TIFR CSE 2012 | Part A | Question: 2

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?

(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 )

• A. $Mr.M$ is guilty.
• B. $Mr.M$ is not guilty.
• C. From these facts one cannot conclude that $Mr.M$ is guilty.
• D. There is a witness who is lying.
• E. No witness is lying.

Answer 1

$\newcommand{m}{\text{Mr. } M}$If $\m$ is guilty, then if we pick a witness, we know that the witness won't lie unless he is afraid. If the witness is afraid, it may lie or it may lot lie (nothing is guaranteed).

However, unless we know what the victim said in the court (whether he said that $\m$ was guilty or not guilty), we can't say anything about $\m$.

All we know is that we've a witness who is afraid, so he may or may not lie in the court. We haven't been told anything about what actually happened in the court proceeding.

So, we can't logically conclude anything about $\m$ being guilty or not guilty.

Thus, options (a) and (b) are False.

Furthermore, that witness who was afraid, he may or may not lie. Since he is afraid, we know that he "can" lie, but we're not guaranteed that he will lie.

Thus, options (d) and (e) are False too.

This leaves option c, and as we have seen earlier, we cannot conclude anything about $\m$ being guilty or not guilty.

Hence, option (c) is the correct answer. 

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Although not necessary, the logic equivalent of the given statement will be:

$\Big[G\Rightarrow\neg\exists x:(W(x)\wedge L(x)\wedge\neg A(x))\Big]\equiv$

$\Big[G\Rightarrow\forall x:(W(x)\Rightarrow(\neg A(x)\Rightarrow\neg L(x)))\Big]$

Comments

I am getting the same. Why unless condition is checked after implication? It is like

If(M is guilty){

 if(witness is not afraid)

    No witness is lying

}

plz correct me. idea taken from:https://gateoverflow.in/812/gate2002-1-8

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 Expression which is given in answer is correct.

"$Q \; unless\; (\sim P)$" means "$P\rightarrow Q$" 

1) If Mr.M is guilty, then (no witness is lying) unless (he is afraid).

Here , (no witness is lying) unless (he is afraid) can be written as :-

$\forall x \; [(\sim A(x))\; \rightarrow [W(x)\rightarrow \;(\sim\;L(x)) ]]$

Now, "If Mr.M is guilty, then (no witness is lying) unless (he is afraid)." can be written as :-

$G \; \rightarrow \; [\forall x \; [(\sim A(x))\; \rightarrow [W(x)\rightarrow \;(\sim\;L(x)) ]]]$

Now, if we simplify it then it will be :-

$G \; \rightarrow \; [\forall x \; [( A(x))\; \vee (\sim W(x)) \vee \;(\sim\;L(x)) ]]$

$G \; \rightarrow \; [\forall x \; [ (\sim W(x))\vee[( A(x))\; \vee \;(\sim\;L(x))] ]]$

$G \; \rightarrow \; [\forall x \; [ (W(x))\rightarrow [(\sim ( A(x))\; \rightarrow \;(\sim\;L(x))] ]]$. It is same expression which is given in answer.

"If Mr.M is guilty, then (no witness is lying) unless (he is afraid)" can also be written as :-

$G \; \rightarrow \; [\forall x \; [ (W(x))\wedge [(\sim ( A(x))]\; \rightarrow \;(\sim\;L(x))]]$. It will also give the same expression after simplification.

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$\forall x \; [(\sim A(x))\; \rightarrow [W(x)\rightarrow \;(\sim\;L(x)) ]]$

@ankitgupta.1729

Sir, how you are interpretating this $[W(x)\rightarrow \;(\sim\;L(x)) ]]$ from the given statement.

$ \&$

$G \; \rightarrow \; [\forall x \; [ (W(x))\wedge [(\sim ( A(x))]\; \rightarrow \;(\sim\;L(x))]]$

How we are getting this ?

$ \&$

How these expression's might help us in concluding the answer ?

 

Answer 2

It is one those questions which are easier to answer if you use propositional and predicate logic.

Given the argument:

1. G ⟹ ∀x (W(x) ∧ ¬A(x) ⟹ ¬L(x))

2. ∃x (W(x) ∧ A(x))

Since statement 2 is true, W(x) ∧ ¬A(x) in statement 1 is false. So, W(x) ∧ ¬A(x) ⟹ ¬L(x) is always true for any x no matter L(x) is true or false for any value of x. Also, since ∀x (W(x) ∧ ¬A(x) ⟹ ¬L(x)) is true, therefore, statement 1 is true no matter G is true or false.

Hence, option C is correct.

Comments

Can you explain your statement

"statement 1 is true no matter G is true or false."

How is possible for statement 1 to be true irrespective of G.

p--> q  is true if q , p is true or p is false.

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I am getting the same. Why unless condition is checked after implication? It is like

If(M is guilty){

 if(witness is not afraid)

    No witness is lying

}

plz correct me. idea taken from:https://gateoverflow.in/812/gate2002-1-8

---

Nice Explanation!!

Answer 3

"((If Mr.M is guilty, then no witness is lying) unless he is afraid)" :
=G->¬∃x(W(x)∧L(x)) unless ∃x(A(x))
=¬∃x(A(x))->G->¬∃x(W(x)∧L(x))
=(¬∃x(A(x))∧G)->¬∃x(W(x)∧L(x))
=¬∃x((A(x))∧G)->(W(x)∧L(x)))
=¬∃x((A(x))∧G)->W(x)) ∧ ¬∃x((A(x))∧G)->L(x))

"There is a witness who is afraid" :
=∃x(W(x)∧A(x))
=∃xW(x) ∧ ∃xA(x)

¬∃x(((A(x))∧G)->W(x)) ...(1)
¬∃x(((A(x))∧G)->L(x))   ...(2)
∃xW(x)                                      ...(3)
∃xA(x)                                        ...(4)

using (1) and (3) we get
¬∃x((A(x))∧G))        ...(5)

No further simplication happen
Hence Ans is C

Answer 4

if p then q unless r =   $(p ∧¬r)⟹q$

If Mr. M is guilty then no witness is lying unless he is afraid.=  

 ∃x(G ∧¬A(x))⟹¬∃x:(W(x)∧L(x))

∧

There is a witness who is afraid.  =

∃x:(W(x)∧A(x))

Right or not??

Answer 5

We have a witness but unless he is afraid but witness is afraid this eliminates a  as well as b too similiarly d and e because we don't have exact information that witness lie or not so eventually we don't have enough information belong to this case so C would be correct