Q:

How could we prove that $(p \vee q)  \wedge( \neg p \vee r )\rightarrow (q \vee r)$ is a tautology?

How could we prove that $(p \vee q)  \wedge( \neg p \vee r )\rightarrow (q \vee r)$ is a tautology? 
I am more interested in the algebraic method. We can use all the rules of inferences except the Resolution.

A:

We have
\begin{align*}
  (p \vee q)  \wedge( \neg p \vee r )\rightarrow (q \vee r)
   &\equiv (p \wedge \neg p) \vee (q \wedge \neg p) \vee (q \wedge r) \vee (p \wedge r) \to q \vee r\\
  &\equiv \neg \bigl((q \wedge \neg p) \vee (q \wedge r) \vee (p \wedge r)\bigr) \vee q \vee r\\
  &\equiv \bigl((\neg q \vee p) \wedge (\neg q \vee \neg r) \wedge (\neg p \vee \neg r)\bigr) \vee q \vee r\\
  &\equiv  (\neg q \vee p \vee q \vee r) \wedge (\neg q \vee \neg r \vee q \vee r) \wedge (\neg p \vee \neg r \vee q \vee r)\\
  &\equiv \top \wedge \top \wedge \top \equiv \top.
\end{align*}

