To obtain the Disjunctive Normal Form (DNF) of the logical expression given, we will first simplify the expression and then transform it into DNF.

Given Expression

The expression is: $p \vee (\neg p \rightarrow (q \vee (q \rightarrow \neg r)))$

Step 1: Simplify the Expression

First, recall that the implication $A \rightarrow B$ is equivalent to $\neg A \vee B$.

So, simplify the expression:

1. Simplify $\neg p \rightarrow (q \vee (q \rightarrow \neg r))$: $\neg p \rightarrow (q \vee (\neg q \vee \neg r))$ This simplifies to: $\neg p \rightarrow (q \vee \neg q \vee \neg r)$ Since $q \vee \neg q$ is always true (a tautology), we have: $q \vee \neg q \vee \neg r \equiv \text{True}$ Thus: $\neg p \rightarrow \text{True} \equiv \text{True}$

2. The expression now simplifies to: $p \vee \text{True}$ Which simplifies to: $\text{True}$

Step 2: Disjunctive Normal Form (DNF)

The expression simplifies to $\text{True}$, which is a tautology.

In terms of Disjunctive Normal Form, a tautology (which is always true) can be written as just "True" or as a trivial DNF, such as: $\text{True} = (p \vee \neg p)$ Here, $p \vee \neg p$ represents the simplest possible DNF, as it covers all possible truth values of $p$.

So, the Disjunctive Normal Form (DNF) of the given expression is: $\text{True}$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the Conjunctive Normal Form (CNF) of the same expression?
2. How do you convert a more complex logical expression into its DNF?
3. Can you find the DNF for $\neg (p \wedge q)$?
4. What is the importance of DNF in Boolean algebra and circuit design?
5. How can you verify that an expression is a tautology using truth tables?

Tip: When simplifying logical expressions, always check for tautologies, contradictions, and implications, as they can often simplify the process significantly.