proof by contradiction - making more than one assumption I have a pet peeve with a proof by contradiction. Such proofs begin by an assumption which is shown to lead to a contradiction thereby leading us to conclude that our initial assumption must have been wrong.
However, it often happens that we make more than one assumption in the course of the proof by contradiction. My question is: How can we know which assumption is responsible for the contradiction?
See amWhys answer in deducing $\lnot B \implies \lnot A$ from $A \implies B$ - she makes two assumptions: not B and A.
 A: If you assume $A$, $B$, $C$, and $D$ and manage to derive a contradiction, then you have a proof by contradiction of each of the following theorems (which are all logically equivalent):
$$ A\land B \land C \implies \neg D $$
$$ A\land B \land D \implies \neg C $$
$$ A\land C \land D \implies \neg B $$
$$ B\land C \land D \implies \neg A $$
Often it will only be one of these you're actually interested in, but you're free to choose to state any or all of them as a conclusion once you've done the work.
A: Usually, we structure a proof so that the assumption we want to show leads to a contradiction is the last assumption made prior to reaching the contradiction.
So we may at times nest assumptions, and the innermost assumption leading to the most immediate ensuing contradiction is the assumption we can then "throw out" (negate).
So, for example, often times we commit ourselves to certain assumptions (e.g., hypotheses, premises, or the antecedent of a conditional we want to prove, etc.), and then follow by "assuming for the sake of contradiction that $p$...". We then seek a contradiction that follows from having just assumed $p$: it may contradict an assumption/hypothesis which we had previously committed ourselves to, earlier in the proof. If/When we arrive at a contradiction, we can then use negation introduction to conclude $\lnot p$, and the assumption $p$ is then considered "discharged." $\lnot p$ then becomes a valid premise which can be used later in the proof. 
If you look at the structure of the proof you linked to, you'll see that assumptions are literally "nested," to help identify the logic of the proof, and also the scope of the two assumptions.
A: That's the way implication works, the only way to break
$$ A \implies B$$
is if you have
$$ \mbox{True} \implies \mbox{False}$$
So what do we do, we assume $\mbox{Not } B$ is true and $A$ is also true.  If this leads to a contradiction, then the implication is true.
