Let's construct the truth table for the logical statement:

(p → ¬p) ∨ (p → q)

Steps:

1. Break the statement into components:
   • $p$
   • $\neg p$ (negation of $p$)
   • $p \rightarrow \neg p$ (implication)
   • $p \rightarrow q$ (implication)
   • $(p \rightarrow \neg p) \lor (p \rightarrow q)$ (disjunction)

Truth table:

$p$	$q$	$\neg p$	$p \rightarrow \neg p$	$p \rightarrow q$	$(p \rightarrow \neg p) \lor (p \rightarrow q)$
T	T	F	F	T	T
T	F	F	F	F	F
F	T	T	T	T	T
F	F	T	T	T	T

Explanation:

1. $p \rightarrow \neg p$: This is true only when $p$ is false, as a true proposition cannot imply its own negation.
2. $p \rightarrow q$: This is true when either $p$ is false, or when both $p$ and $q$ are true.
3. $(p \rightarrow \neg p) \lor (p \rightarrow q)$: The disjunction of these two implications is true whenever at least one of the implications is true.

Result:

The truth table shows that the overall statement $(p \rightarrow \neg p) \lor (p \rightarrow q)$ is only false when $p = T$ and $q = F$.

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Do you want further clarification or details on any step?

Here are five related questions:

1. What is the truth table for $\neg(p \lor q)$?
2. How does the truth table change if we swap $\neg p$ with $\neg q$?
3. How do we construct a truth table for three variables, like $p$, $q$, and $r$?
4. Can you construct a truth table for $p \rightarrow (q \rightarrow p)$?
5. What is the difference between implication and bi-conditional statements?

Tip: In logical implications, $p \rightarrow q$ is always true if $p$ is false, regardless of $q$.