# Formulate the problem using the given predicates.

Statement : If Mr.M is guilty, then no witness is lying unless he is afraid

Also given in question : There is a witness who is afraid.

Given predicates :

• G−Mr.M is guilty
• W(x)−x is a witness
• L(x)−x is lying
• A(x)−x is afraid

Answer according to me is : G⟹∀x(¬L(x)⟹¬A(x))

However according to this answer site , the answer is : \begin{array}{c} G \implies \lnot \exists x: \Bigl (W(x) \land L(x) \land \lnot A(x) \Bigr )\\[1em] \equiv\\[1em] G \implies \forall x: \Biggl (W(x) \implies \Bigl ( \lnot A(x) \implies \lnot L(x) \Bigr ) \Biggr ) \end{array}

If I am wrong, kindly explain where my logic went wrong?

• In your answer you have forgotten the part that $x$ is a witness. Jun 19, 2017 at 15:22
• Try step-by-step "no witness is lying unless he is afraid"; the part: ""no witness is lying" must be: $\lnot \exists x \ (W(x) \land L(x))$ that is equivalent to: $\forall x \ (W(x) \to \lnot L(x))$. Jun 19, 2017 at 15:37
• "Unless" is tricky: can be $\lor$. If so, we can write: $∀x \ [W(x)→(¬L(x) \lor A(x))]$ that in turn is equivalent to: $∀x \ [W(x)→(¬A(x) \to ¬L(x))]$. Jun 19, 2017 at 15:39
• "Going backward" we have the equivalent: $∀x [¬W(x) \lor A(x) \lor ¬L(x))]$ i.e. $∀x ¬[W(x) \land ¬A(x) \land L(x))]$ i.e $¬∃x \ [W(x) \land ¬A(x) \land L(x))]$. Jun 19, 2017 at 15:44
• It is the unless part that is confusing me,because I studied that p => q can be said as not p unless q and if applied to given statement, I think it is in this form.
– momo
Jun 19, 2017 at 16:08

1. You forgot to say that you are talking about witnesses. So, you need to add a $W(x)$ predicate
2. You need to say that for all witnesses $x$: '$x$ is not lying unless $x$ is afraid', which is equivalent to '$x$ is not lying if $x$ is not afraid' (whenever you see 'unless', just substitute 'of not'!) , which is symbolized as $\neg A(x) \to \neg L(x)$ You have $\neg L(x) \to \neg A(x)$, which is just the other way around
• @momo Yeah, 'unless' is always tricky! :) The article is right: 'not $p$ unless $q$' is $p \to q$. But I was right too: if you substitite 'if not' for 'unless', you get: 'not $p$ unless q' = 'not $p$ if not $q$' = $\neg q \to \neg p$ = $p \to q$. Likewise, 'x is not lying unless x is afraid' = 'x is not lying if not x is afraid' = $\neg A(x) \to \neg L(x)$ = $L(x) \to A(x)$ Jun 19, 2017 at 16:56