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The image shows four logical equivalence problems, where each one includes a logical formula and a corresponding equivalence to be evaluated. Let's go through each case:

**(a)** \[
p \vee (p \land (p \vee q)) \equiv p
\]
We need to simplify the left-hand side of the equation and see if it is equivalent to \(p\).

1. By distribution:
   \[
   p \land (p \vee q) \equiv p
   \]
   This is because \(p \land (p \vee q)\) will always result in \(p\).
   
2. Now the expression becomes:
   \[
   p \vee p \equiv p
   \]
   which is a tautology, as \(p \vee p\) simplifies to \(p\). Hence, (a) is true.

**(b)** \[
\overline{(p \vee q) \land r} \vee \overline{q} \equiv q \land r
\]
This is a bit more complex and involves De Morgan's laws and distribution:

1. Apply De Morgan’s Law to the negations:
   \[
   \overline{(p \vee q) \land r} \equiv \overline{(p \vee q)} \vee \overline{r}
   \]
   So we have:
   \[
   (\overline{p} \land \overline{q}) \vee \overline{r} \vee \overline{q}
   \]

2. Combine the terms:
   \[
   (\overline{p} \vee \overline{r}) \land (\overline{q} \vee \overline{r})
   \]
   Without further simplification, this doesn't directly simplify to \(q \land r\), so (b) might be false. Further detailed steps could be done to prove this rigorously.

**(c)** \[
p \vee q \vee (\overline{p} \land \overline{q} \land r) \equiv p \vee q \vee r
\]
Let's analyze this:

1. The expression \( \overline{p} \land \overline{q} \land r \) adds \(r\) only when both \(p\) and \(q\) are false.

2. So the left-hand side is adding \(r\) to the disjunction only if both \(p\) and \(q\) are false. However, since the entire expression is \(p \vee q \vee r\), the condition \(p \vee q\) already makes the final result \(p \vee q \vee r\), regardless of whether both \(p\) and \(q\) are false. Therefore, the two sides are equivalent. Hence, (c) is true.

**(d)** \[
(\overline{p \vee q} \Rightarrow (p \land q \land r)) \equiv p \land q
\]
First, simplify the implication \(\overline{p \vee q} \Rightarrow (p \land q \land r)\):

1. The negation \(\overline{p \vee q}\) simplifies to \(\overline{p} \land \overline{q}\).

2. The implication \((\overline{p} \land \overline{q}) \Rightarrow (p \land q \land r)\) means that if both \(p\) and \(q\) are false, then the right side should hold true. However, \(p \land q \land r\) cannot be true if \(p\) and \(q\) are both false. So, the implication will always be true.

Now compare this to \(p \land q\), which is only true if both \(p\) and \(q\) are true. Therefore, (d) might be false since the conditions on the left side don't directly match \(p \land q\).

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Would you like to dive deeper into any specific step or need further clarification? Let me know if you'd like details on particular logical laws!

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Here are 5 related questions for further practice:
1. How can De Morgan's laws be applied to simplify logical expressions?
2. Can you explain how the distributive property works in propositional logic?
3. What are the key differences between conjunctions and disjunctions in logic?
4. How does implication behave when the premise is false?
5. What is a tautology in propositional logic?

**Tip**: Always start by simplifying the innermost parentheses in a logical expression to make the overall simplification easier.

Logical equivalence problems involving the formulas provided in the image.

Explore the logical equivalence problems, including simplifications using De Morgan's law, tautology, and implications. This is suitable for university-level students studying discrete mathematics or logic.

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