suppose the conditional p→q is true and the conditional q→r is false. must the conditional p→r also be false? explain. choose the correct answer below. a. no, p→r is not false. if p→q is true and q→r is false, then the hypothesis of p→r is false. this means p→r is true. b. no, p→r is not false. if q→r is false, then q is true and r is false. this means that either p is true and r is false or p and r are both false. in either case, p→r is true. c. yes, p→r is false. if q→r is false, then q is true and r is false. this means that either p is true and r is false or p and r are both false. in either case, p→r is false. d. no, p→r is not false. the two conditionals p→q and q→r can be combined to form the single conditional p→r. this means p→r is true. e. no, p→r is not necessarily false. if p→q is true and q→r is false, then the conclusion of p→r is false. this means p→r could be either true or false. f. no, p→r is not necessarily false. if q→r is false, then q is true and r is false. this means that either p is true and r is false, so that p→r is false, or p and r are both false, so that p→r is true. therefore, p→r could be either true or false.

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# Brief Explanations:
To determine if $ p \to r $ must be false, we analyze the truth values of conditionals. A conditional $ q \to r $ is false only when $ q $ is true and $ r $ is false (by the definition of a conditional: $ A \to B $ is false iff $ A $ is true and $ B $ is false).

Given $ p \to q $ is true, let's consider cases for $ p $:
- Case 1: If $ p $ is true, then since $ p \to q $ is true, $ q $ must be true (because a true conditional with a true hypothesis has a true conclusion). We already know $ q $ is true (from $ q \to r $ being false) and $ r $ is false. So $ p $ is true and $ r $ is false. But wait, for $ p \to r $, a true hypothesis ($ p $) and false conclusion ($ r $) would make $ p \to r $ false? Wait, no—wait, no, let's re - examine. Wait, no, the option B says "in either case, $ p \to r $ is true". Wait, let's re - do the analysis.

Wait, the definition of a conditional $ A \to B $ is: it is false when $ A $ is true and $ B $ is false; otherwise, it is true.

We know $ q \to r $ is false, so $ q = \text{True} $ and $ r=\text{False} $.

Now, $ p \to q $ is true. Let's consider the two possibilities for $ p $:

1. If $ p=\text{True} $: Since $ p \to q $ is true and $ p $ is true, $ q $ must be true (which matches our knowledge from $ q \to r $ being false). Now, for $ p \to r $, $ p $ is true and $ r $ is false. Wait, that would make $ p \to r $ false? But option B says "in either case, $ p \to r $ is true". Wait, I must have made a mistake.

Wait, no—wait, no, let's re - read option B: "If $ q \to r $ is false, then $ q $ is true and $ r $ is false. This means that either $ p $ is true and $ r $ is false or $ p $ and $ r $ are both false. In either case, $ p \to r $ is true."

Wait, when $ p $ is true and $ r $ is false: $ p \to r $ is false (because true $\to$ false is false). But when $ p $ is false and $ r $ is false: $ p \to r $ is true (because false $\to$ false is true). Wait, so there is a contradiction here. Wait, no, maybe my initial analysis of the cases is wrong.

Wait, $ p \to q $ is true. So the possible truth - value combinations for $ p $ and $ q $ are:
   - $ p=\text{False} $, $ q=\text{False} $ (false $\to$ false is true)
   - $ p=\text{False} $, $ q=\text{True} $ (false $\to$ true is true)
   - $ p=\text{True} $, $ q=\text{True} $ (true $\to$ true is true)

But we know from $ q \to r $ being false that $ q = \text{True} $ and $ r=\text{False} $. So we can eliminate the cases where $ q=\text{False} $ (because $ q $ must be true). So the only possible cases for $ p $ and $ q $ (given $ p \to q $ is true and $ q=\text{True} $) are:
   - $ p=\text{False} $, $ q=\text{True} $
   - $ p=\text{True} $, $ q=\text{True} $

Now let's analyze $ p \to r $ for these cases:

- Case 1: $ p=\text{False} $, $ r=\text{False} $: $ p \to r $ is false $\to$ false, which is true.
- Case 2: $ p=\text{True} $, $ r=\text{False} $: $ p \to r $ is true $\to$ false, which is false.

Wait, but option B says "in either case, $ p \to r $ is true". That seems wrong. Wait, maybe I misread the option. Wait, no, let's check the options again.

Wait, option B: "No, $ p \to r $ is not false. If $ q \to r $ is false, then $ q $ is true and $ r $ is false. This means that either $ p $ is true and $ r $ is false or $ p $ and $ r $ are both false. In either case, $ p \to r $ is true."

Wait, when $ p $ is true and $ r $ is false, $ p \to r $ is false. So option B's reasoning is incorrect? But wait, maybe I made a mistake in the truth - value of the conditional. Wait, the truth table for $ A \to B $ is:

| $ A $ | $ B $ | $ A \to B $ |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |

So when $ p $ is true and $ r $ is false, $ p \to r $ is false. When $ p $ is false and $ r $ is false, $ p \to r $ is true. So $ p \to r $ could be either true or false? But that's option F. Wait, option F: "No, $ p \to r $ is not necessarily false. If $ q \to r $ is false, then $ q $ is true and $ r $ is false. This means that either $ p $ is true and $ r $ is false, so that $ p \to r $ is false, or $ p $ and $ r $ are both false, so that $ p \to r $ is true. Therefore, $ p \to r $ could be either true or false."

Wait, but let's go back to the original problem. The question is "Must the conditional $ p \to r $ also be false?". So we need to see if it's always false, or if it can be true or false.

We know $ q \to r $ is false, so $ q = T $, $ r = F $.

$ p \to q $ is true. So $ p $ can be $ T $ (then $ q = T $) or $ p = F $ (then $ q $ can be $ T $ or $ F $, but since $ q = T $, $ p = F $ is allowed).

- If $ p = T $: Then $ p \to r $ is $ T \to F $, which is $ F $.
- If $ p = F $: Then $ p \to r $ is $ F \to F $, which is $ T $.

So $ p \to r $ can be either true or false. So it is not necessarily false. So the correct option is F? Wait, but let's check the options again.

Wait, option B says "in either case, $ p \to r $ is true", which is wrong. Option F says that $ p \to r $ could be either true or false. Let's re - examine the options:

Option F: "No, $ p \to r $ is not necessarily false. If $ q \to r $ is false, then $ q $ is true and $ r $ is false. This means that either $ p $ is true and $ r $ is false, so that $ p \to r $ is false, or $ p $ and $ r $ are both false, so that $ p \to r $ is true. Therefore, $ p \to r $ could be either true or false."

Yes, this is correct. Because when $ p $ is true, $ p \to r $ is false (since $ r $ is false), and when $ p $ is false, $ p \to r $ is true (since false $\to$ false is true). So $ p \to r $ is not necessarily false.

Wait, but earlier I thought option B was a candidate, but it's incorrect. So the correct answer is F.

Wait, but let's check the initial analysis of option B again. Option B says "either $ p $ is true and $ r $ is false or $ p $ and $ r $ are both false. In either case, $ p \to r $ is true." But when $ p $ is true and $ r $ is false, $ p \to r $ is false, so option B is wrong. Option F correctly states that $ p \to r $ could be either true or false.

# Answer:
F. No, $ p \to r $ is not necessarily false. If $ q \to r $ is false, then $ q $ is true and $ r $ is false. This means that either $ p $ is true and $ r $ is false, so that $ p \to r $ is false, or $ p $ and $ r $ are both false, so that $ p \to r $ is true. Therefore, $ p \to r $ could be either true or false.

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