Derive the following rule of inference

Question

This rule is sometimes called resolution:

(p ∨ q) ∧ (~p ∨ r) ⇒ (q ∨ r)

How can it be derived with boolean algebra?

Answer 1

When using boolean algebra you rewrite one expression as a different, but $equivalent$ expression. So, using algebra you don't end up with a statement that is merely implied. So, you typically don't use algebra to demonstrate some implication.

One thing we can do, however, is to derive a conjunction, one conjunct of which is the implied statement. If we can do that, then obviously the implied statement is indeed a logical consequence. So, let's try that:

$(p \lor q) \land (\neg p \lor r) \Leftrightarrow$ (Distribution)

$(p \land \neg p) \lor (p \land r) \lor (q \land \neg p) \lor (q \land r) \Leftrightarrow$ (Complement)

$\bot \lor (p \land r) \lor (q \land \neg p) \lor (q \land r) \Leftrightarrow$ (Identity)

$(p \land r) \lor (q \land \neg p) \lor (q \land r) \Leftrightarrow$ (Distribution)

$[(p \lor q) \land (p \lor \neg p) \land (r \lor q) \land (r \lor \neg p)] \lor (q \land r) \Leftrightarrow$ (Complement)

$[(p \lor q) \land \top \land (r \lor q) \land (r \lor \neg p)] \lor (q \land r) \Leftrightarrow$ (Identity)

$[(p \lor q) \land (r \lor q) \land (r \lor \neg p)] \lor (q \land r) \Leftrightarrow$ (Distribution)

$[p \lor q \lor (q \land r)] \land [r \lor q \lor (q \land r)] \land [r \lor \neg p \lor (q \land r)] \Leftrightarrow$ (Absorption)

$(p \lor q) \land (r \lor q) \land (r \lor \neg p) \Leftrightarrow$ (Commutation)

$(q \lor r) \land (p \lor q) \land (r \lor \neg p)$

So, this statement is equivalent to the original statement, and since $q \lor r$ is one of the conjuncts of this conjunction, it is clearly implied by this statement, and hence by the original statement as well.