Question

Simplify these expressions.

\(\begin{array}{l}{\bf{a) x}} \oplus {\bf{0}}\\{\bf{b) x}} \oplus {\bf{1}}\\{\bf{c) x}} \oplus {\bf{x}}\\{\bf{d) x}} \oplus {\bf{\bar x}}\end{array}\)

Short answer

1. It can simplify\(x \oplus 0\)it to \(x\)
2. It can simplify \(x \oplus 1\)it to \(\bar x\)
3. It can simplify \(x \oplus x\)it to \(0\)
4. It can simplify \(x \oplus \bar x\)it to \(1\)

Step-by-step solution

Definition

The complement of an element:\({\bf{\bar 0 = 1}}\)and\({\bf{\bar 1 = 0}}\)

The Boolean sum\({\bf{ + }}\)or\({\bf{OR}}\)is\({\bf{1}}\)if either term is\({\bf{1}}\).

The Boolean product\({\bf{ \times }}\)or\({\bf{AND}}\)is\({\bf{1}}\)if both terms are\({\bf{1}}\).

The\({\bf{XOR}}\)operator\( \oplus \)is\({\bf{1}}\)if one of the terms is\({\bf{1}}\)(but not both).

Using the Boolean algebra and product

(a)

\(x \oplus 0\)

In Boolean algebra, a variable can only take on the value of \(0\) or\(1\) . Thus \(x\) is \(0\) or\(1\) .

By the definition of Boolean product:

If\(x = 0\), then \(x \oplus 0 = 0 \oplus 0 = 0 = x\)

If\(x = 1\), then \(x \oplus 0 = 1 \oplus 0 = 1 = x\)

Notes that\(x \oplus 0 = x\).

Therefore, you can simplify \(x \oplus 0\) to\(x\).

Using the Boolean algebra and product

(b)

\(x \oplus 1\)

In Boolean algebra, a variable can only take on the value of \(0\) or\(1\) . Thus \(x\) is \(0\) or \(1\) .

By the definition of Boolean product:

If\(x = 0\), then \(x \oplus 1 = 0 \oplus 1 = 1 = \bar 0 = \bar x\)

If\(x = 1\), then \(x \oplus 1 = 1 \oplus 1 = 0 = \bar 1 = \bar x\)

Notes that\(x \oplus 1 = \bar x\).

Therefore, you can simplify \(x \oplus 1\) to\(\bar x\).

 Step 4: Using the Boolean algebra and product

(c)

\(x \oplus x\)

In Boolean algebra, a variable can only take on the value of \(0\) or\(1\) . Thus \(x\) is \(0\) or \(1\) .

By the definition of Boolean product:

If\(x = 0\), then \(x \oplus x = 0 \oplus 0 = 0\)

If\(x = 1\) , then \(x \oplus x = 1 \oplus 1 = 0\)

Notes that\(x \oplus x = 0\).

Therefore, you can simplify \(x \oplus x\) to\(0\).

Using the Boolean algebra and product

(d)

\(x \in \bar x\)

In Boolean algebra, a variable can only take on the value of \(0\) or\(1\) . Thus \(x\) is \(0\) or \(1\) .

By the definition of Boolean product:

If\(x = 0\), then \(x \oplus \bar x = 0 \oplus \bar 0 = 0 \oplus 1 = 1\)

If\(x = 1\), then \(x \oplus \bar x = 1 \oplus \bar 1 = 1 \oplus 0 = 1\)

Notes that\(x \oplus \bar x = 1\).

Therefore, you can simplify \(x \oplus \bar x\) to\(1\).