Question 1042178
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You have left something out.
<pre>
M -> B  |   ~M & ~B
</pre>

Doesn't necessarily follow.  If you have not B, then you can infer not M by Modus Tollens, but as it stands the conclusion is invalid.

In other words, the problem would have to read: "If the murder happened in the hotel room, then there are bloodstains somewhere in the room. <i><b>There are no bloodstains in the room</b></i>.  It follows that it is not the case that the murder happened in the hotel room and there are not bloodstains somewhere in the room."  Then your proof would look like:
<pre>
1.  M -> B  
2.  ~B          |   ~M & ~B

3.  ~M              1, 2 Modus Tollens
4.  :. ~M & ~B      2, 3 Conjuction Introduction
</pre>

John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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