Let's break down the logic propositions step by step and evaluate them based on the given values of $p = T$, $q = T$, and $r = F$.

Proposition 1: $(q \lor r) \rightarrow \neg p$

• $q \lor r$:

  • $q = T$ and $r = F$, so $q \lor r = T$.

• $\neg p$:

  • $p = T$, so $\neg p = F$.

• $(q \lor r) \rightarrow \neg p$:

  • $T \rightarrow F$ is false (since the only false case for implication is when the first part is true and the second part is false).

So, Proposition 1 evaluates to false.

Proposition 2: $\neg r \leftrightarrow (p \land q)$

• $\neg r$:

  • $r = F$, so $\neg r = T$.

• $p \land q$:

  • $p = T$ and $q = T$, so $p \land q = T$.

• $\neg r \leftrightarrow (p \land q)$:

  • $T \leftrightarrow T$ is true (since the biconditional is true when both sides have the same truth value).

So, Proposition 2 evaluates to true.

Conclusion:

• Proposition 1 is false.
• Proposition 2 is true.

Now, let's match this with the given options:

• Option C correctly states that $(q \lor r) \rightarrow \neg p = F$ and $\neg r \leftrightarrow (p \land q) = T$.

So, the correct answer is C.

Would you like more details or further clarification? Here are some related questions you might find useful:

1. What is the difference between implication and biconditional statements?
2. How do truth tables work for logical operators like $\lor$, $\land$, and $\rightarrow$?
3. Can a proposition be true and false under different interpretations?
4. What is the only case where an implication is false?
5. How can we represent logical operations using set theory?

Tip: When evaluating implications, remember that the implication is false only if the premise is true and the conclusion is false.