4

Combine the $\neg B$ with premise 3 to get $A$, which with premise 1 gives you $\neg C$, and thus the contradiction with $C$ you are looking for.


3

Because ∀x(K(Al, x) → (x = Bill)) also allows the possibilty that Al doesn't know any x.


2

Unpacking the book's reasoning a bit, when $5\mid x$ is true, then $5\mid xy$ is true can be expanded to $5\mid x$ implies $5\mid xy$, and since I did not say anything about $5\mid y$ just then, the implication is true when $5\mid y$ is true and it also is true when $5\mid y$ is false. Or to put it another way, when someone says "suppose $5\mid x$ ...


2

The case $5|x$ and $5|y$ is already covered by the other cases, so it is not necessary to consider it separately. The important thing is to make sure that the cases you consider are exhaustive.


2

From $\neg R \to P$ you can indeed infer $\neg P \to R$, thought this is typically considered Contraposition rather than Modus Tollens, which would infer $R$ from $\neg R \to P$ and $\neg P$ And no, this is not a fallacy. I think you might be thinking of the Denying the Antecedent Fallacy, which would try to infer $\neg P$ from $\neg R \to P$ and $R$ ... ...


2

It is an implication. If Al knows x, then x is Bill. But without the first part we do not actually know if Al knows Bill, just that if Al does know somebody, it must be Bill.


2

The statement $\Gamma\vDash \phi$ has the meaning "in all models where the formulas in $\Gamma$ are true, also the formula $\phi$ is true." If we want to show that $\Gamma\vDash\phi$, two methods seem reasonable: If we show that $\phi$ is a direct consequence of (some of) the formulas in $\Gamma$, then we have proved that $\Gamma\vDash\phi$. If we show ...


2

The 'correct' transformation depends on what rules you have .... Here is a transformation that uses pretty elementary equivalence principles: $$(A \wedge \neg B) \wedge (A \vee \neg C)$$ $$\overset{Commutation}{=}$$ $$(\neg B \land A) \wedge (A \vee \neg C)$$ $$\overset{Association}{=}$$ $$\neg B \land (A \wedge (A \vee \neg C))$$ $$\overset{...


2

I know from your previous question that you have shown $\neg \neg p \vdash p$ So, since you can easily show that $p, p \to \neg q \vdash \neg q$, you can use the fact that $\neg \neg p \vdash p$ to show that $p, \neg \neg (p \to \neg q) \vdash \neg q$. Thus, by the Deduction Theorem, you have $p \vdash \neg \neg (p \to \neg q) \to \neg q$. Likewise, since ...


1

If the proposition $Q$ is true, then $P \rightarrow Q$ is true regardless of whether $P$ is true. In particular, $K(\text{Al}, \text{Bill}) \rightarrow (\text{Bill} = \text{Bill})$ is true regardless of whether Al knows Bill. You need $K(\text{Al}, x) \text{ iff } (x = \text{Bill})$.


1

Starting from the formula : $¬((q \lor ¬p) \land ¬r)$, we have to use the tautological equivalence : $\lnot (\alpha \land \beta) \equiv (\alpha \to \lnot \beta)$ to get : $(q \lor ¬p) \to r$, and then use Material Implication : $(p \to q) \to r$. Same approach for the second one : $(p \lor ¬q) \lor (p \land (q \land ¬r))$. We have to replace $(...


1

Minimum-sized is asking you to find a structure with the smallest possible number of elements in which the sentence is not valid. Just because the relation is given the name $<$ doesn't mean that it has to have the usual properties of an ordering relation. Take a model with 2 elements $a$ and $b$ say and define $x < y$ to hold iff $x \neq y$. Then $a &...


1

Sure, you can 'ignore' the $\lor \neg q$ while working on the rest. You are likewise 'ignoring' the $\to s$. It's ok: boolean logic laws can be applied to component statement that, as such, ignore everything else about that statement, i.e.just leave the rest alone. Yes, you can treat the $p \land q$ as a single statement that, when distibuted over $r \lor \...


1

The goal is a conditional, so do a conditional proof: assume $B$, and try to get to $\neg C \to D$ Ok, but that new goal is itself a conditional, so do a second conditional proof inside the first one: assume $\neg C$, and try to got to $D$ Now, given the premise, it is clear that you can get to $D$ if only you can prove $B \to \neg C$. So, the new goal is ...


1

$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}$ Always keep your mind on the goal. You are attempting to deduce $\neg C$ using a subproof with an assumption of $C$. The goal of this subproof is a contradiction. The means of achieving that goal is using the third premise and $\vee$-elimination. $$\fitch{~~1.~A\to\neg C\\~~2.~\neg (B\wedge\...


1

The strategy here appears to be the deliberate introduction of a contradiction by assuming $(¬S∧−J)∧S$. As far as I can tell, one could use this strategy to derive $S→Q$ where $Q$ stands for any statement. Yes. That is a typical use of the Rule of Explosion. $$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{\neg S}{\fitch{S}{\neg S\...


1

I'll start from the end and answer your question - no, that's not the logical consequence. I will now explain why. Your theorem concerning the relation between entailmaent is correct. Let's write it again. $ T \models A \Leftrightarrow T\cup \{ \neg A\} $ is not satisfiable This is logically equivalent to: $ \neg ( T \models A ) \Leftrightarrow \neg ( T\...


1

Another trick: The Consensus Theorem says: $XY+X'Z=XY+X'Z+YZ$ which can be generalized to: Nested Consensus $WXY+WX'Z=WXY+WX'Z+WYZ$ Applying this to your statement: $$A'B'C+A'C'D'+AC'D+A'BC'$$ $$\overset{Consensus: AC'D+A'BC' = AC'D+A'BC'+BC'D}{=}$$ $$A'B'C+A'C'D'+AC'D+A'BC'+BC'D$$ $$\overset{Consensus: A'C'D'+BC'D = A'C'D'+BC'D+A'BC'}{=}$$ $$A'B'...


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