QuestionSimplify the Boolean expression $Y=\bar{A} \bar{B} C+A \bar{B} \bar{C}+\bar{A} \bar{B} \bar{C}$ .

Studdy Solution

STEP 1

Assumptions1. The Boolean Algebra expression is $Y=\bar{A} \bar{B} C+A \bar{B} \bar{C}+\bar{A} \bar{B} \bar{C}$ . . We are asked to simplify the expression in either SOP (Sum of Products) form or POS (Product ofums) form.

STEP 2

First, we notice that there are common terms in the expression. We can factor out these common terms to simplify the expression. The common terms are $\bar{A} \bar{B}$ .
$Y=\bar{A} \bar{B} (C+\bar{C})+A \bar{B} \bar{C}$

STEP 3

Now, we know that in Boolean Algebra, $C+\bar{C}$ is always equal to1. This is because it represents the sum of a variable and its negation, which is always true.
$Y=\bar{A} \bar{B} \cdot1 + A \bar{B} \bar{C}$

STEP 4

The multiplication of any term with1 is the term itself. So, we can simplify the expression further.
$Y=\bar{A} \bar{B} + A \bar{B} \bar{C}$

STEP 5

Next, we notice that the term $\bar{B}$ is common in both terms. We can factor this out.
$Y=\bar{B} (\bar{A} + A \bar{C})$

STEP 6

Again, we know that in Boolean Algebra, $\bar{A} + A$ is always equal to1. This is because it represents the sum of a variable and its negation, which is always true.
$Y=\bar{B} (1 + \bar{C})$

STEP 7

In Boolean Algebra, the sum of any term with1 is always1. This is because it represents the sum of a variable and the truth value, which is always true.
$Y=\bar{B} \cdot1$

STEP 8

Finally, the multiplication of any term with1 is the term itself. So, we can simplify the expression to its simplest form.
$Y=\bar{B}$ So, the simplified Boolean Algebra expression is $Y=\bar{B}$ .

Was this helpful?

QuestionSimplify the Boolean expression $Y=\bar{A} \bar{B} C+A \bar{B} \bar{C}+\bar{A} \bar{B} \bar{C}$ .

Studdy Solution

STEP 1

Assumptions1. The Boolean Algebra expression is $Y=\bar{A} \bar{B} C+A \bar{B} \bar{C}+\bar{A} \bar{B} \bar{C}$ . . We are asked to simplify the expression in either SOP (Sum of Products) form or POS (Product ofums) form.

STEP 2

First, we notice that there are common terms in the expression. We can factor out these common terms to simplify the expression. The common terms are $\bar{A} \bar{B}$ .
$Y=\bar{A} \bar{B} (C+\bar{C})+A \bar{B} \bar{C}$

STEP 3

Now, we know that in Boolean Algebra, $C+\bar{C}$ is always equal to1. This is because it represents the sum of a variable and its negation, which is always true.
$Y=\bar{A} \bar{B} \cdot1 + A \bar{B} \bar{C}$

STEP 4

The multiplication of any term with1 is the term itself. So, we can simplify the expression further.
$Y=\bar{A} \bar{B} + A \bar{B} \bar{C}$

STEP 5

Next, we notice that the term $\bar{B}$ is common in both terms. We can factor this out.
$Y=\bar{B} (\bar{A} + A \bar{C})$

STEP 6

Again, we know that in Boolean Algebra, $\bar{A} + A$ is always equal to1. This is because it represents the sum of a variable and its negation, which is always true.
$Y=\bar{B} (1 + \bar{C})$

STEP 7

In Boolean Algebra, the sum of any term with1 is always1. This is because it represents the sum of a variable and the truth value, which is always true.
$Y=\bar{B} \cdot1$

STEP 8

Finally, the multiplication of any term with1 is the term itself. So, we can simplify the expression to its simplest form.
$Y=\bar{B}$ So, the simplified Boolean Algebra expression is $Y=\bar{B}$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionSimplify the Boolean expression Y=AˉBˉC+ABˉCˉ+AˉBˉCˉY=\bar{A} \bar{B} C+A \bar{B} \bar{C}+\bar{A} \bar{B} \bar{C}Y=AˉBˉC+ABˉCˉ+AˉBˉCˉ.

Studdy Solution

STEP 1

Assumptions1. The Boolean Algebra expression is Y=AˉBˉC+ABˉCˉ+AˉBˉCˉY=\bar{A} \bar{B} C+A \bar{B} \bar{C}+\bar{A} \bar{B} \bar{C}Y=AˉBˉC+ABˉCˉ+AˉBˉCˉ. . We are asked to simplify the expression in either SOP (Sum of Products) form or POS (Product ofums) form.

STEP 2

First, we notice that there are common terms in the expression. We can factor out these common terms to simplify the expression. The common terms are AˉBˉ\bar{A} \bar{B}AˉBˉ.Y=AˉBˉ(C+Cˉ)+ABˉCˉY=\bar{A} \bar{B} (C+\bar{C})+A \bar{B} \bar{C}Y=AˉBˉ(C+Cˉ)+ABˉCˉ

STEP 3

Now, we know that in Boolean Algebra, C+CˉC+\bar{C}C+Cˉ is always equal to1. This is because it represents the sum of a variable and its negation, which is always true.Y=AˉBˉ⋅1+ABˉCˉY=\bar{A} \bar{B} \cdot1 + A \bar{B} \bar{C}Y=AˉBˉ⋅1+ABˉCˉ

STEP 4

The multiplication of any term with1 is the term itself. So, we can simplify the expression further.Y=AˉBˉ+ABˉCˉY=\bar{A} \bar{B} + A \bar{B} \bar{C}Y=AˉBˉ+ABˉCˉ

STEP 5

Next, we notice that the term Bˉ\bar{B}Bˉ is common in both terms. We can factor this out.Y=Bˉ(Aˉ+ACˉ)Y=\bar{B} (\bar{A} + A \bar{C})Y=Bˉ(Aˉ+ACˉ)

STEP 6

Again, we know that in Boolean Algebra, Aˉ+A\bar{A} + AAˉ+A is always equal to1. This is because it represents the sum of a variable and its negation, which is always true.Y=Bˉ(1+Cˉ)Y=\bar{B} (1 + \bar{C})Y=Bˉ(1+Cˉ)

STEP 7

In Boolean Algebra, the sum of any term with1 is always1. This is because it represents the sum of a variable and the truth value, which is always true.Y=Bˉ⋅1Y=\bar{B} \cdot1Y=Bˉ⋅1

STEP 8

Finally, the multiplication of any term with1 is the term itself. So, we can simplify the expression to its simplest form.Y=BˉY=\bar{B}Y=BˉSo, the simplified Boolean Algebra expression is Y=BˉY=\bar{B}Y=Bˉ.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionSimplify the Boolean expression $Y=\bar{A} \bar{B} C+A \bar{B} \bar{C}+\bar{A} \bar{B} \bar{C}$ .

Assumptions1. The Boolean Algebra expression is $Y=\bar{A} \bar{B} C+A \bar{B} \bar{C}+\bar{A} \bar{B} \bar{C}$ . . We are asked to simplify the expression in either SOP (Sum of Products) form or POS (Product ofums) form.

First, we notice that there are common terms in the expression. We can factor out these common terms to simplify the expression. The common terms are $\bar{A} \bar{B}$ .
$Y=\bar{A} \bar{B} (C+\bar{C})+A \bar{B} \bar{C}$

Now, we know that in Boolean Algebra, $C+\bar{C}$ is always equal to1. This is because it represents the sum of a variable and its negation, which is always true.
$Y=\bar{A} \bar{B} \cdot1 + A \bar{B} \bar{C}$

The multiplication of any term with1 is the term itself. So, we can simplify the expression further.
$Y=\bar{A} \bar{B} + A \bar{B} \bar{C}$

Next, we notice that the term $\bar{B}$ is common in both terms. We can factor this out.
$Y=\bar{B} (\bar{A} + A \bar{C})$

Again, we know that in Boolean Algebra, $\bar{A} + A$ is always equal to1. This is because it represents the sum of a variable and its negation, which is always true.
$Y=\bar{B} (1 + \bar{C})$

In Boolean Algebra, the sum of any term with1 is always1. This is because it represents the sum of a variable and the truth value, which is always true.
$Y=\bar{B} \cdot1$

Finally, the multiplication of any term with1 is the term itself. So, we can simplify the expression to its simplest form.
$Y=\bar{B}$ So, the simplified Boolean Algebra expression is $Y=\bar{B}$ .