Answer
See the step-by-step answer.
Work Step by Step
For each of these, I will give a proof via a chain of logical equivalences (eq), and a standard proof (pf). For the problem you do not need to do both. Rather, it is here to show you two ways to solve the same problem. This might depend on which class you are taking.
a) (eq) Show $(p\wedge q)\Rightarrow q$ is a tautology (a true statement). Let us first note that for two atomic formula $a$ and $b$, there is the equivalence $a\Rightarrow b\equiv\neg a\vee b$. Thus,
$(p\wedge q)\Rightarrow q\equiv\neg(p\wedge q)\vee q$
Now we can use one of De Morgan's laws, namely $\neg(a\wedge b)\equiv \neg a\vee \neg b$. This yields
$\neg(p\wedge q)\vee q\equiv (\neg p\vee \neg q)\vee q$
Now by associativity of disjunction ($\vee$), the relation $a\vee (b\vee c)\equiv (a\vee b)\vee c$ holds. Using this, we get
$(\neg p\vee\neg q)\vee q\equiv \neg p\vee (\neg q\vee q)$
However, this is the Law of Excluded Middle, i.e. $a\vee \neg a\equiv \mathrm{T}$. By this and commutivity of disjunction ($\vee$),
$\neg p\vee (\neg q\vee q)\equiv \neg p\vee\mathrm{T}$
Yet, by the laws of disjunction ($\vee$), we have that $a\vee\mathrm{T} \equiv\mathrm{T}$. So our final logical equivalence is
$\neg p\vee\mathrm{T}\equiv\mathrm{T}$
Thus, through a chain of equivalences, we have shown that $(p\wedge q)\Rightarrow q\equiv\mathrm{T}$, so it is a tautology as required.
a) (pf) The proof is rather simple:
In order to prove an implication, we assume the hypothesis and try to prove the conclusion. Thus, suppose $p\wedge q$ is true, then by the definition of conjunction ($\wedge$), both $p$ and $q$ are true. Therefore, $q$ is true. Hence, we have shown the conclusion assuming the hypothesis, so $(p\wedge q)\Rightarrow q$ is true.
b) (eq) Show that $p\Rightarrow(p\vee q)$ is a tautology. Using the same rule for implication above,
$p\Rightarrow (p\vee q)\equiv \neg p\vee (p\vee q)$
Thus, by associativity of disjunction ($\vee$), we have
$\neg p\vee (p\vee q)\equiv (\neg p\vee p)\vee q$
However, by the Law of Excluded Middle ($a\vee \neg a\equiv\mathrm{T}$) and commutivity of disjunction,
$(\neg p\vee p)\vee q\equiv \mathrm{T}\vee q$
Finally, since $a\vee b\equiv b\vee a$ and $a\vee\mathrm{T}\equiv\mathrm{T}$, this yields
$\mathrm{T}\vee q\equiv\mathrm{T}$
Thus, $p\Rightarrow (p\vee q)\equiv\mathrm{T}$, as desired.
b) (pf) The proof is just as easy:
As before, to prove an implication, we assume the hypothesis and show the conclusion. Hence, assume $p$ is true, then since one of $p$ or $q$ is true (namely $p$), we have that $p\vee q$ holds. Thus, since we have arrived at the conclusion, this proves that $p\Rightarrow(p\vee q)$.
For the next two problems, unless I use a new logical equivalence, I will simply list the equivalences:
c) (eq) Show $\neg p\Rightarrow(p\Rightarrow q)$ is a tautology:
By the Law of Double Negation and the rule for implication, i.e. $\neg\neg a\equiv a$, we have
$\neg p\Rightarrow (p\Rightarrow q)\equiv p\vee (p\Rightarrow q)$
$p\vee (p\Rightarrow q)\equiv p\vee (\neg p\vee q)$
$p\vee (\neg p\vee q)\equiv (p\vee\neg p)\vee q$
$(p\vee\neg p)\vee q\equiv\mathrm{T}\vee q$
$\mathrm{T}\vee q\equiv\mathrm{T}$
So $\neg p\Rightarrow (p\Rightarrow q)\equiv\mathrm{T}$.
c) (pf) This proof is a little more involved:
To prove the implication, suppose $\neg p$ is true. Now we wish to prove $p\Rightarrow q$, so assume $p$ is true, then we have that $\neg p\wedge p$ is true. However, this the Law of Contradiction, i.e. $\neg p\wedge p\equiv\mathrm{F}$, but this means that $\mathrm{T}\equiv\mathrm{F}$ since we know that $\neg p\wedge p$ is true. Hence, we can use the Principle of Explosion, which says that if we have shown $\mathrm{T}\equiv\mathrm{F}$, then everything is true, because truth and falsehood are the same. Thus, $q$ is true. Since we assumed $p$ and proved $q$, we know $p\Rightarrow q$ is true. However, since we assumed $\neg p$ and proved $p\Rightarrow q$, then we know that $\neg p\Rightarrow (p\Rightarrow q)$ is true.
d) (eq) Show $(p\wedge q)\Rightarrow (p\Rightarrow q)$ is a tautology.
$(p\wedge q)\Rightarrow (p\Rightarrow q)\equiv \neg(p\wedge q)\vee(p\Rightarrow q)$
$\neg(p\wedge q)\vee(p\Rightarrow q)\equiv (\neg p\vee\neg q)\vee (p\Rightarrow q)$
$(\neg p\vee\neg q)\vee (p\Rightarrow q)\equiv (\neg p\vee\neg q)\vee (\neg p\vee q)$
$(\neg p\vee\neg q)\vee (\neg p\vee q)\equiv (\neg p\vee\neg q)\vee (q\vee\neg p)$
$(\neg p\vee\neg q)\vee (q\vee\neg p)\equiv((\neg p\vee\neg q)\vee q)\vee\neg p$
$((\neg p\vee\neg q)\vee q)\vee\neg p\equiv(\neg p\vee(\neg q\vee q))\vee\neg p$
$(\neg p\vee(\neg q\vee q))\vee\neg p\equiv (\neg p\vee\mathrm{T})\vee\neg p$
$(\neg p\vee\mathrm{T})\vee\neg p\equiv\mathrm{T}\vee\neg p$
$\mathrm{T}\vee\neg p\equiv \mathrm{T}$
So $(p\wedge q)\Rightarrow (p\Rightarrow q)$ is a tautology.
d) (pf) To prove the first implication, suppose $p\wedge q$ is true, then $p$ is true and $q$ is true. Now, to show the second implication, suppose $p$ is true, then since $q$ is true, by the last sentence, we have shown $p\Rightarrow q$. However, since we assumed $p\wedge q$ and proved $p\Rightarrow q$, we have that $(p\wedge q)\Rightarrow (p\Rightarrow q)$.
e) (eq) Show that $\neg (p\Rightarrow q)\Rightarrow p$ is a tautology.
$\neg (p\Rightarrow q)\Rightarrow p\equiv\neg(\neg p\vee q)\Rightarrow p$
$\neg(\neg p\vee q)\Rightarrow p\equiv (\neg p\vee q)\vee p$
$(\neg p\vee q)\vee p\equiv (q \vee \neg p)\vee p$
$(q \vee \neg p)\vee p\equiv q\vee (\neg p\vee p)$
$q\vee (\neg p\vee p)\equiv q\vee\mathrm{T}$
$q\vee\mathrm{T}\equiv\mathrm{T}$
So $\neg (p\Rightarrow q)\Rightarrow p$ is, in fact, a tautology.
e) (pf) To prove the implication, assume $\neg(p\Rightarrow q)$ is true. Now note that $\mathrm{T}$ always holds, and that $\mathrm{T}\equiv p\vee\neg p$. (Here we need to use case analysis since we do not know which is true, $p$ or $\neg p$). By case analysis:
1) Suppose $p$ is true.
2) Suppose $\neg p$ is true, then at least one of $\neg p$ or $q$ is true, so $\neg p\vee q$ is true. But remember that $\neg p\vee q\equiv p\Rightarrow q$, so the implication $p\Rightarrow q$ is true. However, now we have that $\neg(p\Rightarrow q)$ and $p\Rightarrow q$ is true. By the Law of Contradiction, we have shown that $\mathrm{T}\equiv\mathrm{F}$. So by the Principle of Explosion, we can assume whatever we want. In particular, $p$ is true.
Since for all cases we showed that $p$ is true, we have shown that $\neg (p\Rightarrow q)\Rightarrow p$ is a tautology.
f) (eq) Show that $\neg(p\Rightarrow q)\Rightarrow\neg q$ is a tautology.
$\neg(p\Rightarrow q)\Rightarrow\neg q\equiv(p\Rightarrow q)\vee\neg q$
$(p\Rightarrow q)\vee\neg q\equiv (\neg p\vee q)\vee\neg q$
$(\neg p\vee q)\vee\neg q\equiv \neg p\vee (q\vee\neg q)$
$\neg p\vee (q\vee\neg q)\equiv\neg p\vee\mathrm{T}$
$\neg p\vee\mathrm{T}\equiv\mathrm{T}$
f) (pf) To prove the implication, suppose $\neg (p\Rightarrow q)$ is true, then since $\mathrm{T}\equiv\neg q\vee q$, we can use case analysis:
1) Suppose $\neg q$ is true.
2) Suppose $q$ is true, then at least one of $\neg p$ or $q$ is true, so $\neg p\vee q$ is true. However, $\neg p\vee q\equiv p\Rightarrow q$, so the implication $p\Rightarrow q$ is true. Now we have reached a contradiction since $\neg(p\Rightarrow q)$ is true as well. By the Principle of Explosion, we can assume $\neg q$ to be true.
Since we have shown that $\neg q$ is true in both cases, then $\neg (p\Rightarrow q)\Rightarrow\neg q$ is true.
