Question

In Exercises 35–42,Use the laws in Definition \(1\) to show that the stated properties hold in every Boolean algebra.

Show that in a Boolean algebra, if \(x \vee y{\bf{ = }}0\), then \(x{\bf{ = }}0\) and \(y{\bf{ = }}0\), and that if \(x \wedge y{\bf{ = }}1\), then \(x{\bf{ = }}1\) and \(y{\bf{ = }}1\).

Short answer

If it uses \(x \vee y = 0\), then it will get \(x = 0\) and \(y = 0\)

If it uses \(x \wedge y = 1\), then it will get \(x = 1\) and \(y = 1\)

Step-by-step solution

Definition

Identity laws

\(\begin{aligned}x \vee 0 = x\\x \wedge 1 = x\end{aligned}\)

Associative laws

\(\begin{aligned}x \vee \left( {y \vee z} \right) = \left( {x \vee y} \right) \vee z\\x \wedge \left( {y \wedge z} \right) = \left( {x \wedge y} \right) \wedge z\end{aligned}\)

Distributive laws

\(\begin{aligned}x \vee \left( {y \wedge z} \right) &= \left( {x \vee y} \right) \wedge \left( {x \vee z} \right)\\x \wedge \left( {y \vee z} \right) &= \left( {x \wedge y} \right) \vee \left( {x \wedge z} \right)\end{aligned}\)

Idempotent laws

\(\begin{aligned}x \vee x &= x\\x \wedge x &= x\end{aligned}\)

Using the identity law, associative law and idempotent law

Using \(x \vee y = 0\)

Let \(x \vee y{\bf{ = }}0\)

\(\begin{aligned}x &= x \vee 0\\ &= x \vee \left( {x \vee y} \right)\\ &= \left( {x \vee x} \right) \vee y\\ &= x \vee y\\ &= 0\end{aligned}\)

So, it gets \(x{\bf{ = }}0\)

Using the identity law, associative law and idempotent law

Using \(x \vee y = 0\)

\(\begin{aligned}y &= y \vee 0\\ &= y \vee \left( {x \vee y} \right)\\ &= y \vee \left( {y \vee x} \right)\\ &= \left( {y \vee y} \right) \vee x\end{aligned}\)

\(\begin{aligned} &= y \vee x\\ &=x \vee y\\ &= 0\end{aligned}\)

Therefore, if it uses \(x \vee y{\bf{ = }}0\), then it will get \(x{\bf{ = }}0\) and \(y{\bf{ = }}0\).

Using the identity law, associative law and idempotent law

Using \(x \wedge y{\bf{ = }}1\)

To proof: If \(x \wedge y = 1\), then \(x = 1\) and \(y = 1\)

Let \(x \wedge y = 1\)

\(\begin{aligned}x &= x \wedge 1\\ &= x \wedge \left( {x \wedge y} \right)\\ &= \left( {x \wedge x} \right) \wedge y\\ &= x \wedge y\\ &= 1\end{aligned}\)

Using the identity law, associative law and idempotent law

Using \(x \wedge y = 1\)

\(\begin{aligned}y &= y \wedge 1\\ &= y \wedge \left( {x \wedge y} \right)\\ &= y \wedge \left( {y \wedge x} \right)\\ &= \left( {y \wedge y} \right) \wedge x\end{aligned}\)

\(\begin{aligned} &= y \wedge x\\ &= x \wedge y\\ &= 1\end{aligned}\)

Therefore, if it uses\(x \wedge y{\bf{ = }}1\), then it will get \(x{\bf{ = }}1\) and \(y{\bf{ = }}1\).