Simplify the following logical expression using the Laws of Logic.

Question

( ¬P ∨ ¬Q ∨ R) → ¬ ( Q ∨ R )

( ¬P ∨ ¬Q ∨ R) → ( ¬Q ∧ ¬R )

¬ ( ¬P ∨ ¬Q ∨ R ) ∨ ( ¬Q ∧ ¬R )

( P ∧ Q ∧ ¬R ) ∨ ( ¬Q ∧ ¬R )

P ∧ ( Q ∧ ¬R ) ∨ ( ¬Q ∧ ¬R )

I think it can be simplify again. Can someone help me with this?

Answer 1

((P ∧ Q) ∨ ¬Q) ∧ ¬R

(P ∨ ¬Q) ∧ ¬R

Answer 2

First, notice that your last expression $P \land (Q \land \neg R) \lor (\neg Q \land \neg R)$ is of the form $\phi \land \psi \lor \varphi$, which is ambiguous, so you should not take that step from line 4 to line 5

Starting at line 4 however, we can do:

$$(P \land Q \land \neg R) \lor (\neg Q \land \neg R) \Leftrightarrow \text{ (Distribution)}$$

$$((P \land Q) \lor \neg Q) \land \neg R \Leftrightarrow \text{ (Reduction)}$$

$$(P \lor \neg Q) \land R$$

If you don't have or are not allowed to use Reduction, then do:

$$((P \land Q) \lor \neg Q) \land \neg R \Leftrightarrow \text{ (Distribution)}$$

$$((P \lor \neg Q) \land (Q \lor \neg Q)) \land R \text{ (Complement)}$$

$$((P \lor \neg Q) \land \top) \land R \text{ (Identity)}$$

$$(P \lor \neg Q) \land R$$