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Q30E.

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Found in: Page 35

### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

# Show that ${\mathbf{\left(}}{\mathbit{p}}{\mathbf{\vee }}{\mathbit{q}}{\mathbf{\right)}}{\mathbf{\wedge }}{\mathbf{\left(}}{\mathbf{¬}}{\mathbit{p}}{\mathbf{\vee }}{\mathbit{r}}{\mathbf{\right)}}{\mathbf{\to }}{\mathbf{\left(}}{\mathbit{q}}{\mathbf{\vee }}{\mathbit{r}}{\mathbf{\right)}}$ is a tautology

${\left(}{p}{\vee }{q}{\right)}{\wedge }{\left(}{¬}{p}{\vee }{r}{\right)}{\to }{\left(}{q}{\vee }{r}{\right)}$ is a tautology.

See the step by step solution

## Step1:Definition of Tautology

Tautology results in true.

## Step 2: The given statement is a tautology

We write the given statement,

$$\begin{array}{l}(p \vee q) \wedge (\neg p \vee r) \to (q \vee r)\;\, = (p \vee q) \wedge (\neg (\neg p \vee r) \vee (q \vee r))\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = (p \vee q) \wedge (p \wedge \neg r) \vee (q \vee r)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = (p \vee q) \wedge \neg (p \wedge \neg r) \wedge \neg (q \vee r)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = (p \vee q) \wedge (\neg p \vee r) \wedge (\neg q \vee \neg r)\end{array}$$

$$\begin{array}{l}(p \vee q) \wedge (\neg p \vee r) \to (q \vee r)\,\; = (p \vee \neg p) \wedge (q \vee \neg q) \vee (\neg r \vee r)......\left( {Associativity and commutativity} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = T \wedge T \wedge T\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = T\end{array}$$

Hence,$$(p \vee q) \wedge (\neg p \vee r) \to (q \vee r)$$is a tautology.

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