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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?

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  • $\begingroup$ In your answer you have forgotten the part that $x$ is a witness. $\endgroup$
    – Mauro ALLEGRANZA
    Jun 19, 2017 at 15:22
  • $\begingroup$ 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: $\forall x \ (W(x) \to \lnot L(x))$. $\endgroup$
    – Mauro ALLEGRANZA
    Jun 19, 2017 at 15:37
  • $\begingroup$ "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: $∀x \ [W(x)→(¬A(x) \to ¬L(x))]$. $\endgroup$
    – Mauro ALLEGRANZA
    Jun 19, 2017 at 15:39
  • $\begingroup$ "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 $¬∃x \ [W(x) \land ¬A(x) \land L(x))]$. $\endgroup$
    – Mauro ALLEGRANZA
    Jun 19, 2017 at 15:44
  • $\begingroup$ 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. $\endgroup$
    – momo
    Jun 19, 2017 at 16:08

1 Answer 1

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You made two mistakes:

  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

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  • $\begingroup$ I have this confusion regarding unless. According to your answer, 'x is not lying unless x is afraid' is in the form q=>p. But according to this article link the statement is in the form 'not p unless q' i.e. p => q. $\endgroup$
    – momo
    Jun 19, 2017 at 16:33
  • $\begingroup$ @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' = $\neg A(x) \to \neg L(x)$ = $L(x) \to A(x)$ $\endgroup$
    – Bram28
    Jun 19, 2017 at 16:56
  • $\begingroup$ oh okay! I understand now. Thanks a lot! $\endgroup$
    – momo
    Jun 20, 2017 at 0:19
  • $\begingroup$ @momo You're welcome. Glad I could help! :) $\endgroup$
    – Bram28
    Jun 20, 2017 at 0:45

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