Could you please help me with the following puzzle:
Consider the following puzzle:
Suppose there are two box makers: Knight and Knave. Knight always writes true statements on his box, while Knave always writes false ones.
(ed: Each box was made by either a knight or a knave, and each one has a note written by its maker -- comments from MJD)
Suppose there are three boxes: A, B, and C. One of the box contains a bomb. The boxes have the following note:
A: There is a bomb in this box.
B: The bomb is not in this box.
C: At most one of these three boxes was made by Knight.
Suppose your task is to avoid choosing a box that contains bomb. Which one should you choose?
My conclusion is that we should choose box C. I derive the conclusion from:
1) Assume that the note in box C is correct.
It means there can only be one box that has correct note i.e. the box C itself. The two other boxes have incorrect notes which mean the bomb will be on box B.
2) Assume that the note in box C is wrong.
This means there will be two (or three boxes) that have correct note. But not all three boxes are correct, because we already assume box C has incorrect note. So, only box A and box B that have the correct note. In this case, it means the bomb is in box A.
So, for both case, the safe choice would be box C. Is this a correct logic reasoning in math?
PS: Additionally, is this a correct way to answer this question? Or is there a more formal way (mathematically)?
Thanks a lot for the help.
For comparison, here is "a more formal way" to write down your perfectly correct reasoning.
Call the three boxes $\;a,b,c\;$, and one of them has the bomb: call that one $\;d\;$ (for dynamite). Write $\;T(x)\;$ for "box $\;x\;$ was made by a Knight" ($\;T\;$ for Truth).
Using this notation, we are given that \begin{align} T(a) & \;\equiv\; a = d \tag{0} \\ T(b) & \;\equiv\; b \not= d \tag{1} \\ T(c) & \;\equiv\; \text{at most one of }T(a),T(b),T(c)\text{ is true} \tag{2} \\ \end{align}
Looking at the shape of these formulae, $(0)$ and $(1)$ can just be substituted into $(2)$, so we must work on simplifying $(2)$.
Now an expression of the form $\;\text{at most one of }P,Q,R\text{ is true}\;$ is not easy to manipulate. And because $\;T(c)\;$ occurs twice in $(2)$, it seems best to make a case distinction on $\;T(c)\;$.
If $\;T(c)\;$, then $(2)$ simplifies to \begin{align} & T(c) \;\equiv\; \text{at most one of }T(a),T(b),T(c)\text{ is true} \tag{2} \\ \equiv & \qquad \text{"using assumption $\;T(c)\;$"} \\ & \text{true} \;\equiv\; \text{at most one of }T(a),T(b),\text{true}\text{ is true} \\ \equiv & \qquad \text{"simplify"} \\ & \lnot T(a) \land \lnot T(b) \\ \equiv & \qquad \text{"using $(0)$ and $(1)$"} \\ & a \ne d \land b = d \\ \equiv & \qquad \text{"substitute $\;d:=b\;$ in left hand side; using $\;a \not= b\;$"} \\ & b = d \tag{3} \\ \end{align}
If $\;\lnot T(c)\;$, then $(2)$ simplifies to \begin{align} & T(c) \;\equiv\; \text{at most one of }T(a),T(b),T(c)\text{ is true} \tag{2} \\ \equiv & \qquad \text{"using assumption $\;\lnot T(c)\;$"} \\ & \text{false} \;\equiv\; \text{at most one of }T(a),T(b),\text{false}\text{ is true} \\ \equiv & \qquad \text{"simplify"} \\ & T(a) \land T(b) \\ \equiv & \qquad \text{"using $(0)$ and $(1)$"} \\ & a = d \land b \not= d \\ \equiv & \qquad \text{"substitute $\;d:=a\;$ in right hand side; using $\;b \not= a\;$"} \\ & a = d \tag{4} \\ \end{align} Now we see that both $(3)$ and $(4)$ imply $\;c \not= d\;$ (using $\;b \not= c\;$ and $\;a \not= c\;$, respectively). In other words, in both cases we can be sure that box $\;c\;$ does not contain the bomb.
The advantage of this style of doing proofs, is that often the shape of the formulae help to direct the search for the proof-- and this also helps in presenting the proof, so that it is easy to follow for the readers.