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Six PLK secret agents are arrested and interrogated for several days. An agent tells the truth when all his statements are true, otherwise he lies. A report is drawn up containing the following conclusions:

  1. If agent $3$ tells the truth and agent $4$ lies then the agent $5$ tells the truth.
  2. If agent $2$ tells the truth then the same is true for agent $1$ or agent $3$.
  3. Agent $1$ and $2$ tell the truth when agent $3$ lies or agent $4$ tells the truth.
  4. If the agent $5$ tells the truth, the agent $6$ does the same.
  5. Agent $1$ or agent $2$ lie (edit).
  6. If agent $4$ lies and the agents $5$ and $6$ tell the truth, then the agent $1$ lies or the agent $3$ tells the truth.
  7. The agent $1$ tell the truth only if the agent $3$ or the agent $6$ lie.

a) Translate the conclusions with simple propositions $a_1, a_2, ..., a_6$ where $a_i$ is defined as 'the agent $i$ tell the truth' and logic connectors .

b) According to the report, who is telling the truth and who lies?

c) Should the interrogation be extended to at least one officer to determine whether his statements are all true or at least one of them is false? Explain.

Could anyone be able to give me a complete answer to this question?


EDIT: I have added the statement number 5.

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  • 1
    $\begingroup$ Did you create this puzzle yourself, or find it somewhere? If the latter, please edit your question to include a source. $\endgroup$ – Rand al'Thor Jun 6 '17 at 14:30
  • 5
    $\begingroup$ Is this homework for a Logic class? $\endgroup$ – Sconibulus Jun 6 '17 at 14:30
  • $\begingroup$ @Sconibulus Sorry, but it is not an homework. This is for personal knowing. $\endgroup$ – J.Doe Jun 6 '17 at 14:47
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a)

  1. $(a_3\land¬a_4)\Rightarrow a_5$
  2. $a_2\Rightarrow (a_1\lor a_3)$
  3. $(a_1\land a_2)\Leftarrow(a_4\lor¬a_3)$
  4. $a_5\Rightarrow a_6$
  5. $a_1\land a_2\Rightarrow\perp$
  6. $(a_5\land a_6\land¬a_4)\Rightarrow(a_3\lor¬a_1)$
  7. $a_1\Rightarrow¬(a_3\land a_6)$

(All symbols are standard logic notation as defined here: $\land$ = and, $\lor$ = or, ¬ = not, $\Rightarrow$ = implies, $\Leftarrow$ = is implied by, $\perp$ = falsehood/contradiction.)

b)

Statement 5 can be rewritten as $a_2\Rightarrow¬a_1$. Putting this together with statement 2, we find

$a_2\Rightarrow (¬a_1\land(a_1\lor a_3))\Rightarrow a_3.$

Putting statements 5 and 3 together gives

$(a_4\lor¬a_3)\Rightarrow(a_1\land a_2)\Rightarrow\perp$, i.e. $¬(a_4\lor¬a_3)$ is true, i.e. $a_3\land¬a_4$ is true.

In other words:

  • $a_3$ is true;

  • $a_4$ is false.

By modus ponens, this together with statement 1 implies that

  • $a_5$ is true.

By modus ponens again, this together with statement 4 implies that

  • $a_6$ is true.

Finally, the contrapositive of statement 7 is

$(a_3\land a_6)\Rightarrow¬a_1$.

Thus by modus ponens again,

  • $a_1$ is false.

Now we can run through all seven given statements and check that they're all valid; it seems that $a_2$'s truth or falsehood is unverifiable from the statements given.

c)

No, there don't seem to be any contradictions. Unless we were supposed to be able to work out the truth or falsehood of every $a_i$, in which case yes, some of the officers' statements must be wrong.

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  • $\begingroup$ I modified the question. It misses one statement. Sorry for that... $\endgroup$ – J.Doe Jun 6 '17 at 14:48
  • $\begingroup$ I can't answer the b). I got the same thing as you for a), but indeed b) is tricky. $\endgroup$ – J.Doe Jun 6 '17 at 14:50
  • $\begingroup$ @J.Doe Modified my answer accordingly ... but, given the new statement 5, what's the point in having statement 3 at all? $\endgroup$ – Rand al'Thor Jun 6 '17 at 14:50
  • $\begingroup$ Omg, I am so stupid. The statement 3 is 'Agent 1 and 2 tell the truth when agent 3 lies or agent 4 tells the truth.' and not 'if agent 1 and 2 tell the truth then agent 3 lies or agent 4 tells the truth.' Sorry I translate that from french. $\endgroup$ – J.Doe Jun 6 '17 at 14:52
  • $\begingroup$ @J.Doe OK, I think I've done b) and c). There could be a mistake somewhere though, so do check my logic. $\endgroup$ – Rand al'Thor Jun 6 '17 at 14:58

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