To simplify the logical expression $((\neg p \lor r) \land (p \lor q))$, we can apply the distributive laws of logic and other simplification rules such as the identity, dominance, absorption, and negation laws. Here’s the step-by-step simplification:

Expression:

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

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Step 1: Expand using the distributive property

We distribute each term in $(\neg p \lor r)$ across $(p \lor q)$: [ [(\neg p \lor r) \land (p \lor q)] = [(\neg p \land (p \lor q)) \lor (r \land (p \lor q))] ]

This gives us two separate components:

1. $(\neg p \land (p \lor q))$
2. $(r \land (p \lor q))$

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Step 2: Simplify the first term $(\neg p \land (p \lor q))$

Distribute $\neg p$ across $(p \lor q)$: $(\neg p \land (p \lor q)) = (\neg p \land p) \lor (\neg p \land q)$

• $(\neg p \land p) = \text{False}$ (Contradiction Law)
• $(\neg p \land q) = (\neg p \land q)$

Thus: $(\neg p \land (p \lor q)) = (\neg p \land q)$

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Step 3: Simplify the second term $(r \land (p \lor q))$

Distribute $r$ across $(p \lor q)$: $(r \land (p \lor q)) = (r \land p) \lor (r \land q)$

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Step 4: Combine the results

Now we combine the simplified terms: $(\neg p \lor r) \land (p \lor q) = (\neg p \land q) \lor (r \land p) \lor (r \land q)$

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Step 5: Factor and simplify further if possible

The expression can be left in its simplified form: $(\neg p \land q) \lor (r \land p) \lor (r \land q)$

Alternatively, you could write it as: $(r \land (p \lor q)) \lor (\neg p \land q)$

This is the simplest form.

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Final Answer:

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

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Do you want me to break down any specific step further? 😊

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Related Questions:

1. What are the primary laws of logic used in this simplification?
2. Can you explain why $(\neg p \land p)$ simplifies to $\text{False}$?
3. How does distributive property in logic compare to distributive property in algebra?
4. How would you express this simplified formula as a truth table?
5. Can this expression be further reduced using Karnaugh maps?

Tip:

When simplifying logical expressions, always identify terms that result in contradictions ($p \land \neg p$) or tautologies ($p \lor \neg p$) to quickly eliminate unnecessary components.