Kenneth Rosen Edition 7 Exercise 1.2 Question 35 (Page No. 24)

Question

A detective has interviewed four witnesses to a crime. From the stories of the witnesses the detective has concluded that if the butler is telling the truth then so is the cook; the cook and the gardener cannot both be telling the truth; the gardener and the handyman are not both lying; and if the handyman is telling the truth then the cook is lying. For each of the four persons can the detective determine whether that person is telling the truth or lying ? Explain your reasoning.

Comments

@Arjun @Habib @Bikram can you please explain why the solution (Buttler:F, Cook:F , Gardner:T ,handyman: F) is not valid.

 

Original answer says G and H cannot be determined.

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@tusharp

It is valid.

'G' and 'H' can't be determined means we can't fix the truth values for 'G' and 'H'.

Possible assignment of truth values is :-

1) $G = T , H = F , C = F , B= F$

2) $G = F , H = T , C = F , B= F$

3) $G =T , H = T , C = F , B= F$

Answer 1

Figured out the answer myself.

I will use B,C,G,H

The problem says (~B + C).(~C + ~G).(G + H).(~H + ~C) using + for OR and . for AND

Taking B to be true we get :

C.(~C + ~G).(G + H).(~H + ~C) which gives C.~G.(G + H).~H which is a contradiction. So B is false. Taking B to be false the main problem reduces to :

(~C + ~G).(G + H).(~H + ~C)

Now let us take C to be true. Taking C to be true will give ~G.(G + H).~H which is a contradiction. So C has to be false.

At the end we are left with (G + H). So we cannot say whether handyman or gardener or both are telling the truth.

Comments

@Rounak can you explain C.(~C + ~G).(G + H).(~H + ~C) which gives C.~G.(G + H).~H which is a contradiction. So B is false. little in depth ?

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@Ankit Shah,

C.(~C + ~G).(G + H).(~H + ~C) has to be true. For it to be true, all the conjuncts (things separated by . operator in the expression) must be true. Thus we conclude C is true. If C is true, then ~C is false. If ~C is false, then (~C + ~G) is equal to ~G, because A OR false = A. Same argument for (~H + ~C) = ~H.

Now we have C.~G.(G + H).~H

If both ~G and ~H are true then (G + H) will be false OR false, which is false. But (G + H) has to be true because it is one of the conjuncts. Hence the contradiction.

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@ Rounak Agarwal, check this via truth table, though quite long

B	C	G	H	F1=B->C	F2=(C.G)'	F3=G+H	F4=H->C'	F=F1.F2.F3.F4
0	0	0	0	1	1	0	1	0
0	0	0	1	1	1	1	1	1
0	0	1	0	1	1	1	1	1
0	0	1	1	1	1	1	1	1
0	1	0	0	1	1	0	1	0
0	1	0	1	1	1	1	0	0
0	1	1	0	1	0	1	1	0
0	1	1	1	1	0	1	0	0
1	0	0	0	0	1	0	1	0
1	0	0	1	0	1	1	1	0
1	0	1	0	0	1	1	1	0
1	0	1	1	0	1	1	1	0
1	1	0	0	1	1	0	1	0
1	1	0	1	1	1	1	0	0
1	1	1	0	1	0	1	1	0
1	1	1	1	1	0	1	0	0

F=B'C'(G'H+GH'+GH)

F=B'C'(G+H)  ....So B & C are lying, either or both among G & H are telling the truth..

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we can also solve this using minimization,

After translates given english statement into logical statements. We get following expression,

(~B + C).(~C + ~G).(G + H).(~H + ~C)

=($\bar{B}\bar{C}+\bar{B}\bar{G}+C\bar{G}$).($G\bar{H}+G\bar{C}+H\bar{C}$) 

=($\bar{B}\bar{C}+C\bar{G}$).($G\bar{H}+H\bar{C}$)

=$\bar{B}\bar{C}G\bar{H}+\bar{B}H\bar{C}$

=$\bar{B}\bar{C}(G\bar{H}+H)$

=$\bar{B}\bar{C}(G+H)$

Now, we can make this expression equivalent to True if , 

B=F, C=F and atleast one of G and H is T. 

Hence, we can conclude B and C is lying and can't say about G and H that who's telling truth. it can be G or H or both. 

 

 

 

Answer 2

P1: Butler is telling truth
P2: Cook is telling truth
P3: Gardener is telling truth
P4: HandyMan is telling the truth

• if the butler is telling the truth then so is the cook;
  • P1 => P2 i.e ~P1 or P2 is true -(1)
•  the cook and the gardener cannot both be telling the truth;
  • ~(P2.P3)-(2)
• the gardener and the handyman are not both lying;
  • P3 or P4 is true - (3)
•  if the handyman is telling the truth then the cook is lying
  • P4 => ~P2 i.e ~p4 or ~p2 is true i.e ~(P4.P2) is true - (4)

  
When P1 is true:
- P2 must be true (By 1) -(5)
- P3 must be false (By 2) -(6)
- P4 must be false (By 4 and 5)
When P1 is false
- P2 can be true or false (By 1)

•  P1 is false P2 is true
  • P3 must be false - (By 2)
  • P4 must be false - (By 4)
  • But this means p3 or p4 is false. This will contradict (3). So this is not possible.
•  P1 is false P2 is false.
  • P3 or P4 must be true. (P3,P4 = TT,TF,FT)

So there are the following possible solutions
P1,P2,P3,P4 = {T,T,F,F} and {F,F,T,F}, {F,F,T,T}, {F,F,F,T}

Hence we can't uniquely determine the values of P1,P2,P3,P4

Comments

I have edited the answer. I made a mistake in translating one of the statement into proposition. Please verify

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You made an error in translating the second piece of information. It says "The cook and the gardener cannot both be telling the truth". You translated it to "Exactly one of cook and gardener is telling the truth". But it is also possible that both of them are lying. I had made the same error because of which I did not get the answer. Also, after incorrect translation, you used incorrect symbol as well. You have chosen P2 for cook and P3 for gardener, but you used P1 and P3 in symbolic representation.

The correct answer is that butler and cook are lying and we can say nothing about the other two.

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I have edited my answer. Please see if it is correct now

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Janaky, your edited answer is incorrect as well. We can in fact determine that butler and cook are lying.

I will use B,C,G,H

The problem says (~B + C).(~C + ~G).(G + H).(~H + ~C) using + for OR and . for AND

Taking B to be true we get :

C.(~C + ~G).(G + H).(~H + ~C) which gives C.~G.(G + H).~H which is a contradiction. So B is false. Taking B to be false the main problem reduces to :

(~C + ~G).(G + H).(~H + ~C)

Now let us take C to be true. Taking C to be true will give ~G.(G + H).~H which is a contradiction. So C has to be false.

At the end we are left with (G + H). So we cannot say whether handyman or gardener or both are telling the truth.