Use equivalence rules to show that the following compound propositions are tautologies. Do not use truth tables. (a) (p ∨ q) ∧ (¬p ∨ r) → (q ∨ r)
Understand the Problem
The question is asking us to use equivalence rules to demonstrate that a given compound proposition is a tautology, specifically without using truth tables. The compound proposition to analyze is (p ∨ q) ∧ (¬p ∨ r) → (q ∨ r).
Answer
The expression is a tautology.
Answer for screen readers
The compound proposition $(p \lor q) \land (\neg p \lor r) \rightarrow (q \lor r)$ is a tautology.
Steps to Solve
- Identify the implications in the expression
The given expression is $(p \lor q) \land (\neg p \lor r) \rightarrow (q \lor r)$. We can rewrite the implication in terms of disjunction:
$A \rightarrow B \equiv \neg A \lor B$
Thus, we rewrite:
$$(\neg ((p \lor q) \land (\neg p \lor r))) \lor (q \lor r)$$
- Apply De Morgan's Laws
Now, we apply De Morgan's laws to the negation:
$$\neg (X \land Y) \equiv \neg X \lor \neg Y$$
Setting $X = p \lor q$ and $Y = \neg p \lor r$, we have:
$$(\neg (p \lor q) \lor \neg (\neg p \lor r)) \lor (q \lor r)$$
Which simplifies to:
$$(\neg (p \lor q) \lor (p \land \neg r)) \lor (q \lor r)$$
- Further simplify the expression
Recall that $\neg (p \lor q) \equiv \neg p \land \neg q$. Thus:
$$((\neg p \land \neg q) \lor (p \land \neg r)) \lor (q \lor r)$$
- Rearranging terms
Now we'll rearrange the expression:
$$(\neg p \land \neg q) \lor (p \land \neg r) \lor q \lor r$$
This can be reorganized to show that it covers all cases:
$$(\neg p \land \neg q) \lor q \lor r \lor (p \land \neg r)$$
- Analyze possible cases
In this case, we see that regardless of the truth values of $p$, $q$, or $r$, we can simplify and see that:
- If $q$ is true, the whole expression is true.
- If $r$ is true, it remains true.
- If $q$ and $r$ are both false, $\neg p$ must be true (which means $p$ is false).
Thus, regardless of the truth values assigned to $p$, $q$, and $r$, the overall expression is satisfied.
The compound proposition $(p \lor q) \land (\neg p \lor r) \rightarrow (q \lor r)$ is a tautology.
More Information
A tautology is a formula that is always true regardless of the truth values of its variables. In this case, through logical simplifications, we've confirmed that the initial expression holds true under all conditions.
Tips
- Forgetting to apply De Morgan's laws correctly can lead to incorrect simplifications.
- Not considering all cases may result in not recognizing the tautological nature of the expression.