To show that $(p \land \neg q \implies r) \equiv (p \land \neg r \implies q)$, we can approach this problem using logical equivalences to manipulate and compare both sides of the equivalence.

Steps using Logical Equivalences:

We start with the expression on the left-hand side: $p \land \neg q \implies r$ By definition of implication ($A \implies B \equiv \neg A \lor B$), we can rewrite the implication: $p \land \neg q \implies r \equiv \neg(p \land \neg q) \lor r$ Now, apply De Morgan's law to $\neg(p \land \neg q)$: $\neg(p \land \neg q) \equiv \neg p \lor q$ So, the expression becomes: $\neg p \lor q \lor r$ This simplifies to: $\neg p \lor (q \lor r)$ Thus, we have: $p \land \neg q \implies r \equiv \neg p \lor (q \lor r)$

Now let's look at the right-hand side of the equivalence: $p \land \neg r \implies q$ Again, using the definition of implication: $p \land \neg r \implies q \equiv \neg(p \land \neg r) \lor q$ Apply De Morgan's law again to $\neg(p \land \neg r)$: $\neg(p \land \neg r) \equiv \neg p \lor r$ So, the expression becomes: $\neg p \lor r \lor q$ This simplifies to: $\neg p \lor (r \lor q)$ And since $r \lor q$ is equivalent to $q \lor r$, this is the same as: $\neg p \lor (q \lor r)$ Thus, we have: $p \land \neg r \implies q \equiv \neg p \lor (q \lor r)$

Since both sides of the equivalence reduce to $\neg p \lor (q \lor r)$, we can conclude: $(p \land \neg q \implies r) \equiv (p \land \neg r \implies q)$

Conclusion:

We have shown that $(p \land \neg q \implies r) \equiv (p \land \neg r \implies q)$ by reducing both sides to the same logical expression, $\neg p \lor (q \lor r)$.

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Would you like a more detailed breakdown of any steps or have any questions? Here are 5 related questions for further exploration:

1. How does De Morgan's law work for more complex logical expressions?
2. Can you show the proof using a truth table instead of logical equivalences?
3. What is the role of distributive properties in logical equivalence?
4. How would the equivalence change if we added negations to both $p$ and $r$?
5. How does this logical equivalence apply to proving $p \implies (q \lor r)$ directly?

Tip: When using logical equivalences, remember that De Morgan's laws and the definition of implication ($A \implies B \equiv \neg A \lor B$) are powerful tools for simplifying complex logical statements.