To simplify the given Boolean expression, we will break it down into smaller, manageable parts and apply the fundamental laws of Boolean algebra. The expression is:

\(\left( {x + y} \right)\left( {x + \bar y} \right) + \overline {\left( {x\bar y} \right) + \bar x} \)

Boolean Expression Simplification

We will simplify this expression step-by-step, focusing on each major component.

First Part Simplification: \(\left( {x + y} \right)\left( {x + \bar y} \right)\)

Let's simplify the first part of the Boolean expression, which is \(\left( {x + y} \right)\left( {x + \bar y} \right)\).

We can use the distributive law, which states that \(A \cdot (B + C) = A \cdot B + A \cdot C\), or a specific Boolean identity known as the Consensus Theorem variant, \((A+B)(A+C) = A + BC\).

Applying the identity \((A+B)(A+C) = A + BC\), where \(A = x\), \(B = y\), and \(C = \bar y\):

• \(\left( {x + y} \right)\left( {x + \bar y} \right) = x + y\bar y\)

According to the Complement Law of Boolean algebra, \(y\bar y = 0\) (a variable ANDed with its complement always results in 0).

• So, \(x + y\bar y = x + 0\)
• And \(x + 0 = x\) (Identity Law: a variable ORed with 0 is the variable itself).

Therefore, the first part simplifies to \(x\).

Intermediate Result 1: \(\left( {x + y} \right)\left( {x + \bar y} \right) = x\)

Second Part Simplification: \(\overline {\left( {x\bar y} \right) + \bar x} \)

Now, let's simplify the second part of the Boolean expression, which is \(\overline {\left( {x\bar y} \right) + \bar x} \).

We will use De Morgan's Law, which states that \(\overline{A+B} = \bar A \cdot \bar B\).

Applying De Morgan's Law to \(\overline {\left( {x\bar y} \right) + \bar x} \):

• Let \(A = x\bar y\) and \(B = \bar x\).
• So, \(\overline {\left( {x\bar y} \right) + \bar x} = \overline {\left( {x\bar y} \right)} \cdot \overline {\left( {\bar x} \right)}\)

Now, let's simplify each term in the product:

1. Simplify \(\overline {\left( {x\bar y} \right)}\):
   • Apply De Morgan's Law again: \(\overline{AB} = \bar A + \bar B\).
   • So, \(\overline {\left( {x\bar y} \right)} = \bar x + \overline {\bar y}\)
   • Using the Double Negation Law (\(\overline {\bar A} = A\)), \(\overline {\bar y} = y\).
   • Therefore, \(\overline {\left( {x\bar y} \right)} = \bar x + y\)
2. Simplify \(\overline {\left( {\bar x} \right)}\):
   • Using the Double Negation Law (\(\overline {\bar A} = A\)), \(\overline {\left( {\bar x} \right)} = x\).

Substitute these simplified terms back into the expression:

• \(\overline {\left( {x\bar y} \right) + \bar x} = \left( {\bar x + y} \right) \cdot x\)

Now, apply the distributive law \(A \cdot (B + C) = A \cdot B + A \cdot C\):

• \(\left( {\bar x + y} \right) \cdot x = \bar x \cdot x + y \cdot x\)

According to the Complement Law, \(\bar x \cdot x = 0\).

• So, \(\bar x \cdot x + y \cdot x = 0 + yx\)
• And \(0 + yx = yx\) (Identity Law)
• This can also be written as \(xy\) (Commutative Law).

Therefore, the second part simplifies to \(xy\).

Intermediate Result 2: \(\overline {\left( {x\bar y} \right) + \bar x} = xy\)

Combining the Simplified Parts

Now we combine the simplified results from the first and second parts.

The original expression was \(\left( {x + y} \right)\left( {x + \bar y} \right) + \overline {\left( {x\bar y} \right) + \bar x} \).

Substituting Intermediate Result 1 (\(x\)) and Intermediate Result 2 (\(xy\)):

• The expression becomes \(x + xy\)

This form can be simplified using the Absorption Law, which states that \(A + AB = A\).

• Here, let \(A = x\) and \(B = y\).
• So, \(x + xy = x\)

Thus, the entire Boolean expression simplifies to \(x\).

Key Boolean Algebra Laws Used

Here's a summary of the Boolean algebra laws applied in this simplification:

Final Result

After applying various Boolean algebra laws systematically, the given expression simplifies to \(x\).

The simplified Boolean expression is \(x\).