: Apply the consensus theorem: AB +A′C=AB +B′C+AC.
AB +A′C=AB +B′C+AC
= (A+B′)(B′+C)(A+C)
Therefore, the simplified form of the Boolean expression is (A+B′)(B′+
C)(A+C).
Question 30
Question
Let A,B, and Cbe three Boolean variables. Simplify the following Boolean
expression:
(A+B)·(A·C+B·C)
Solution
To simplify the given Boolean expression, we will use the rules of Boolean algebra
such as distribution, absorption, and identity properties.
Step 1: Distribute the OR operation over the AND operation
(A+B)·(A·C+B·C) = A·(A·C+B·C) + B·(A·C+B·C)
Step 2: Apply the Distributive Law
=A·A·C+A·B·C+B·A·C+B·B·C
Step 3: Simplify the first and last terms using the Absorption Law
=A·C+A·B·C+B·A·C+B·C
Step 4: Simplify using the Identity Law
=A·C+A·B·C+A·C+B·C
Step 5: Combine like terms
=A·C+A·C+A·B·C+B·C
Step 6: Simplify further
=A·C+A·B·C+B·C
Step 7: Factor out common terms
=A·(C+B·C) + B·C
Step 8: Apply the Absorption Law
=A·1+B·C
Step 9: Simplify using the Identity Law
=A+B·C
Therefore, the simplified form of the given Boolean expression is A+B·C.
19
Question 31
Question
Simplify the following Boolean expression: (A′B+C)(A+B′C) + AB +AC
Solution
To simplify the given Boolean expression, we will first expand the terms using
the distributive property, then simplify using Boolean algebra laws.
Step 1: Expand the expression
(A′B+C)(A+B′C) + AB +AC =A′B·A+A′B·B′C+C·A+C·B′C+AB +AC
=A′B·A+A′B·B′C+C·A+C·B′C+AB +AC
Step 2: Apply the AND absorption law
A′B·A+A′B·B′C+C·A+C·B′C+AB +AC =A′B+A′B·C+C·A+C·B′C+AB +AC
=A′B+C(A′+B′C) + C·B′C+AB +AC
Step 3: Apply the distributive law
A′B+C(A′+B′C) + C·B′C+AB +AC =A′B+CA′+CB′C+AB +AC
=A′B+A′C+B′C+AB +AC
Step 4: Apply the consensus theorem
A′B+A′C+B′C+AB +AC =A′B+AB +A′C+AC +B′C
=B+A′C+AC +B′C
=B+A′C+C(B+B′)
=B+A′C+C
Therefore, the simplified form of the given Boolean expression is B+A′C+C.
Question 32
Question
Simplify the following Boolean expression: (A+B)·(A+B).
Solution
To simplify the given Boolean expression, we will use the distributive property
of Boolean algebra.
20
A B A +B(A+B)·(A+B)
0 0 1 0
0 1 0 0
1 0 1 1
1 1 1 1
(A+B)·(A+B) = A·(A+B) + B·(A+B) (Distributive Property)
=A·A+A·B+B·A+B·B(Distributive Property)
=A+A·B+B·A+ 0 (Complement Property)
=A+A·B+B·A(Identity Property)
=A+AB +AB (Commutative Property)
=A+AB (Idempotent Law)
=A(1 + B) (Distributive Property)
=A·1 (Complement Property)
=A(Identity Property)
Therefore, (A+B)·(A+B) simplifies to A.
Question 33
Question
Simplify the following Boolean expression: (A+B)(A+B+C).
Solution
Step 1: Use the distributive property to expand the expression.
(A+B)(A+B+C) = A(A+B+C) + B(A+B+C)
Step 2: Apply the distributive property again to simplify each term.
A(A+B+C) = AA +AB +AC =A+AB +AC
B(A+B+C) = BA +BB +BC =AB +BC
Step 3: Combine the simplified terms.
A+AB +AC +AB +BC =A+AB +AB +AC +BC
Step 4: Use the idempotent law (X+X=X) and the absorption law
(X+XY =X) to simplify further.
A+AB +AB +AC +BC =A+AB +AC
21
=A(1 + B) + AC =A+AC
Step 5: Apply absorption once again to get the final simplified expression.
A+AC =A(1 + C) = A
Therefore, the simplified expression is A.
Question 34
Question
Simplify the following Boolean expression: (A+BC)(AB +AC).
Solution
To simplify the given Boolean expression (A+BC)(AB +AC), we will use
the distributive property and the fact that X+X=Xfor any variable Xin
Boolean algebra.
Step 1: Apply the distributive property to expand the expression.
(A+BC)(AB +AC) = AAB +AAC +BCAB +BCAC
Step 2: Simplify the expression by combining like terms.
AAB +AAC +BCAB +BCAC =A+AC +AB(C+C)
Step 3: Use the fact that X+X=Xin Boolean algebra to simplify further.
A+AC +AB(C+C) = A+AC +AB
Step 4: Apply the distributive property in reverse to factor out a common
term.
A+AC +AB =A(1 + C) + AB =A+AB =A(1 + B) = A
Therefore, the simplified form of the Boolean expression (A+BC)(AB +AC)
is A.
Question 35
Question
Simplify the following Boolean expression:
F=A′B′C+AB′C+ABC +ABC′
22
Question 4
Question
Let A,B, and Cbe Boolean variables. Simplify the following Boolean expres-
sion:
(A+B·C)·(A·B+C)
Solution
To simplify the given Boolean expression, we will use basic Boolean algebra
rules and laws.
Step 1: Distribute the terms
(A+B·C)·(A·B+C) = A·A·B+A·C+B·C·A·B+B·C·C
Step 2: Apply the complement law
A·A= 0
Step 3: Simplify the expression
0·B+A·C+0+B·C·0 = A·C+B·0 = A·C
Step 4: Final answer Therefore, the simplified form of (A+B·C)·(A·B+C)
is A·C.
Question 5
Question
Simplify the following Boolean expression using Boolean algebra laws:
F=A′B′C+AB′C′+ABC +AB′C
Solution
To simplify the given Boolean expression F=A′B′C+AB′C′+ABC +AB′C,
we will use the laws of Boolean algebra.
Step 1: Apply absorption law
AB′C′+AB′C=AB′C′
Now we have:
F=A′B′C+AB′C′+ABC +AB′C′
Step 2: Apply absorption law again
ABC +AB′C′=AB(C+C′) = AB
3
Now we have:
F=A′B′C+AB′C′+AB
Step 3: Apply consensus theorem (or theorem of consensus) law
A′B′C+AB =A′C+AB
Now we have:
F=A′C+AB′C′
Step 4: No further simplification is possible, hence the simplified
expression is
F=A′C+AB′C′
Question 6
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: Use the distributive law to expand the expression.
(A+B+C)(A+B+C)(A+B+C) = A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)
Step 2: Further simplify each term by distributing.
=AA +AB +AC +BA +AB +BB +BC +CA +BC +CC
Since XX =Xfor any Boolean variable X, we simplify AB,AB,BC, and
CC terms:
=A+A+AC +AB +B+0+BC +CA +0+C
Step 3: Combine like terms.
=A+AC +AB +B+BC +CA +C
Step 4: Use the consensus theorem, XY +XZ +Y Z =XY +XZ, to
simplify AC +AB +BC.
=A(C+B) + BC
Step 5: Apply the distributive law once again to obtain the final simplified
expression.
=A+BC
Therefore, (A+B+C)(A+B+C)(A+B+C) simplifies to A+BC.
4
Question 7
Question
Simplify the following Boolean expression using algebraic manipulation:
F=A′B′C′+A′BC′+AB′C+ABC
Solution
Step 1: Apply the distributive law to factor out C′:
F=A′B′C′(1 + A) + AB′C(1 + A)
Step 2: Apply the distributive law again to factor out A:
F=A′C′(A′+B′) + AB′C(A+ 1)
Step 3: Use the complement laws (XX′= 0 and X+X′= 1) to simplify
the terms:
F=A′C′+AB′C
Step 4: Apply the distributive law to factor out C:
F=C(A′+AB′)
Step 5: Apply the absorption law (X+XY =X) to simplify the expression:
F=C(A′+B′)
So, the simplified Boolean expression is F=C(A′+B′).
Question 8
Question
Simplify the Boolean expression: (A+B+C)(A+D)(B+C+D)(A+B+C+D).
Solution
Step 1: Use the distributive law to expand the given expression.
(A+B+C)(A+D)(B+C+D)(A+B+C+D)
= (A2+AD +AB +BD +AC +CD +BC +BD)(A+B+C+D)
Step 2: Simplify the expression using the complement law (AA′= 0 and
A+A′= 1).
(A2+AD +AB +BD +AC +CD +BC +BD)(A+B+C+D)
= (AD +AB +BD +AC +CD +BC)(A+B+C+D)
5
Step 3: Expand the expression further.
(AD +AB +BD +AC +CD +BC)(A+B+C+D)
=ADA +ADB +ADC +ADB +ACD +ACC +BCD +BCB
Step 4: Simplify using the idempotent law (AA =A).
ADA +ADB +ADC +ADB +ACD +ACC +BCD +BCB
= 0 + ADB +ADC +ADB +ACD +0+BCD + 0
Step 5: Simplify further by combining terms.
0 + ADB +ADC +ADB +ACD +0+BCD + 0
=ADB +ADB +ACD +BCD
=AD(B+B+C) + CD(A+B)
=AD +CD
Therefore, the simplified form of the Boolean expression is AD +CD.
Question 9
Question
Simplify the following Boolean algebra expression: (A+B+C)(A′+B+C)(A+
B′+C)(A+B+C′).
Solution
To simplify the given Boolean algebra expression, we will use the properties of
Boolean algebra including the distributive law, complement law, identity law,
and simplification rules.
Step 1: Apply the distributive law: (A+B+C)(A′+B+C)(A+B′+
C)(A+B+C′)
(A+B+C)(A′+B+C)(A+B′+C)(A+B+C′)=(A+B)(A+B′+C)(A+B+C′)
Step 2: Apply the distributive law again: (A+B)(A+B′+C)(A+B+C′)
(A+B)(A+B′+C)(A+B+C′) = AA +AB′+AC +BBA +BBB′+BBC′
(A+B)(A+B′+C)(A+B+C′) = A+AB′+AC +0+B′+ 0
Step 3: Simplify the expression: A+AB′+AC +B′
A+AB′+AC +B′=A(1 + B′) + B′(1 + A)
A+AB′+AC +B′=A+B′
Therefore, the simplified Boolean algebra expression is A+B′.
6
Question 10
Question
Simplify the following Boolean expression using laws of Boolean Algebra: (A+
B)·(A′+B)·(A+B′).
Solution
Step 1: Apply the distributive law X·(Y+Z) = X·Y+X·Z.
= (A+B)·(A+B′)+(A+B)·(A′+B)
Step 2: Apply the distributive law again.
=AA +AB′+BA +BB′+AA′+AB
Step 3: Simplify by applying the idempotent law XX =Xand the identity
law X+X′= 1.
=A+AB′+AB +B+A′+AB
Step 4: Apply the idempotent law and simplify further.
=A+B+A′
Step 5: Apply the idempotent law one more time to get the final simplified
expression.
=A+B+A
Step 6: Apply the idempotent law for the last time to obtain the simplest
form of the expression.
=A+B
Therefore, (A+B)·(A′+B)·(A+B′) simplifies to A+B.
Question 11
Question
Simplify the Boolean expression F=A′B+AC +BC′.
Solution
To simplify the Boolean expression F=A′B+AC +BC′, we will use Boolean
algebra laws and rules to simplify the expression step by step.
Step 1: Apply the distributive law to factor out A.
F=A′B+AC +BC′=A(B′+C) + BC′
Step 2: Apply the distributive law again to factor out B.
F=A(B′+C) + BC′=AB′+AC +BC′
Step 3: Apply the consensus theorem, AB +A′C+BC =AB +A′C, where
AB is the consensus term.
F=AB′+AC +BC′=AB′+A′C
7
Therefore, the simplified form of the Boolean expression F=A′B+AC+BC′
is F=AB′+A′C.
Question 12
Question
Simplify the Boolean expression (A+B)·(A+B·C) using Boolean algebra
laws.
Solution
Step 1: Apply the Distributive Law: X·(Y+Z) = X·Y+X·Z
(A+B)·(A+B·C) = A·A+A·B·C+B·A+B·B·C
=A+AB ·C+A·B+ 0
=A+AB ·C+A·B
Step 2: Apply the Idempotent Law: X+X=X
A+AB ·C+A·B=A+AB ·C
Therefore, the simplified form of the Boolean expression (A+B)·(A+B·C)
is A+AB ·C.
Question 13
Question
Simplify the following Boolean expression: (A+B)(A+B+C) using Boolean
algebra laws.
Solution
To simplify the expression (A+B)(A+B+C), we will use the distributive law
and De Morgan’s law.
Step 1: Apply distributive law:
(A+B)(A+B+C) = A(A+B+C) + B(A+B+C)
Step 2: Apply distributive law again:
=AA +AB +AC +BA +BB +BC
Step 3: Apply the complement law (AA = 0 and BB = 0):
= 0 + AB +AC +BA +0+BC
8
Step 4: Rearrange the terms:
=AC +BC +AB +BA
Step 5: Apply the absorption law (AC +BC =C):
=C+AB+BA
Step 6: Apply the commutative law:
=C+BA +AB
Therefore, the simplified form of the given Boolean expression is C+BA +
AB.
Question 14
Question
Simplify the Boolean expression (A+B+C)(A′+C′)(A′+B).
Solution
We will simplify the given Boolean expression step by step using the laws of
Boolean Algebra.
Step 1: Apply the Distributive Law: X(Y+Z) = XY +XZ.
(A+B+C)(A′+C′)(A′+B)=(A+B+C)(A′A′+A′C′+AB +BC′+AC′+BC)
= (A+B+C)(0 + A′C′+AB +BC′+AC′+BC)
=A′C′(A+B+C) + AB(A+B+C) + BC′(A+B+C)
=A′C′A+A′C′B+A′C′C+ABA +ABB +ABC +BCA +BCB +BCC
=0+0+0+AB +0+0+0+0+0
=AB
Step 2: Apply the Idempotent Law: X+X=X.
AB =AB
Therefore, the simplified Boolean expression is AB.
Question 15
Question
Simplify the Boolean expression F=A′B′C+A′BC +AB′C′+ABC′using
Boolean algebra laws and theorems.
9
Solution
To simplify the given Boolean expression F=A′B′C+A′BC +AB′C′+ABC′,
we will use the laws and theorems of Boolean algebra.
Step 1: Apply the absorption law: XY +XZ =X(Y+Z).
F=A′B′C+A′BC +AB′C′+ABC′
=A′B′C+A′BC +A′B(C′+C′) + AB′C′(Applying absorption law)
=A′B′C+A′BC +A′B+AB′C′
=A′B′C+A′BC +A′B+A′B′C′(Reordering terms)
Step 2: Apply the consensus theorem: XY +X′Z+Y Z =XY +X′Z.
F=A′B′C+A′BC +A′B+A′B′C′
=A′B′C+A′BC +A′B+A′B′C(Applying consensus theorem)
Therefore, the simplified Boolean expression is F=A′B′C+A′BC +A′B+
A′B′C.
Question 16
Question
Simplify the following expression using Boolean Algebra:
F=A′B′C+AB′C+ABC′+ABC
Solution
To simplify the expression F=A′B′C+AB′C+ABC′+ABC, we will use the
properties of Boolean Algebra.
Step 1: Apply the absorption law XY +XY ′=X.
F=A′B′C+AB′C+ABC′+ABC
=A′B′C+AB′C+A(B+BC′)
Step 2: Apply the absorption law XY +X′Z=XY +X′Z+Y Z twice.
F=A′B′C+AB′C+A(B+BC′)
=A′B′C+AB′C+AB +ABC′+ABC
Step 3: Apply the consensus theorem (X+Y)(X+Z)(Y+Z)=(X+
Y)(Y+Z).
F=A′B′C+AB′C+AB +ABC′+ABC
=A′B′C+AB′C+AB +ABC
10
Step 4: Apply the absorption law XY +X′Z=XY +X′Z+Y Z.
F=A′B′C+AB′C+AB +ABC
=A+AB +AC
Step 5: Apply the consensus theorem (X+Y)(X+Z)(Y+Z)=(X+
Y)(Y+Z).
F=A+AB +AC
=A(1 + B) + AC
Step 6: Apply the identity law X+X′Y=X+Y.
F=A+AC
=A(1 + C)
Thus, the simplified expression for F=A′B′C+AB′C+ABC′+ABC is
F=A(1 + C).
Question 17
Question
Simplify the following Boolean expression using Boolean algebra rules:
F=A′B+AB′+ABC′+BC
Solution
To simplify the Boolean expression F=A′B+AB′+ABC′+BC, we will use
the rules of Boolean algebra to manipulate the expression.
Step 1: Apply the absorption law (X+XY =X) to A′B+AB′.
A′B+AB′=A′B+AB′+ABC′+BC
Step 2: Apply the consensus theorem (X+X′Y=X+Y) to A′B+ABC′.
A′B+AB′+ABC′=A′B+BC
Step 3: Apply the consensus theorem to A′B+BC.
A′B+BC =A′B+A′C+BC
Step 4: Apply the absorption law to A′B+A′C.
A′B+A′C+BC =A′B+A′C+BC +BC
Step 5: Apply the idempotent law (X+X=X) to BC +BC.
A′B+A′C+BC +BC =A′B+A′C+BC
Step 6: The simplified expression is A′B+A′C+BC.
11
Question 18
Question
Given the Boolean expression (A∧B)∨(¬A∧C)∨(B∧ ¬C), simplify the
expression using Boolean algebra laws.
Solution
To simplify the given Boolean expression, we will use the laws of Boolean alge-
bra, including the commutative law, associative law, distributive law, identity
law, complement law, etc.
Step 1: Apply the distributive law to the given expression.
(A∧B)∨(¬A∧C)∨(B∧ ¬C)=(A∨ ¬A)∧(A∨C)∧(B∨C)
Step 2: Apply the complement law to simplify (A∨ ¬A).
(A∨ ¬A)∧(A∨C)∧(B∨C) = 1∧(A∨C)∧(B∨C)
Step 3: Apply the identity law to simplify 1∧(A∨C).
1∧(A∨C)∧(B∨C) = A∨C∧(B∨C)
Step 4: Apply the absorption law to simplify (A∨C)∧(B∨C).
A∨C∧(B∨C) = A∨C
Therefore, the simplified Boolean expression is A∨C.
Question 19
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+C)
Solution
To simplify the given Boolean expression, we will use the distributive property
and Boolean algebra laws.
Step 1: Apply the distributive property to expand the expression.
(A+B+C)(A+B+C)(A+B+C)
= (A+B+C)(A+B+C)(A+B+C)
=A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)
Step 2: Apply the Absorption Law XY +XY =X.
=A(A+B+C) + B(A+B+C) + C(A+B+C)
=A+B+C
Therefore, the simplified form of the given Boolean expression is A+B+C.
12
Question 20
Question
Simplify the following Boolean expression: (A+B)(A+B)(A+B).
Solution
To simplify the given Boolean expression, we will use the Boolean algebra laws,
such as the distributive law, the complement law, and the idempotent law.
Step 1: Apply the distributive law to expand the expression.
(A+B)(A+B)(A+B)=(AA +AB +BA +BB)(A+B)
Step 2: Simplify the expression by applying the idempotent law (X+X=
X) and the complement law (XX = 0).
(0 + AB +BA +B)(A+B) = (AB +BA +B)(A+B)
Step 3: Apply the distributive law to expand the expression further.
(AB +BA +B)(A+B) = ABA +ABB +BA +BB
Step 4: Simplify the expression by applying the idempotent law and the
complement law.
ABA +0+BA + 0 = AB +BA
Step 5: Rearrange the terms using the commutative law (XY =Y X).
AB +BA =AB +AB
Step 6: Apply the idempotent law to simplify the expression.
AB +AB =AB
Therefore, (A+B)(A+B)(A+B) simplifies to AB.
Question 21
Question
Simplify the following Boolean expression using algebraic manipulation:
F=A′B+AB′+AB
13
Solution
Step 1: Apply the commutative property of OR (+) to rearrange the terms.
F=A′B+AB′+AB
=A′B+AB +AB′
=A′B+AB +BA (Step 1)
Step 2: Apply the idempotent law to simplify AB +BA.
F=A′B+AB +BA
=A′B+A(B+B)
=A′B+A(1)
=A′B+A(Step 2)
Step 3: Apply the absorption law to simplify A′B+A.
F=A′B+A
=A+A′B
=A(1 + B)
=A(Step 3)
Therefore, the simplified expression is F=A.
Question 22
Question
Simplify the following Boolean expression using Boolean algebra laws: (A+B+
C)(A′+B′C) + A′B′.
Solution
To simplify the given Boolean expression, we will use the laws of Boolean algebra
to manipulate the expression step by step.
Step 1: Apply the distributive law: (A+B+C)(A′+B′C) + A′B′.
Step 2: Use the distributive property: AA′+AB′C+BA′+BB′C+CA′+
CB′C+A′B′.
Step 3: Simplify the terms involving complement pairs: 0 + AB′C+0+
0+0+0+A′B′.
Step 4: Further simplify the expression: AB′C+A′B′.
Step 5: Apply the complement law: AB′C+A′B′=AB′C+A′B′(C+C′).
Step 6: Use the distributive property: AB′C+A′B′C+A′B′C′.
Step 7: Combine terms with Cusing the consensus theorem: AB′C+
A′B′C+A′B′C′=AB′C+A′B′C′.
Therefore, the simplified Boolean expression is AB′C+A′B′C′.
14
Question 23
Question
Simplify the following Boolean expression: (A+B)(A+B)(A+B).
Solution
To simplify the given Boolean expression, we will use the distributive property
and De Morgan’s laws.
Step 1: Apply the distributive property to expand the expression.
(A+B)(A+B)(A+B)
= (A+B)((A·B)+(A·B)+(B·A)+(B·B)) (Expanding)
= (A+B)(AB +AB +AB + 0) (Complement law: B·B= 0)
= (A+B)(AB +AB)
Step 2: Apply the distributive property again to simplify further.
(A+B)(AB +AB)
=A(AB +AB) + B(AB +AB) (Distributive property)
=AAB +AAB +BAB +BAB (Expanding)
=AB + 0B+AB + 0 (Complement law: A·A= 0 and B·A= 0)
=AB +AB
=AB
Therefore, the simplified form of the given Boolean expression is AB.
Question 24
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+C).
Solution
To simplify the given Boolean expression (A+B+C)(A+B+C)(A+B+C),
we can expand the expression using the distributive property and then simplify
it using the rules of Boolean Algebra.
Step 1: Expand the expression by distributing:
(A+B+C)(A+B+C)(A+B+C)
=A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)
15
Step 2: Apply the distributive property to each term:
=AAA +AAB +AAC +ABA +ABB +ABC +ACA +ACB +ACC+
+BAA +BAB +BAC +BBA +BBB +BBC +BCA +BCB +BCC+
+CAA +CAB +CAC +CBA +CBB +CBC +CCA +CCB +CCC
Step 3: Simplify the expression using the rules of Boolean Algebra. Note
that XX =Xand XX = 0:
= 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0
= 0
Therefore, the simplified Boolean expression is 0.
Question 25
Question
Simplify the Boolean expression (A+BC)(AB′C+A′B).
Solution
To simplify the given Boolean expression, we will use the basic laws of Boolean
algebra: commutative law, associative law, distributive law, identity law, com-
plement law, and absorption law.
Step 1: Apply the distributive law (A+BC)(AB′C+A′B).
=AAB′C+AA′B+ABCB′C+ABCB
=A′B+A′B+ABC +ABC
=A′B+ABC
Therefore, (A+BC)(AB′C+A′B) simplifies to A′B+ABC.
Question 26
Question
Simplify the Boolean expression F=AB +AC +BC +A′B′C′.
16
Solution
To simplify the given Boolean expression, we will use the laws of Boolean alge-
bra.
Step 1: Apply the absorption law X+XY =Xto simplify AB +AC.
AB +AC =A(B+C)
Step 2: Apply the absorption law again to simplify A(B+C).
A(B+C) = A
Step 3: Using the identity X′X= 0, simplify A′B′C′.
A′B′C′= 0
Step 4: Rewrite the simplified expressions and combine them to get the
final simplified expression for F.
F=AB +AC +BC +A′B′C′=A+BC + 0 = A+BC
Therefore, the simplified Boolean expression is F=A+BC.
Question 27
Question
Simplify the following Boolean expression: (A+B)(A+C)+(A+B).
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
algebra.
Step 1: Apply the distributive law:
(A+B)(A+C)+(A+B) = A(A+C) + B(A+C) + A+B.
Step 2: Apply the distributive law again:
A(A+C) + B(A+C) + A+B=AA +AC +BA +BC +A+B.
Step 3: Use the idempotent law (AA =A) and zero property (XA = 0 if
Xis a variable and not complemented):
AA +AC +BA +BC +A+B=A+AC +BA +BC +A+B.
Step 4: Apply the idempotent law once more:
A+AC +BA +BC +A+B=A+BA +BC +B.
Step 5: Apply the absorption law (X+XY =X):
A+BA +BC +B=A+B(1 + C) = A+B.
Therefore, (A+B)(A+C)+(A+B) simplifies to A+B.
17
Question 28
Question
Simplify the Boolean expression (A+B)(A+B+C)(A+B+C).
Solution
Step 1: Apply the distributive property to expand the expression.
(A+B)(A+B+C)(A+B+C) = AA+BA+BA+BB+BC+AB+BC+BB+AB+ABC
Step 2: Simplify the terms involving complement pairs.
0+0+0+0+0+AB +BC +0+AB +ABC
Step 3: Simplify the remaining terms.
AB +BC +AB +ABC =AB(1 + C) + AB(1 + C)
Step 4: Use the distributive property to factor out a common term.
AB(1 + C) + AB(1 + C) = AB +AB
Step 5: Apply the absorption law X+X′Y=X+Y.
AB +AB =A+B
Therefore, the simplified Boolean expression is A+B.
Question 29
Question
Simplify the Boolean expression: AB +A′B+A′C+BC +B′C
Solution
To simplify the Boolean expression AB +A′B+A′C+BC +B′C, we can use
Boolean algebra laws and properties to simplify the expression step by step.
Step 1: Apply the absorption law A+AB =A.
AB +A′B+A′C+BC +B′C=AB +A′B+A′C+BC +B′C+ 0
=AB +A′B+A′C+ABC +B′C
=AB(1 + C) + A′C(1 + B)
=AB +A′C
18
Step 2: Apply the consensus theorem: AB +A′C=AB +B′C+AC.
AB +A′C=AB +B′C+AC
= (A+B′)(B′+C)(A+C)
Therefore, the simplified form of the Boolean expression is (A+B′)(B′+
C)(A+C).
Question 30
Question
Let A,B, and Cbe three Boolean variables. Simplify the following Boolean
expression:
(A+B)·(A·C+B·C)
Solution
To simplify the given Boolean expression, we will use the rules of Boolean algebra
such as distribution, absorption, and identity properties.
Step 1: Distribute the OR operation over the AND operation
(A+B)·(A·C+B·C) = A·(A·C+B·C) + B·(A·C+B·C)
Step 2: Apply the Distributive Law
=A·A·C+A·B·C+B·A·C+B·B·C
Step 3: Simplify the first and last terms using the Absorption Law
=A·C+A·B·C+B·A·C+B·C
Step 4: Simplify using the Identity Law
=A·C+A·B·C+A·C+B·C
Step 5: Combine like terms
=A·C+A·C+A·B·C+B·C
Step 6: Simplify further
=A·C+A·B·C+B·C
Step 7: Factor out common terms
=A·(C+B·C) + B·C
Step 8: Apply the Absorption Law
=A·1+B·C
Step 9: Simplify using the Identity Law
=A+B·C
Therefore, the simplified form of the given Boolean expression is A+B·C.
19
Question 31
Question
Simplify the following Boolean expression: (A′B+C)(A+B′C) + AB +AC
Solution
To simplify the given Boolean expression, we will first expand the terms using
the distributive property, then simplify using Boolean algebra laws.
Step 1: Expand the expression
(A′B+C)(A+B′C) + AB +AC =A′B·A+A′B·B′C+C·A+C·B′C+AB +AC
=A′B·A+A′B·B′C+C·A+C·B′C+AB +AC
Step 2: Apply the AND absorption law
A′B·A+A′B·B′C+C·A+C·B′C+AB +AC =A′B+A′B·C+C·A+C·B′C+AB +AC
=A′B+C(A′+B′C) + C·B′C+AB +AC
Step 3: Apply the distributive law
A′B+C(A′+B′C) + C·B′C+AB +AC =A′B+CA′+CB′C+AB +AC
=A′B+A′C+B′C+AB +AC
Step 4: Apply the consensus theorem
A′B+A′C+B′C+AB +AC =A′B+AB +A′C+AC +B′C
=B+A′C+AC +B′C
=B+A′C+C(B+B′)
=B+A′C+C
Therefore, the simplified form of the given Boolean expression is B+A′C+C.
Question 32
Question
Simplify the following Boolean expression: (A+B)·(A+B).
Solution
To simplify the given Boolean expression, we will use the distributive property
of Boolean algebra.
20
A B A +B(A+B)·(A+B)
0 0 1 0
0 1 0 0
1 0 1 1
1 1 1 1
(A+B)·(A+B) = A·(A+B) + B·(A+B) (Distributive Property)
=A·A+A·B+B·A+B·B(Distributive Property)
=A+A·B+B·A+ 0 (Complement Property)
=A+A·B+B·A(Identity Property)
=A+AB +AB (Commutative Property)
=A+AB (Idempotent Law)
=A(1 + B) (Distributive Property)
=A·1 (Complement Property)
=A(Identity Property)
Therefore, (A+B)·(A+B) simplifies to A.
Question 33
Question
Simplify the following Boolean expression: (A+B)(A+B+C).
Solution
Step 1: Use the distributive property to expand the expression.
(A+B)(A+B+C) = A(A+B+C) + B(A+B+C)
Step 2: Apply the distributive property again to simplify each term.
A(A+B+C) = AA +AB +AC =A+AB +AC
B(A+B+C) = BA +BB +BC =AB +BC
Step 3: Combine the simplified terms.
A+AB +AC +AB +BC =A+AB +AB +AC +BC
Step 4: Use the idempotent law (X+X=X) and the absorption law
(X+XY =X) to simplify further.
A+AB +AB +AC +BC =A+AB +AC
21
=A(1 + B) + AC =A+AC
Step 5: Apply absorption once again to get the final simplified expression.
A+AC =A(1 + C) = A
Therefore, the simplified expression is A.
Question 34
Question
Simplify the following Boolean expression: (A+BC)(AB +AC).
Solution
To simplify the given Boolean expression (A+BC)(AB +AC), we will use
the distributive property and the fact that X+X=Xfor any variable Xin
Boolean algebra.
Step 1: Apply the distributive property to expand the expression.
(A+BC)(AB +AC) = AAB +AAC +BCAB +BCAC
Step 2: Simplify the expression by combining like terms.
AAB +AAC +BCAB +BCAC =A+AC +AB(C+C)
Step 3: Use the fact that X+X=Xin Boolean algebra to simplify further.
A+AC +AB(C+C) = A+AC +AB
Step 4: Apply the distributive property in reverse to factor out a common
term.
A+AC +AB =A(1 + C) + AB =A+AB =A(1 + B) = A
Therefore, the simplified form of the Boolean expression (A+BC)(AB +AC)
is A.
Question 35
Question
Simplify the following Boolean expression:
F=A′B′C+AB′C+ABC +ABC′
22
Question 4
Question
Let A,B, and Cbe Boolean variables. Simplify the following Boolean expres-
sion:
(A+B·C)·(A·B+C)
Solution
To simplify the given Boolean expression, we will use basic Boolean algebra
rules and laws.
Step 1: Distribute the terms
(A+B·C)·(A·B+C) = A·A·B+A·C+B·C·A·B+B·C·C
Step 2: Apply the complement law
A·A= 0
Step 3: Simplify the expression
0·B+A·C+0+B·C·0 = A·C+B·0 = A·C
Step 4: Final answer Therefore, the simplified form of (A+B·C)·(A·B+C)
is A·C.
Question 5
Question
Simplify the following Boolean expression using Boolean algebra laws:
F=A′B′C+AB′C′+ABC +AB′C
Solution
To simplify the given Boolean expression F=A′B′C+AB′C′+ABC +AB′C,
we will use the laws of Boolean algebra.
Step 1: Apply absorption law
AB′C′+AB′C=AB′C′
Now we have:
F=A′B′C+AB′C′+ABC +AB′C′
Step 2: Apply absorption law again
ABC +AB′C′=AB(C+C′) = AB
3
Now we have:
F=A′B′C+AB′C′+AB
Step 3: Apply consensus theorem (or theorem of consensus) law
A′B′C+AB =A′C+AB
Now we have:
F=A′C+AB′C′
Step 4: No further simplification is possible, hence the simplified
expression is
F=A′C+AB′C′
Question 6
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: Use the distributive law to expand the expression.
(A+B+C)(A+B+C)(A+B+C) = A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)
Step 2: Further simplify each term by distributing.
=AA +AB +AC +BA +AB +BB +BC +CA +BC +CC
Since XX =Xfor any Boolean variable X, we simplify AB,AB,BC, and
CC terms:
=A+A+AC +AB +B+0+BC +CA +0+C
Step 3: Combine like terms.
=A+AC +AB +B+BC +CA +C
Step 4: Use the consensus theorem, XY +XZ +Y Z =XY +XZ, to
simplify AC +AB +BC.
=A(C+B) + BC
Step 5: Apply the distributive law once again to obtain the final simplified
expression.
=A+BC
Therefore, (A+B+C)(A+B+C)(A+B+C) simplifies to A+BC.
4
Question 7
Question
Simplify the following Boolean expression using algebraic manipulation:
F=A′B′C′+A′BC′+AB′C+ABC
Solution
Step 1: Apply the distributive law to factor out C′:
F=A′B′C′(1 + A) + AB′C(1 + A)
Step 2: Apply the distributive law again to factor out A:
F=A′C′(A′+B′) + AB′C(A+ 1)
Step 3: Use the complement laws (XX′= 0 and X+X′= 1) to simplify
the terms:
F=A′C′+AB′C
Step 4: Apply the distributive law to factor out C:
F=C(A′+AB′)
Step 5: Apply the absorption law (X+XY =X) to simplify the expression:
F=C(A′+B′)
So, the simplified Boolean expression is F=C(A′+B′).
Question 8
Question
Simplify the Boolean expression: (A+B+C)(A+D)(B+C+D)(A+B+C+D).
Solution
Step 1: Use the distributive law to expand the given expression.
(A+B+C)(A+D)(B+C+D)(A+B+C+D)
= (A2+AD +AB +BD +AC +CD +BC +BD)(A+B+C+D)
Step 2: Simplify the expression using the complement law (AA′= 0 and
A+A′= 1).
(A2+AD +AB +BD +AC +CD +BC +BD)(A+B+C+D)
= (AD +AB +BD +AC +CD +BC)(A+B+C+D)
5
Step 3: Expand the expression further.
(AD +AB +BD +AC +CD +BC)(A+B+C+D)
=ADA +ADB +ADC +ADB +ACD +ACC +BCD +BCB
Step 4: Simplify using the idempotent law (AA =A).
ADA +ADB +ADC +ADB +ACD +ACC +BCD +BCB
= 0 + ADB +ADC +ADB +ACD +0+BCD + 0
Step 5: Simplify further by combining terms.
0 + ADB +ADC +ADB +ACD +0+BCD + 0
=ADB +ADB +ACD +BCD
=AD(B+B+C) + CD(A+B)
=AD +CD
Therefore, the simplified form of the Boolean expression is AD +CD.
Question 9
Question
Simplify the following Boolean algebra expression: (A+B+C)(A′+B+C)(A+
B′+C)(A+B+C′).
Solution
To simplify the given Boolean algebra expression, we will use the properties of
Boolean algebra including the distributive law, complement law, identity law,
and simplification rules.
Step 1: Apply the distributive law: (A+B+C)(A′+B+C)(A+B′+
C)(A+B+C′)
(A+B+C)(A′+B+C)(A+B′+C)(A+B+C′)=(A+B)(A+B′+C)(A+B+C′)
Step 2: Apply the distributive law again: (A+B)(A+B′+C)(A+B+C′)
(A+B)(A+B′+C)(A+B+C′) = AA +AB′+AC +BBA +BBB′+BBC′
(A+B)(A+B′+C)(A+B+C′) = A+AB′+AC +0+B′+ 0
Step 3: Simplify the expression: A+AB′+AC +B′
A+AB′+AC +B′=A(1 + B′) + B′(1 + A)
A+AB′+AC +B′=A+B′
Therefore, the simplified Boolean algebra expression is A+B′.
6
Question 10
Question
Simplify the following Boolean expression using laws of Boolean Algebra: (A+
B)·(A′+B)·(A+B′).
Solution
Step 1: Apply the distributive law X·(Y+Z) = X·Y+X·Z.
= (A+B)·(A+B′)+(A+B)·(A′+B)
Step 2: Apply the distributive law again.
=AA +AB′+BA +BB′+AA′+AB
Step 3: Simplify by applying the idempotent law XX =Xand the identity
law X+X′= 1.
=A+AB′+AB +B+A′+AB
Step 4: Apply the idempotent law and simplify further.
=A+B+A′
Step 5: Apply the idempotent law one more time to get the final simplified
expression.
=A+B+A
Step 6: Apply the idempotent law for the last time to obtain the simplest
form of the expression.
=A+B
Therefore, (A+B)·(A′+B)·(A+B′) simplifies to A+B.
Question 11
Question
Simplify the Boolean expression F=A′B+AC +BC′.
Solution
To simplify the Boolean expression F=A′B+AC +BC′, we will use Boolean
algebra laws and rules to simplify the expression step by step.
Step 1: Apply the distributive law to factor out A.
F=A′B+AC +BC′=A(B′+C) + BC′
Step 2: Apply the distributive law again to factor out B.
F=A(B′+C) + BC′=AB′+AC +BC′
Step 3: Apply the consensus theorem, AB +A′C+BC =AB +A′C, where
AB is the consensus term.
F=AB′+AC +BC′=AB′+A′C
7
Therefore, the simplified form of the Boolean expression F=A′B+AC+BC′
is F=AB′+A′C.
Question 12
Question
Simplify the Boolean expression (A+B)·(A+B·C) using Boolean algebra
laws.
Solution
Step 1: Apply the Distributive Law: X·(Y+Z) = X·Y+X·Z
(A+B)·(A+B·C) = A·A+A·B·C+B·A+B·B·C
=A+AB ·C+A·B+ 0
=A+AB ·C+A·B
Step 2: Apply the Idempotent Law: X+X=X
A+AB ·C+A·B=A+AB ·C
Therefore, the simplified form of the Boolean expression (A+B)·(A+B·C)
is A+AB ·C.
Question 13
Question
Simplify the following Boolean expression: (A+B)(A+B+C) using Boolean
algebra laws.
Solution
To simplify the expression (A+B)(A+B+C), we will use the distributive law
and De Morgan’s law.
Step 1: Apply distributive law:
(A+B)(A+B+C) = A(A+B+C) + B(A+B+C)
Step 2: Apply distributive law again:
=AA +AB +AC +BA +BB +BC
Step 3: Apply the complement law (AA = 0 and BB = 0):
= 0 + AB +AC +BA +0+BC
8
Step 4: Rearrange the terms:
=AC +BC +AB +BA
Step 5: Apply the absorption law (AC +BC =C):
=C+AB+BA
Step 6: Apply the commutative law:
=C+BA +AB
Therefore, the simplified form of the given Boolean expression is C+BA +
AB.
Question 14
Question
Simplify the Boolean expression (A+B+C)(A′+C′)(A′+B).
Solution
We will simplify the given Boolean expression step by step using the laws of
Boolean Algebra.
Step 1: Apply the Distributive Law: X(Y+Z) = XY +XZ.
(A+B+C)(A′+C′)(A′+B)=(A+B+C)(A′A′+A′C′+AB +BC′+AC′+BC)
= (A+B+C)(0 + A′C′+AB +BC′+AC′+BC)
=A′C′(A+B+C) + AB(A+B+C) + BC′(A+B+C)
=A′C′A+A′C′B+A′C′C+ABA +ABB +ABC +BCA +BCB +BCC
=0+0+0+AB +0+0+0+0+0
=AB
Step 2: Apply the Idempotent Law: X+X=X.
AB =AB
Therefore, the simplified Boolean expression is AB.
Question 15
Question
Simplify the Boolean expression F=A′B′C+A′BC +AB′C′+ABC′using
Boolean algebra laws and theorems.
9
Solution
To simplify the given Boolean expression F=A′B′C+A′BC +AB′C′+ABC′,
we will use the laws and theorems of Boolean algebra.
Step 1: Apply the absorption law: XY +XZ =X(Y+Z).
F=A′B′C+A′BC +AB′C′+ABC′
=A′B′C+A′BC +A′B(C′+C′) + AB′C′(Applying absorption law)
=A′B′C+A′BC +A′B+AB′C′
=A′B′C+A′BC +A′B+A′B′C′(Reordering terms)
Step 2: Apply the consensus theorem: XY +X′Z+Y Z =XY +X′Z.
F=A′B′C+A′BC +A′B+A′B′C′
=A′B′C+A′BC +A′B+A′B′C(Applying consensus theorem)
Therefore, the simplified Boolean expression is F=A′B′C+A′BC +A′B+
A′B′C.
Question 16
Question
Simplify the following expression using Boolean Algebra:
F=A′B′C+AB′C+ABC′+ABC
Solution
To simplify the expression F=A′B′C+AB′C+ABC′+ABC, we will use the
properties of Boolean Algebra.
Step 1: Apply the absorption law XY +XY ′=X.
F=A′B′C+AB′C+ABC′+ABC
=A′B′C+AB′C+A(B+BC′)
Step 2: Apply the absorption law XY +X′Z=XY +X′Z+Y Z twice.
F=A′B′C+AB′C+A(B+BC′)
=A′B′C+AB′C+AB +ABC′+ABC
Step 3: Apply the consensus theorem (X+Y)(X+Z)(Y+Z)=(X+
Y)(Y+Z).
F=A′B′C+AB′C+AB +ABC′+ABC
=A′B′C+AB′C+AB +ABC
10
Step 4: Apply the absorption law XY +X′Z=XY +X′Z+Y Z.
F=A′B′C+AB′C+AB +ABC
=A+AB +AC
Step 5: Apply the consensus theorem (X+Y)(X+Z)(Y+Z)=(X+
Y)(Y+Z).
F=A+AB +AC
=A(1 + B) + AC
Step 6: Apply the identity law X+X′Y=X+Y.
F=A+AC
=A(1 + C)
Thus, the simplified expression for F=A′B′C+AB′C+ABC′+ABC is
F=A(1 + C).
Question 17
Question
Simplify the following Boolean expression using Boolean algebra rules:
F=A′B+AB′+ABC′+BC
Solution
To simplify the Boolean expression F=A′B+AB′+ABC′+BC, we will use
the rules of Boolean algebra to manipulate the expression.
Step 1: Apply the absorption law (X+XY =X) to A′B+AB′.
A′B+AB′=A′B+AB′+ABC′+BC
Step 2: Apply the consensus theorem (X+X′Y=X+Y) to A′B+ABC′.
A′B+AB′+ABC′=A′B+BC
Step 3: Apply the consensus theorem to A′B+BC.
A′B+BC =A′B+A′C+BC
Step 4: Apply the absorption law to A′B+A′C.
A′B+A′C+BC =A′B+A′C+BC +BC
Step 5: Apply the idempotent law (X+X=X) to BC +BC.
A′B+A′C+BC +BC =A′B+A′C+BC
Step 6: The simplified expression is A′B+A′C+BC.
11
Question 18
Question
Given the Boolean expression (A∧B)∨(¬A∧C)∨(B∧ ¬C), simplify the
expression using Boolean algebra laws.
Solution
To simplify the given Boolean expression, we will use the laws of Boolean alge-
bra, including the commutative law, associative law, distributive law, identity
law, complement law, etc.
Step 1: Apply the distributive law to the given expression.
(A∧B)∨(¬A∧C)∨(B∧ ¬C)=(A∨ ¬A)∧(A∨C)∧(B∨C)
Step 2: Apply the complement law to simplify (A∨ ¬A).
(A∨ ¬A)∧(A∨C)∧(B∨C) = 1∧(A∨C)∧(B∨C)
Step 3: Apply the identity law to simplify 1∧(A∨C).
1∧(A∨C)∧(B∨C) = A∨C∧(B∨C)
Step 4: Apply the absorption law to simplify (A∨C)∧(B∨C).
A∨C∧(B∨C) = A∨C
Therefore, the simplified Boolean expression is A∨C.
Question 19
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+C)
Solution
To simplify the given Boolean expression, we will use the distributive property
and Boolean algebra laws.
Step 1: Apply the distributive property to expand the expression.
(A+B+C)(A+B+C)(A+B+C)
= (A+B+C)(A+B+C)(A+B+C)
=A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)
Step 2: Apply the Absorption Law XY +XY =X.
=A(A+B+C) + B(A+B+C) + C(A+B+C)
=A+B+C
Therefore, the simplified form of the given Boolean expression is A+B+C.
12
Question 20
Question
Simplify the following Boolean expression: (A+B)(A+B)(A+B).
Solution
To simplify the given Boolean expression, we will use the Boolean algebra laws,
such as the distributive law, the complement law, and the idempotent law.
Step 1: Apply the distributive law to expand the expression.
(A+B)(A+B)(A+B)=(AA +AB +BA +BB)(A+B)
Step 2: Simplify the expression by applying the idempotent law (X+X=
X) and the complement law (XX = 0).
(0 + AB +BA +B)(A+B) = (AB +BA +B)(A+B)
Step 3: Apply the distributive law to expand the expression further.
(AB +BA +B)(A+B) = ABA +ABB +BA +BB
Step 4: Simplify the expression by applying the idempotent law and the
complement law.
ABA +0+BA + 0 = AB +BA
Step 5: Rearrange the terms using the commutative law (XY =Y X).
AB +BA =AB +AB
Step 6: Apply the idempotent law to simplify the expression.
AB +AB =AB
Therefore, (A+B)(A+B)(A+B) simplifies to AB.
Question 21
Question
Simplify the following Boolean expression using algebraic manipulation:
F=A′B+AB′+AB
13
Solution
Step 1: Apply the commutative property of OR (+) to rearrange the terms.
F=A′B+AB′+AB
=A′B+AB +AB′
=A′B+AB +BA (Step 1)
Step 2: Apply the idempotent law to simplify AB +BA.
F=A′B+AB +BA
=A′B+A(B+B)
=A′B+A(1)
=A′B+A(Step 2)
Step 3: Apply the absorption law to simplify A′B+A.
F=A′B+A
=A+A′B
=A(1 + B)
=A(Step 3)
Therefore, the simplified expression is F=A.
Question 22
Question
Simplify the following Boolean expression using Boolean algebra laws: (A+B+
C)(A′+B′C) + A′B′.
Solution
To simplify the given Boolean expression, we will use the laws of Boolean algebra
to manipulate the expression step by step.
Step 1: Apply the distributive law: (A+B+C)(A′+B′C) + A′B′.
Step 2: Use the distributive property: AA′+AB′C+BA′+BB′C+CA′+
CB′C+A′B′.
Step 3: Simplify the terms involving complement pairs: 0 + AB′C+0+
0+0+0+A′B′.
Step 4: Further simplify the expression: AB′C+A′B′.
Step 5: Apply the complement law: AB′C+A′B′=AB′C+A′B′(C+C′).
Step 6: Use the distributive property: AB′C+A′B′C+A′B′C′.
Step 7: Combine terms with Cusing the consensus theorem: AB′C+
A′B′C+A′B′C′=AB′C+A′B′C′.
Therefore, the simplified Boolean expression is AB′C+A′B′C′.
14
Question 23
Question
Simplify the following Boolean expression: (A+B)(A+B)(A+B).
Solution
To simplify the given Boolean expression, we will use the distributive property
and De Morgan’s laws.
Step 1: Apply the distributive property to expand the expression.
(A+B)(A+B)(A+B)
= (A+B)((A·B)+(A·B)+(B·A)+(B·B)) (Expanding)
= (A+B)(AB +AB +AB + 0) (Complement law: B·B= 0)
= (A+B)(AB +AB)
Step 2: Apply the distributive property again to simplify further.
(A+B)(AB +AB)
=A(AB +AB) + B(AB +AB) (Distributive property)
=AAB +AAB +BAB +BAB (Expanding)
=AB + 0B+AB + 0 (Complement law: A·A= 0 and B·A= 0)
=AB +AB
=AB
Therefore, the simplified form of the given Boolean expression is AB.
Question 24
Question
Simplify the following Boolean expression: (A+B+C)(A+B+C)(A+B+C).
Solution
To simplify the given Boolean expression (A+B+C)(A+B+C)(A+B+C),
we can expand the expression using the distributive property and then simplify
it using the rules of Boolean Algebra.
Step 1: Expand the expression by distributing:
(A+B+C)(A+B+C)(A+B+C)
=A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)
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Step 2: Apply the distributive property to each term:
=AAA +AAB +AAC +ABA +ABB +ABC +ACA +ACB +ACC+
+BAA +BAB +BAC +BBA +BBB +BBC +BCA +BCB +BCC+
+CAA +CAB +CAC +CBA +CBB +CBC +CCA +CCB +CCC
Step 3: Simplify the expression using the rules of Boolean Algebra. Note
that XX =Xand XX = 0:
= 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0
= 0
Therefore, the simplified Boolean expression is 0.
Question 25
Question
Simplify the Boolean expression (A+BC)(AB′C+A′B).
Solution
To simplify the given Boolean expression, we will use the basic laws of Boolean
algebra: commutative law, associative law, distributive law, identity law, com-
plement law, and absorption law.
Step 1: Apply the distributive law (A+BC)(AB′C+A′B).
=AAB′C+AA′B+ABCB′C+ABCB
=A′B+A′B+ABC +ABC
=A′B+ABC
Therefore, (A+BC)(AB′C+A′B) simplifies to A′B+ABC.
Question 26
Question
Simplify the Boolean expression F=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 X+XY =Xto simplify AB +AC.
AB +AC =A(B+C)
Step 2: Apply the absorption law again to simplify A(B+C).
A(B+C) = A
Step 3: Using the identity X′X= 0, simplify A′B′C′.
A′B′C′= 0
Step 4: Rewrite the simplified expressions and combine them to get the
final simplified expression for F.
F=AB +AC +BC +A′B′C′=A+BC + 0 = A+BC
Therefore, the simplified Boolean expression is F=A+BC.
Question 27
Question
Simplify the following Boolean expression: (A+B)(A+C)+(A+B).
Solution
To simplify the given Boolean expression, we will use the properties of Boolean
algebra.
Step 1: Apply the distributive law:
(A+B)(A+C)+(A+B) = A(A+C) + B(A+C) + A+B.
Step 2: Apply the distributive law again:
A(A+C) + B(A+C) + A+B=AA +AC +BA +BC +A+B.
Step 3: Use the idempotent law (AA =A) and zero property (XA = 0 if
Xis a variable and not complemented):
AA +AC +BA +BC +A+B=A+AC +BA +BC +A+B.
Step 4: Apply the idempotent law once more:
A+AC +BA +BC +A+B=A+BA +BC +B.
Step 5: Apply the absorption law (X+XY =X):
A+BA +BC +B=A+B(1 + C) = A+B.
Therefore, (A+B)(A+C)+(A+B) simplifies to A+B.
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Question 28
Question
Simplify the Boolean expression (A+B)(A+B+C)(A+B+C).
Solution
Step 1: Apply the distributive property to expand the expression.
(A+B)(A+B+C)(A+B+C) = AA+BA+BA+BB+BC+AB+BC+BB+AB+ABC
Step 2: Simplify the terms involving complement pairs.
0+0+0+0+0+AB +BC +0+AB +ABC
Step 3: Simplify the remaining terms.
AB +BC +AB +ABC =AB(1 + C) + AB(1 + C)
Step 4: Use the distributive property to factor out a common term.
AB(1 + C) + AB(1 + C) = AB +AB
Step 5: Apply the absorption law X+X′Y=X+Y.
AB +AB =A+B
Therefore, the simplified Boolean expression is A+B.
Question 29
Question
Simplify the Boolean expression: AB +A′B+A′C+BC +B′C
Solution
To simplify the Boolean expression AB +A′B+A′C+BC +B′C, we can use
Boolean algebra laws and properties to simplify the expression step by step.
Step 1: Apply the absorption law A+AB =A.
AB +A′B+A′C+BC +B′C=AB +A′B+A′C+BC +B′C+ 0
=AB +A′B+A′C+ABC +B′C
=AB(1 + C) + A′C(1 + B)
=AB +A′C
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Step 2: Apply the consensus theorem: AB +A′C=AB +B′C+AC.
AB +A′C=AB +B′C+AC
= (A+B′)(B′+C)(A+C)
Therefore, the simplified form of the Boolean expression is (A+B′)(B′+
C)(A+C).
Question 30
Question
Let A,B, and Cbe three Boolean variables. Simplify the following Boolean
expression:
(A+B)·(A·C+B·C)
Solution
To simplify the given Boolean expression, we will use the rules of Boolean algebra
such as distribution, absorption, and identity properties.
Step 1: Distribute the OR operation over the AND operation
(A+B)·(A·C+B·C) = A·(A·C+B·C) + B·(A·C+B·C)
Step 2: Apply the Distributive Law
=A·A·C+A·B·C+B·A·C+B·B·C
Step 3: Simplify the first and last terms using the Absorption Law
=A·C+A·B·C+B·A·C+B·C
Step 4: Simplify using the Identity Law
=A·C+A·B·C+A·C+B·C
Step 5: Combine like terms
=A·C+A·C+A·B·C+B·C
Step 6: Simplify further
=A·C+A·B·C+B·C
Step 7: Factor out common terms
=A·(C+B·C) + B·C
Step 8: Apply the Absorption Law
=A·1+B·C
Step 9: Simplify using the Identity Law
=A+B·C
Therefore, the simplified form of the given Boolean expression is A+B·C.
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Question 31
Question
Simplify the following Boolean expression: (A′B+C)(A+B′C) + AB +AC
Solution
To simplify the given Boolean expression, we will first expand the terms using
the distributive property, then simplify using Boolean algebra laws.
Step 1: Expand the expression
(A′B+C)(A+B′C) + AB +AC =A′B·A+A′B·B′C+C·A+C·B′C+AB +AC
=A′B·A+A′B·B′C+C·A+C·B′C+AB +AC
Step 2: Apply the AND absorption law
A′B·A+A′B·B′C+C·A+C·B′C+AB +AC =A′B+A′B·C+C·A+C·B′C+AB +AC
=A′B+C(A′+B′C) + C·B′C+AB +AC
Step 3: Apply the distributive law
A′B+C(A′+B′C) + C·B′C+AB +AC =A′B+CA′+CB′C+AB +AC
=A′B+A′C+B′C+AB +AC
Step 4: Apply the consensus theorem
A′B+A′C+B′C+AB +AC =A′B+AB +A′C+AC +B′C
=B+A′C+AC +B′C
=B+A′C+C(B+B′)
=B+A′C+C
Therefore, the simplified form of the given Boolean expression is B+A′C+C.
Question 32
Question
Simplify the following Boolean expression: (A+B)·(A+B).
Solution
To simplify the given Boolean expression, we will use the distributive property
of Boolean algebra.
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A B A +B(A+B)·(A+B)
0 0 1 0
0 1 0 0
1 0 1 1
1 1 1 1
(A+B)·(A+B) = A·(A+B) + B·(A+B) (Distributive Property)
=A·A+A·B+B·A+B·B(Distributive Property)
=A+A·B+B·A+ 0 (Complement Property)
=A+A·B+B·A(Identity Property)
=A+AB +AB (Commutative Property)
=A+AB (Idempotent Law)
=A(1 + B) (Distributive Property)
=A·1 (Complement Property)
=A(Identity Property)
Therefore, (A+B)·(A+B) simplifies to A.
Question 33
Question
Simplify the following Boolean expression: (A+B)(A+B+C).
Solution
Step 1: Use the distributive property to expand the expression.
(A+B)(A+B+C) = A(A+B+C) + B(A+B+C)
Step 2: Apply the distributive property again to simplify each term.
A(A+B+C) = AA +AB +AC =A+AB +AC
B(A+B+C) = BA +BB +BC =AB +BC
Step 3: Combine the simplified terms.
A+AB +AC +AB +BC =A+AB +AB +AC +BC
Step 4: Use the idempotent law (X+X=X) and the absorption law
(X+XY =X) to simplify further.
A+AB +AB +AC +BC =A+AB +AC
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=A(1 + B) + AC =A+AC
Step 5: Apply absorption once again to get the final simplified expression.
A+AC =A(1 + C) = A
Therefore, the simplified expression is A.
Question 34
Question
Simplify the following Boolean expression: (A+BC)(AB +AC).
Solution
To simplify the given Boolean expression (A+BC)(AB +AC), we will use
the distributive property and the fact that X+X=Xfor any variable Xin
Boolean algebra.
Step 1: Apply the distributive property to expand the expression.
(A+BC)(AB +AC) = AAB +AAC +BCAB +BCAC
Step 2: Simplify the expression by combining like terms.
AAB +AAC +BCAB +BCAC =A+AC +AB(C+C)
Step 3: Use the fact that X+X=Xin Boolean algebra to simplify further.
A+AC +AB(C+C) = A+AC +AB
Step 4: Apply the distributive property in reverse to factor out a common
term.
A+AC +AB =A(1 + C) + AB =A+AB =A(1 + B) = A
Therefore, the simplified form of the Boolean expression (A+BC)(AB +AC)
is A.
Question 35
Question
Simplify the following Boolean expression:
F=A′B′C+AB′C+ABC +ABC′
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Solution
To simplify the given Boolean expression F=A′B′C+AB′C+ABC +ABC′,
we will use Boolean algebra laws and rules.
1. Use the absorption law to simplify the expression:
F=A′B′C+AB′C+ABC +ABC′
=B′C(A′+A) + BC(A+A′)
=B′C+BC
=C(B′+B)
=C
Therefore, the simplified form of the Boolean expression is F=C.
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