Knight Knave puzzle with three boxes 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.
A: 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.
