Mathematical logic problem on Mr. M

Question

Although this question is posted here,still i am posting this because the solution present there is copied from here and also i want to know flaw in my approach.

Question

If Mr.M is guilty, then no witness is lying unless he is afraid. There is a witness who is afraid. Which of the following statements is true?

1. Mr.M is guilty.
2. Mr.M is not guilty.
3. From these facts one cannot conclude that Mr.M is guilty.
4. There is a witness who is lying.
5. No witness is lying.

Hint: Formulate the problem using the following predicates

$G$ − Mr.M is guilty

$W(x)$ − $x$ is a witness

$L(x)$ − $x$ is lying

$A(x)$ − $x$ is afraid

My Approach

As we know that $p \rightarrow q \equiv \text{q unless p'}$

I can write the entire sentence into logical equation as

$(G\rightarrow (W(x) \wedge L(x))) \text{unless } A(x)$

$\equiv A(x) ^{'} \rightarrow (G\rightarrow (W(x) \wedge L(x)))$

Now it is given that $A(x)$ is true hence LHS becomes false and hene entire statement becomes true and hence it is independent of RHS .So option $3$ is true.

Am i doing right? please help

Answer 1

Your formulation lacks quantifiers. The given statements are

1. $G\rightarrow \bigg( \forall x\colon W(x) \rightarrow (\neg A(x) \rightarrow \neg L(x)) \bigg)$
2. $\exists x\colon W(x)\wedge A(x)$.

Using $P\to Q \equiv \neg P \vee Q$ we may write the first statement as $$G\rightarrow \bigg( \forall x\colon \neg W(x) \vee A(x) \vee \neg L(x) \bigg).$$

Define $Q \equiv \forall x\colon \neg W(x) \vee A(x) \vee \neg L(x)$, so that we have $G\to Q$. Given $G\to Q$ the only conclusion about $G$ we can make is that when $Q$ is false, $G$ has to be false (since $G\to Q\equiv \neg Q\to\neg G$). Hence, we can never conclude that $G$ is true from this. But we also can not show that $Q$ is false, since our only known person $x$ does not falsify $Q$. Hence, we can not determine the truth value of $G$.

Answer 2

If Mr.M is guilty, then no witness is lying unless he is afraid. There is a witness who is afraid.

  If Mr.M is guilty, then there does not exist a witness who is lying and not afraid.

$$G\to \neg\exists x~(W(x)\wedge L(x)\wedge\neg A(x))$$

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  There exists a witness who is afraid: $$\exists y~(W(y)\wedge A(y))$$

Answer 3

$(G\rightarrow (W(x) \wedge L(x))) \text{unless } A(x)$

No. You're parsing this incorrectly. The being afraid part is just about the witnesses, so the whole statement is a conditional with Mr.M being guilty as the antecedent, and the rest (the bit about the witnesses) as the consequent, so we get:

$G \rightarrow $ "no witness is lying unless he is afraid"

OK, so now we need to translate "no witness is lying unless he is afraid". Again, this is about witnesses, i.e. being a witness is the subject term (that what the claim is about), while the "lying unless afraid" is the predicate term, i.e. what we say about those witnesses. Finally, you need a universal quantifier in here, since this is about all witnesses. That is, we can paraphrase this sentence as:

"all witnesses are not lying unless afraid"

Hence, the general form is:

$\forall x (W(x) \rightarrow$ "$x$ is not lying unless $x$ is afraid")

And now you finally use the trick with the unless (I always think of "P unless Q" as "P if not Q"):

$G \rightarrow (W(x) \rightarrow (\neg A(x) \rightarrow \neg L(x)))$