91Ó°ÊÓ

The XOR (exclusive OR) operator is a fundamental concept in Boolean algebra. It evaluates to true if and only if one of its operands is true, but not both. The symbol for XOR is \(\text{\oplus}\). Here's how it evaluates:

• 1 \(\text{\oplus}\) 1 = 0
• 1 \(\text{\oplus}\) 0 = 1
• 0 \(\text{\oplus}\) 1 = 1
• 0 \(\text{\oplus}\) 0 = 0

In simpler words, the XOR operator outputs 1 only when the inputs are different. For instance, when you XOR 1 and 0, the result is 1. However, XORing 1 and 1, or 0 and 0, will result in 0.

In Boolean algebra, the associative property states that the way in which operands are grouped does not affect the result of the operation. For instance, if we want to prove or disprove the equality \(x \text{\oplus} (y \text{\oplus} z) = (x \text{\oplus} y) \text{\oplus} z\), we need to check if this holds for all possible values of x, y, and z. By evaluating all combinations of x, y, and z, we find that this property holds.\[\begin{align*}(0 \text{\oplus} 0) \text{\oplus} 0 &= 0,\x \text{\oplus} (y \text{\oplus} z) &= (x \text{\oplus} y) \text{\oplus} z = \begin{cases}0,\ \text{if } x=y=z=0 or 1\&, \text{if }(0,1,1)\text{or}(1,0,0)\ \cdots \end{align*}\] The equality is proven true. This demonstrates the associative property of XOR.

The counterexample method is a useful strategy to disprove a statement. To disprove, for instance, \(x + (y \text{\oplus} z) = (x + y) \text{\oplus} (x + z)\), we can look for an instance where this equation doesn't hold. If we set \((x, y, z) = (1, 0, 1)\), the left-hand side (LHS) evaluates as follows: \(1 + (0 \text{\oplus} 1) = 1 + 1 = 2\).\However, the right-hand side (RHS) equates to: \((1 + 0) \text{\oplus} (1 + 1) = 1 \text{\oplus} 2 = 1\). This discrepancy between LHS and RHS demonstrates that the equation does not hold universally, thereby disproving it.

Evaluating Boolean expressions is crucial for understanding and verifying Boolean algebra identities. To rigorously evaluate Boolean expressions, follow these steps:

• Identify the operators and their precedence.
• Substitute all possible values for variables (0 or 1).
• Compute the outcome for each combination methodically.

Consider the expression \((x \text{\oplus} (y+z)) = ((x \text{\oplus} y) + (x \text{\oplus} z))\). Using \((x, y, z) = (0, 1, 1)\), we evaluate:

• Left-hand side: \((0 \text{\oplus} (1+1)) = 0 \text{\oplus} 0 = 0\)
• Right-hand side: \(0 \text{\oplus} 1 + 0 \text{\oplus} 1 = 1 + 1 = 2\)

The discrepancy between both sides reveals that the equation is not valid.