Conversion of a given statement into predicate logic?

Question

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

Hint: Formulate the problem using the following predicates:

$G$: Mr. A is guilty

$W(x)$: $x$ is a witness

$L(x)$: $x$ is lying

$A(x)$: $x$ is afraid

Which of the following statements is true?

• Mr.A is guilty.
• Mr.A is not guilty.
• From these facts one cannot conclude that Mr.A is guilty.
• There is a witness who is lying.
• No witness is lying.

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My try:

I formulated the problem as follows: $$ \neg A(x) \rightarrow \left ( G \rightarrow \neg\exists x \left ( W(x) \wedge L(x) \right ) \right ) $$ Then, I used the Exportation Law and Modus Tollens: $$ \left ( \neg A(x) \wedge G \right ) \rightarrow \left ( \neg \exists x \left ( W(x) \wedge L(x) \right ) \right ) $$ $$ \exists x\left ( W(x) \wedge L(x) \right ) $$ and I drew the conclusion $$ \neg A(x) \rightarrow \neg G $$ OR $$G \rightarrow A(x)$$

The same question has been asked before, but I need to confirm my approach for this problem. Is this correct?

Answer 1

Your conclusion is "If a person is not afraid, then $G$ is guilty" which just intuitively should sound wrong to you.

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I'd say you have two separate sentences. The first is

If Mr.A is guilty, then no witness is lying unless he is afraid

And this sentence starts with an "If". Now, "If $X$, then $Y$" sentences translate usually to $A\implies B$, so I suggest you do the same here. This means the logical statement will start with $G\implies$, not with $\neg A\implies$.

Now, the second part must be the "no witness is lying unless he is afraid". This can be rephrased either as

"if a witness is lying, he is afraid", so $(W(x)\land L(x))\implies A(x)$

or as

"if a witness is not afraid, he is not lying", so $(W(x)\land \neg A(x)\implies \neg L(x)$

(it's easy to see those two phrases are logically equivalent).

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So, the resulting statement is

$$G\implies \forall x:((W(x)\land L(x))\implies A(x))$$

with the second statement of course being $\exists x: W(x)\land A(x)$.

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Now, from jsut the fact that the only statement you have about $G$ is $$G\implies\text{something}$$ it should be clear that there is no way you can prove that $G$ is true (it could always be false, ant that statement will still be true).

However, you can maybe prove $\neg G$ since $A\implies B$ is the same as $\neg B$ implies $\neg A$.

In our case, this means we are interested in the statement

$$\neg(\forall x:((W(x)\land L(x))\implies A(x)))\implies \neg G$$

which simplifies to:

$$\exists x: W(x)\land L(x)\land \neg A(x) \implies \neg G$$

which means "if there exists a lying witness that is not afraid, then mr. A is guilty".

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However, you only know about one witness, and you don't know if they are lying. You know that if they are lying, then mr. A is not guilty, but if they are not lying, mr. A might be guilty or not, you just don't know.