On Homework #7, you proved [p =⇒ (q ∨ r)] ≡ [(p ∧ (¬q) =⇒ r]. Hence, to prove a statement of the form p =⇒ (q ∨ r) directly, you assume that p is true and q is false, then use these assumptions, plus definitions and previously proven results, to show that r is true.
(a) Show that [(p ∧ (¬q) =⇒ r] ≡ [(p ∧ (¬r) =⇒ q]. You may use either a truth tables or strings of previously established logical equivalences. (This result means that you do NOT have to also go through the case where you assume p is true and r is false, then prove q is true.) [3]


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, \(\neg p \lor (q \lor r)\).

---

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.

On Homework #7, you proved [p =⇒ (q ∨ r)] ≡ [(p ∧ (¬q) =⇒ r]. Hence, to prove a statement of the form p =⇒ (q ∨ r) directly, you assume that p is true and q is false, then use these assumptions, plus definitions and previously proven results, to show that r is true. (a) Show that [(p ∧ (¬q) =⇒ r] ≡ [(p ∧ (¬r) =⇒ q]. You may use either a truth tables or strings of previously established logical equivalences. (This result means that you do NOT have to also go through the case where you assume p is true and r is false, then prove q is true.)

Explore how to prove the logical equivalence [(p ∧ ¬q ⇒ r)] ≡ [(p ∧ ¬r ⇒ q)] using logical equivalences and De Morgan's laws. This step-by-step solution is ideal for students studying introductory logic at the college level.

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