# Math Help - is this expression a tautology?

1. ## is this expression a tautology?

Show that this is a tautology without creating a truth table (changing logical expression)

(p v q) ^ (-p v r) implies (q v r)

2. anyone? to prove that this is a tautology then (q v r) must be true right? but how do I proof this?

3. $\left( {p \vee q} \right) \equiv \left( { \sim q \to p} \right)$.
$\left( { \sim p \vee r} \right) \equiv (p \to r)$

Thus from the given:
$\left( {p \vee q} \right) \wedge \left( { \sim p \vee r} \right) \equiv \left( { \sim q \to p} \right) \wedge \left( {p \to r} \right) \equiv \left( { \sim q \to r} \right) \equiv \left( {q \vee r} \right)$

4. what I am confused here is that how do you proof:

(~q -> p) ^ (p -> r) is equivalent to (~q-> r)

5. That is a hypothetical syllogism.
Why were you given these problems to do if you don’t know the basics?

6. Hello, EquinoX!

I found a proof . . . but there must be a better way . . .

Show that this is a tautology without creating a truth table:
. . $(p \lor q) \land (\sim p \lor r) \rightarrow (q \lor r)$
Recall that: . $u \rightarrow v \;\;\equiv\;\;\sim u \lor v$ . (def. implication)

Hence: . $\begin{array}{cccccc}p \lor q & \equiv & \sim p \rightarrow q & & &\\
\sim p \lor r & \equiv & p \rightarrow r & \equiv & \sim r \rightarrow\,\sim p & \text{(contrapositve)}\end{array}$

The left side becomes: . $(\sim r \rightarrow\,\sim p) \land (\sim p \rightarrow q)\;\;\equiv\;\;\sim r \rightarrow q\;\;\;\text{(syllogism)}$

Then: . $\sim r \rightarrow q \;\;\equiv\;\;q \lor r$ . (def. implication)

Ha! . . . Plato beat me to it!