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Section5.2Exercises

2

Prove the commutative property expressed by Equation (5.1.5) and Equation (5.1.6).

Solution

For Equation (5.1.5):

\(x\) \(y\) \(x \cdot y\) \(y \cdot x\)
\(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{0}\) \(\binary{1}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{1}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\)

And for Equation (5.1.6):

\(x\) \(y\) \(x + y\) \(y + x\)
\(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{0}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\)
\(\binary{1}\) \(\binary{0}\) \(\binary{1}\) \(\binary{1}\)
\(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\)
6

Prove the distributive property expressed by Equation (5.1.13) and Equation (5.1.14).

Solution

For Equation (5.1.13):

\(x\) \(y\) \(z\) \(y + z\) \(x \cdot (y + z)\) \(x \cdot y\) \(x \cdot z\) \(x \cdot y + x \cdot z\)
\(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{0}\) \(\binary{0}\) \(\binary{1}\) \(\binary{1}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{0}\) \(\binary{1}\) \(\binary{0}\) \(\binary{1}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{0}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{1}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{1}\) \(\binary{0}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{0}\) \(\binary{1}\) \(\binary{1}\)
\(\binary{1}\) \(\binary{1}\) \(\binary{0}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{0}\) \(\binary{1}\)
\(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\)

And for Equation (5.1.14):

\(x\) \(y\) \(z\) \(y \cdot z\) \(x + y \cdot z\) \(x + y\) \(x + z\) \((x + y) \cdot (x + z)\)
\(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{0}\) \(\binary{0}\) \(\binary{1}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{1}\) \(\binary{0}\)
\(\binary{0}\) \(\binary{1}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{1}\) \(\binary{0}\) \(\binary{0}\)
\(\binary{0}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\)
\(\binary{1}\) \(\binary{0}\) \(\binary{0}\) \(\binary{0}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\)
\(\binary{1}\) \(\binary{0}\) \(\binary{1}\) \(\binary{0}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\)
\(\binary{1}\) \(\binary{1}\) \(\binary{0}\) \(\binary{0}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\)
\(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\) \(\binary{1}\)