Show that this is a tautology without creating a truth table (changing logical expression)
(p v q) ^ (-p v r) implies (q v r)
$\displaystyle \left( {p \vee q} \right) \equiv \left( { \sim q \to p} \right)$.
$\displaystyle \left( { \sim p \vee r} \right) \equiv (p \to r)$
Thus from the given:
$\displaystyle \left( {p \vee q} \right) \wedge \left( { \sim p \vee r} \right) \equiv \left( { \sim q \to p} \right) \wedge \left( {p \to r} \right) \equiv \left( { \sim q \to r} \right) \equiv \left( {q \vee r} \right)$
Hello, EquinoX!
I found a proof . . . but there must be a better way . . .
Recall that: .$\displaystyle u \rightarrow v \;\;\equiv\;\;\sim u \lor v$ . (def. implication)Show that this is a tautology without creating a truth table:
. . $\displaystyle (p \lor q) \land (\sim p \lor r) \rightarrow (q \lor r)$
Hence: .$\displaystyle \begin{array}{cccccc}p \lor q & \equiv & \sim p \rightarrow q & & &\\
\sim p \lor r & \equiv & p \rightarrow r & \equiv & \sim r \rightarrow\,\sim p & \text{(contrapositve)}\end{array}$
The left side becomes: .$\displaystyle (\sim r \rightarrow\,\sim p) \land (\sim p \rightarrow q)\;\;\equiv\;\;\sim r \rightarrow q\;\;\;\text{(syllogism)}$
Then: .$\displaystyle \sim r \rightarrow q \;\;\equiv\;\;q \lor r$ . (def. implication)
Ha! . . . Plato beat me to it!