For the first part, "Exactly $n$ of the statements are false", note that they all contradict each other, so at most one is true, so at least $99$ are false. This guarantees that $1-98$ are false. You can have $99$ being true and $100$ false and all is consistent. If $99$ if false as well we cannot assign a consistent truth value to $100$, so the only assignment is number $99$ true and all others false.
For $100$ sentences saying "At least $n$ of these are false" we cannot have $1$ be false because its negation is all of these are true, so it would have to be true, in conflict with our assumption that it is false, so $1$ is true and at least one of the statements is false. Now we know that statement $100$ is false because one of the statements is true. Statement $2$ cannot be false, because its falsity would require at most one true statement, which is $1$, which would make it true. We continue back and forth, showing the high statement numbers are false and the small ones true until we show $1-50$ are true and $51-100$ are false.
If there were $99$ sentences of the form "At least $n$ of these are false" we proceed just like the $100$ case until we have $1-49$ true and $51-99$ false. Then if $50$ is true there are only $49$ false statements and if it is false there are $50$. Either way is a contradiction and we conclude there is no consistent truth assignment.
Let us start by reconsidering the definition of the dual $\phi^*$ for a given propositional formula $\phi$. It can be summarised by the correspondences:
\begin{align}
\phi \land \psi &\overset*\iff\phi \lor \psi\\
\mathbf F &\overset*\iff \mathbf T
\end{align}
We reuse a fundamental idea of the earlier exercises in the mentioned chapter, namely that $\phi \leftrightarrow \psi$ precisely when $\phi$ and $\psi$ are satisfied by exactly the same lines in a truth table.
Observation 1: A line of a truth table is nothing more than a function $v$ which assigns to each propositional variable a value $\mathbf T$ or $\mathbf F$ and subsequently evaluates the definitions of the connectives for this assignment.
Such a $v$ is known in the literature as a valuation or a boolean interpretation. Abstracting this notion away from the intuitive use in truth tables is a big asset in proving generic theorems. This is because it allows us to discuss rigorously an "arbitrary line of a truth table".
For example, by considering the two possibilities $v(p)=\mathbf T$ and $v(p)=\mathbf F$, it is not hard to see that $v(p\lor \neg p)= \mathbf T$ for every valuation of $\{p\}$. Similarly one can show that $v(p\to q) = v(\neg p \lor q)$ for each valuation of $\{p,q\}$, thus proving equivalence of these statements.
Observation 2: $\phi$ and $\psi$ are equivalent precisely when $v(\phi) =v(\psi)$ for all valuations $v$.
We need one final ingredient before turning to the main result.
Definition: Given a valuation $v$, the valuation $v^*$ is defined by: $$v^*(p) = \begin{cases}\mathbf T&\text{ if $v(p) = F$}\\\mathbf F&\text{ if $v(p) = T$}\end{cases}$$
The desired result will follow from the following:
Theorem: For any formula $\phi$ and valuation $v$, we have: $$v(\phi^*) = v^*(\neg\phi)$$
By assumption $\phi$ will have one of the following forms:
- $\phi = p$
- $\phi = \mathbf F$
- $\phi = \mathbf T$
- $\phi = \neg \tau$
- $\phi = \tau \lor \tau'$
- $\phi = \tau \land \tau'$
I will only prove the assertion for the last possibility $\tau \land \tau'$; the other options are similar or easier. We assume that the result has been proven for $\tau$ and $\tau'$.
Note that this decomposition of $\phi$ into possible forms is known formally as structural induction; it allows us to reason about an arbitrary formula in a structural manner, similar to proving an implication like $(\phi \lor \psi)\to \tau$: we first assume $\phi$ and show $\tau$ from it, and then do the same for $\psi$.
Now for the proof:
\begin{align}
v(\phi^*) &= v((\tau\land\tau')^*)\\
&= v(\tau^* \lor \tau'^*)\\
&= f^\lor(v(\tau^*),v(\tau'^*))\\
&= f^\lor(~v^*(\neg\tau),v^*(\neg\tau')~)\\
&= v^*(\neg \tau \lor \neg\tau')\\
&= v^*(\neg(\tau\land\tau'))\\
&= v^*(\neg\phi)
\end{align}
where $f^\lor$ denotes the definition of $\lor$ in terms of truth values (so $f^\lor(\mathbf F,\mathbf F) = \mathbf F$, otherwise $f^\lor$ yields $\mathbf T$). In the second-to-last step, a De Morgan equivalence is used.
Intuitively, the above result states that for every line of a truth table for $\phi^*$, we can mechanically construct one with the same value for $\neg\phi$. Because $\neg\phi$ and $\neg\psi$ are also equivalent, this means that the constructed valuations agree on $\neg\phi$ and $\neg\psi$. Additionally, it is not hard to see that every valuation on $\neg\phi$ can be constructed in this way, and we can transform them back to valuations of $\phi^*$ since $v^{**}=v$. These ingredients together establish equivalence of $\phi^*$ and $\psi^*$.
Best Answer
Given an English statement $eng$, the statement $\ulcorner eng \urcorner$ is an encoding of $eng$ into propositional logic (in a presumably obvious way). This is so I can avoid having to write out very long and convoluted formulas in propositional logic, but you should make sure that you can see how the English statement could be encoded into propositional logic.
You should consider the $n$th statement to be an atomic proposition $P_n$. We then have a bunch of logical formulas of the form $\phi_n :\equiv (P_n \iff \ulcorner$exactly $n$ of the $P_i$ are false$\urcorner)$. The question is: is there any truth assignment to the atomic propositions such that all the $\phi_n$ are satisfied? In other words, is the statement $\phi_1 \land \cdots \land \phi_{100}$ satisfiable? This is how you should approach part (a). Hint: if all the $\phi_n$ are satisfied by some truth assignment, then let $m$ be the number of statements $P_n$ which are false in the truth assignment. What can you say about $m$?
For part (b), you should do the same, but instead define $\psi_n :\equiv (P_n \iff \ulcorner$at least $n$ of the $P_i$ are false$\urcorner)$. Again, you should consider some truth assignment which might satisfy all of the $\psi_n$, and let $\ell$ be the number of statements which are false in the truth assignment. What can you say about $\ell$?
Part (c) is just part (b) with a different number of formulas, but you should check your logic carefully, as there is a key difference between 99 and 100 which will make itself plain.
As for your example, if statements 50 through 99 are false, then there are at least fifty false statements. But then $P_{50}$ is false.