Question

Prove or disprove these equalities.

\(\begin{array}{l}a)\;x \oplus (y \oplus z){\bf{ = }}(x \oplus y) \oplus z\\b)\;x{\bf{ + }}(y \oplus z){\bf{ = }}(x{\bf{ + }}y) \oplus (x{\bf{ + }}z)\\c)\;x \oplus (y{\bf{ + }}z){\bf{ = }}(x \oplus y){\bf{ + }}(x \oplus z)\end{array}\)

Short answer

1. The given equality \(x \oplus \left( {y \oplus z} \right) = \left( {x \oplus y} \right) \oplus z\) is proved.
2. The given equality \(x + \left( {y \oplus z} \right) = \left( {x + y} \right) \oplus \left( {x + z} \right)\) is proved.
3. The given equality \(x \oplus \left( {y + z} \right) = \left( {x \oplus y} \right) + \left( {x \oplus z} \right)\) is proved.

Step-by-step solution

Definition

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

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

The Boolean product \( \cdot \) or \(AND\) is \(1\) if both terms are \(1\).

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

Step 2:(a) Using the operator

\(x,y\)and \(z\) can both take on the value of \(0\) or \(1\).

The \(XOR\) operator is \(1\) if one of the two elements (but not both) are \(1\).

\(\begin{array}{*{20}{r}}x&y&z&{y \oplus z}&{x \oplus y}&{x \oplus \left( {y \oplus z} \right)}&{\left( {x \oplus y} \right) \oplus z}\\0&0&0&0&0&0&0\\0&0&1&1&0&1&1\\0&1&0&1&1&1&1\\0&1&1&0&1&0&0\\1&0&0&0&1&1&1\\1&0&1&1&1&0&0\\1&1&0&1&0&0&0\\1&1&1&0&0&1&1\end{array}\)

Notes that the last two columns of the table are identical

Therefore, the required result is \(x \oplus \left( {y \oplus z} \right) = \left( {x \oplus y} \right) \oplus z\).

Step 3:(b) Using the operator

\(x\)and \(y\) can both take on the value of \(0\) or \(1\).

The \(XOR\) operator is \(1\) if one of the two elements (but not both) are \(1\).

\(\begin{array}{*{20}{r}}x&y&z&{y \oplus z}&{x + y}&{x + z}&{x + \left( {y \oplus z} \right)}&{\left( {x + y} \right) \oplus \left( {x + z} \right)}\\0&0&0&0&0&0&0&0\\0&0&1&1&0&1&1&1\\0&1&0&1&1&0&1&0\\0&1&1&0&1&1&0&0\\1&0&0&0&1&1&1&0\\1&0&1&1&1&1&1&0\\1&1&0&1&1&1&1&0\\1&1&1&0&1&1&1&0\end{array}\)

Notes that the last two columns of the table are NOT identical.

Therefore, the required result is \(x + \left( {y \oplus z} \right) = \left( {x + y} \right) \oplus \left( {x + z} \right)\).

(c) Using the operator

\(x\)and \(y\) can both take on the value of \(0\) or \(1\).

The \(XOR\) operator is \(1\) if one of the two elements (but not both) are \(1\).

\(\begin{array}{*{20}{r}}x&y&z&{y + z}&{x \oplus y}&{x \oplus z}&{x \oplus \left( {y + z} \right)}&{\left( {x \oplus y} \right) + \left( {x \oplus z} \right)}\\0&0&0&0&0&0&0&0\\0&0&1&1&0&1&1&1\\0&1&0&1&1&0&1&1\\0&1&1&1&1&1&1&1\\1&0&0&0&1&1&1&1\\1&0&1&1&1&0&0&1\\1&1&0&1&0&1&0&1\\1&1&1&1&0&0&0&0\end{array}\)

Notes that the last two columns of the table are NOT identical

Therefore, the required result is \(x \oplus \left( {y + z} \right) = \left( {x \oplus y} \right) + \left( {x \oplus z} \right)\).