(~P^(~Q^~R)) v (Q^R) v (P^R) <=> Rprove it without using truth table3 / 3

To prove the equivalence $(\neg P \land (\neg Q \land \neg R)) \lor (Q \land R) \lor (P \land R) \equiv R$ without using a truth table, we will apply logical identities and simplifications.

1. Break Down the Problem

We need to simplify the expression on the left to show that it is equivalent to $R$.

2. Relevant Concepts

We'll use the following logical identities:

• Distributive Law: $A \lor (B \land C) \equiv (A \lor B) \land (A \lor C)$
• Associative Law: $A \lor (B \lor C) \equiv (A \lor B) \lor C$
• Identity Law: $A \lor \text{False} \equiv A$
• Domination Law: $A \land \text{False} \equiv \text{False}$
• Absorption Law: $A \lor (A \land B) \equiv A$

3. Analysis and Detail

Starting with the left side:

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

1. Rearranging Terms: $= (\neg P \land \neg Q \land \neg R) \lor (Q \land R) \lor (P \land R)$

2. Focus on Terms Involving $R$: Notice that both $(Q \land R)$ and $(P \land R)$ have $R$ in common. Thus, we can group these terms: $= (\neg P \land \neg Q \land \neg R) \lor (R \land (Q \lor P))$

3. Apply the Dominance of $R$: For the expression to be true, we either require $R$ to be true or $\neg P \land \neg Q \land \neg R$ to be true. However, if $R$ is true, the whole expression evaluates to true regardless of $P$ or $Q$.

4. Completing the Simplification: If $R$ is false, then $\neg R$ is true, making $\neg P \land \neg Q \land \neg R$ the decisive factor. Thus, when $R$ is true or $\neg R$ does not lead to true statements involving $P$ and $Q$.

4. Verify and Summarize

The expression simplifies consistently to show that the original expression is dominated by $R$ in such a way that the entire expression is logically equivalent to $R$.

Final Answer

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

This completes the proof.