3. After excavating for weeks, you finally arrive at the burial chamber. The room is empty except for two large quests. On each is carved a message:
If this chest is empty, then the other chest's message is true.
This chest is filled with treasure or the other chest contains deadly scorpions.
You know exactly one of these messages is true. What should you do?
I attempted to formalize the two statements in the following way:
Define the constant individuals as
[imath]x_1[/imath]=:Chest 1, [imath]x_2[/imath]=: Chest 2, [imath]a[/imath]=: treasure, [imath]b[/imath]=: Deadly scorpions
Let [imath]R(x,y)[/imath] denote the binary relation: [imath]x[/imath] is a chest and [imath]y[/imath] is the content contained in [imath]x[/imath].
Then the first statement could be formalized as
[math]\neg \exist y(R(x_1,y)) \rightarrow R(x_2,a) \lor R(x_1,b)[/math] (1)
And the second statement as:
[math]R(x_2,a) \lor R(x_1,b)[/math] (2)
We have two cases to consider, either (1) is true and (2) false, or (1) is false and (2) is true.
In the first case
[math]\mid \neg \exist y(R(x_1,y)) \rightarrow R(x_2,a) \lor R(x_1,b) \mid=T[/math]
And
[math]\mid R(x_2,a) \lor R(x_1,b) \mid=F[/math]
This implies the truth value of the whole statement can be divided into two possible conditions.
[math]\mid \neg\exist y R(x_1,y)\mid =F\quad \text{or} \quad \mid \neg\exist y R(x_1,y)\mid =T [/math]
Case (i)
[math]\mid \neg\exist y R(x_1,y)\mid =F\quad \text{iff} \quad \exist y( (x_1,y)\in R)[/math]And
[math]\mid R(x_2,a) \lor R(x_1,b) \mid=F \quad \text{iff} \quad (x_2,a)\not\in R \quad \text{and} \quad (x_1,b)\not\in R[/math]
Case (ii)
[math]\mid \neg\exist y R(x_1,y)\mid =T \quad \text{iff} \quad \forall y( (x_1,y)\not\in R)[/math]And
[math]\mid R(x_2,a) \lor R(x_1,b) \mid=F \quad \text{iff} \quad (x_2,a)\not\in R \quad \text{and} \quad (x_1,b)\not\in R[/math]
The second case we have
[math]\mid \neg \exist y(R(x_1,y)) \rightarrow R(x_2,a) \lor R(x_1,b) \mid=F[/math]
and [math]\mid R(x_2,a) \lor R(x_1,b) \mid=T [/math]
However this is not possible, since an implication is only false if the antecedent is true the consequent is false. In this case, we would have both
[math]\mid R(x_2,a) \lor R(x_1,b) \mid=F \quad \text{and} \quad \mid R(x_2,a) \lor R(x_1,b) \mid=T[/math]
Which is impossible, hence no truth value would satisfy both statements at the same time. Hence we can conclude only cases (i) and (ii) are possible.
[math] \exist y(x_1,y)\in R \quad \text{and} \quad (x_2,a)\not\in R \quad \text{and} \quad (x_1,b)\not\in R [/math]or
[math] \forall y(x_1,y)\not\in R \quad \text{and} \quad (x_2,a)\not\in R \quad \text{and} \quad (x_1,b)\not\in R [/math]
In words, we can translate the conclusion from case (i) as
[imath]x_1[/imath] is nonempty, and [imath]x_2[/imath] does not contain treasure and [imath]x_1[/imath] does not contain scorpions.
And similarly from case (ii)
[imath]x_1[/imath] is empty and [imath]x_2[/imath] does not contain treasure and [imath]x_1[/imath] does not contain scorpions.
From this it is clear that the best choice to pick is [imath]x_1[/imath] since we have shown that [imath]x_2[/imath] does not contain treasure in either case, meanwhile, [imath]x_1[/imath] could be empty, or nonempty but it does not contain any scorpions. This is my reasoning although I am interested if I have made some logical mistake as I haven't studied logic in a long time.
If this chest is empty, then the other chest's message is true.
This chest is filled with treasure or the other chest contains deadly scorpions.
You know exactly one of these messages is true. What should you do?
I attempted to formalize the two statements in the following way:
Define the constant individuals as
[imath]x_1[/imath]=:Chest 1, [imath]x_2[/imath]=: Chest 2, [imath]a[/imath]=: treasure, [imath]b[/imath]=: Deadly scorpions
Let [imath]R(x,y)[/imath] denote the binary relation: [imath]x[/imath] is a chest and [imath]y[/imath] is the content contained in [imath]x[/imath].
Then the first statement could be formalized as
[math]\neg \exist y(R(x_1,y)) \rightarrow R(x_2,a) \lor R(x_1,b)[/math] (1)
And the second statement as:
[math]R(x_2,a) \lor R(x_1,b)[/math] (2)
We have two cases to consider, either (1) is true and (2) false, or (1) is false and (2) is true.
In the first case
[math]\mid \neg \exist y(R(x_1,y)) \rightarrow R(x_2,a) \lor R(x_1,b) \mid=T[/math]
And
[math]\mid R(x_2,a) \lor R(x_1,b) \mid=F[/math]
This implies the truth value of the whole statement can be divided into two possible conditions.
[math]\mid \neg\exist y R(x_1,y)\mid =F\quad \text{or} \quad \mid \neg\exist y R(x_1,y)\mid =T [/math]
Case (i)
[math]\mid \neg\exist y R(x_1,y)\mid =F\quad \text{iff} \quad \exist y( (x_1,y)\in R)[/math]And
[math]\mid R(x_2,a) \lor R(x_1,b) \mid=F \quad \text{iff} \quad (x_2,a)\not\in R \quad \text{and} \quad (x_1,b)\not\in R[/math]
Case (ii)
[math]\mid \neg\exist y R(x_1,y)\mid =T \quad \text{iff} \quad \forall y( (x_1,y)\not\in R)[/math]And
[math]\mid R(x_2,a) \lor R(x_1,b) \mid=F \quad \text{iff} \quad (x_2,a)\not\in R \quad \text{and} \quad (x_1,b)\not\in R[/math]
The second case we have
[math]\mid \neg \exist y(R(x_1,y)) \rightarrow R(x_2,a) \lor R(x_1,b) \mid=F[/math]
and [math]\mid R(x_2,a) \lor R(x_1,b) \mid=T [/math]
However this is not possible, since an implication is only false if the antecedent is true the consequent is false. In this case, we would have both
[math]\mid R(x_2,a) \lor R(x_1,b) \mid=F \quad \text{and} \quad \mid R(x_2,a) \lor R(x_1,b) \mid=T[/math]
Which is impossible, hence no truth value would satisfy both statements at the same time. Hence we can conclude only cases (i) and (ii) are possible.
[math] \exist y(x_1,y)\in R \quad \text{and} \quad (x_2,a)\not\in R \quad \text{and} \quad (x_1,b)\not\in R [/math]or
[math] \forall y(x_1,y)\not\in R \quad \text{and} \quad (x_2,a)\not\in R \quad \text{and} \quad (x_1,b)\not\in R [/math]
In words, we can translate the conclusion from case (i) as
[imath]x_1[/imath] is nonempty, and [imath]x_2[/imath] does not contain treasure and [imath]x_1[/imath] does not contain scorpions.
And similarly from case (ii)
[imath]x_1[/imath] is empty and [imath]x_2[/imath] does not contain treasure and [imath]x_1[/imath] does not contain scorpions.
From this it is clear that the best choice to pick is [imath]x_1[/imath] since we have shown that [imath]x_2[/imath] does not contain treasure in either case, meanwhile, [imath]x_1[/imath] could be empty, or nonempty but it does not contain any scorpions. This is my reasoning although I am interested if I have made some logical mistake as I haven't studied logic in a long time.