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Chapter 1, #10 (c) : Use a truth table to decide whether $(P \vee Q) \rightarrow (Q \vee P)$ is a tautology or not.
Solution:
$P$ $Q$ $P \vee Q$ $Q \vee P$ $(P \vee Q) \rightarrow (Q \vee P)$
T T T TT
T F T T T
F T T T T
F F F F T
Since the last column is all T, we see that $(P \vee Q) \rightarrow (Q \vee P)$ is a tautology.

Chapter 1 #10 (i): Use a truth table to decide whether $(P \wedge Q) \rightarrow (P \vee R)$ is a tautology or not.
Solution:
$P$$Q$R$P \wedge Q$$P \vee R$$(P \wedge Q) \rightarrow (P \vee R)$
TTTTTT
TTFTTT
TFTFTT
TFFFTT
FTTFTT
FTFFFT
FFTFTT
FFFFFT
Therefore $(P \wedge Q) \rightarrow (P \vee R)$ is a tautology.

Chapter 1 #16 (c) If $P$ and $Q$ are distinct atomic sentences, does the sentence $\neg P \vee Q$ tautologically imply $P \rightarrow Q$?
Solution: For a sentence $S_1$ to tautologically imply a sentence $S_2$ means that the sentence $S_1 \rightarrow S_2$ is a tautology. So for this problem, we must decide whether or not the sentence $(\neg P \vee Q) \rightarrow (P \rightarrow Q)$ is a tautology. We will proceed via truth table:
$P$$Q$$\neg P$$\neg P \vee Q$$P \rightarrow Q$$(\neg P \vee Q) \rightarrow (P \rightarrow Q)$
TTFTTT
TFFFFT
FTTTTT
FFTTTT
Yes. From this table we see that $(\neg P \wedge Q) \rightarrow (P \rightarrow Q)$, which is another way to say that the sentence $\neg P \wedge Q$ tautologically implies $P \rightarrow Q$.

Chapter 1 #17 (b) If $P$ is an atomic sentence, then is $P$ tautologically equivalent to $P \vee \neg P$?
Solution: To check this, we have to decide whether or not the sentence $P \leftrightarrow (P \vee \neg P)$ is a tautology. Recall that $S_1 \leftrightarrow S_2$ was defined by $(S_1 \rightarrow S_2) \wedge (S_2 \rightarrow S_1)$ and construct a truth table:
$P$$\neg P$$P \vee \neg P$$P \rightarrow (P \vee \neg P)$$(P \vee \neg P) \rightarrow P$$P \leftrightarrow (P \vee \neg P)$
TFTTTT
FTTTFF

From this we must conclude that $P \leftrightarrow (P \vee \neg P)$ is not a tautology and hence $P$ and $P \vee \neg P$ are not tautologically equivalent.

Chapter 1 #17 (d) If $P$ is an atomic sentence, then is $P$ tautologically equivalent to $P \rightarrow P$?
Solution: To check this, we have to deicde whether or not the sentence $P \leftrightarrow (P \rightarrow P)$ is a tautology or not. Recall that $S_1 \leftrightarrow S_2$ was defined by $(S_1 \rightarrow S_2) \wedge (S_2 \rightarrow S_1)$ and construct a truth table:
$P$ $P \rightarrow P$$P \rightarrow (P \rightarrow P)$$(P \rightarrow P) \rightarrow P$$P \leftrightarrow (P \rightarrow P)$
TTTTT
FTTFF
From this we must conclude that $P \leftrightarrow (P \rightarrow P)$ is not a tautology and hence $P$ and $P \rightarrow P$ are not tautologically equivalent.