Trying to wrap my head around the idea of Proving Rule of Cases is a valid arument I had a question on my assignment today that asked to "Prove that the Rule of Proof by Cases is a valid argument."  Based of what I've read, Proof by Cases is valid when all cases produce the same outcome.  So for example x^2 >=x.  In cases where x=0, x>0, x<0, all cases are true.  But what of the equation x^3 >= x.  Only 2 cases work out where the statement is true, and that is when x=0 or x>0.  This statement is false when x<0.
So does this mean that Proof of Cases does not work with this equation?  Does this mean that proof is cases is not a valid argument because I was able to find an example where not all cases had the same outcome?  Does this mean when I create my own examples, I have to make sure that all the outcomes are the same? 
 A: First off, you are mistaken in you statement, $x^2\geq x$ is not true in general: it fails when $0<x<1$, for instance for $x=0.5$ where $x^2=0.25$. If you write $x^2-x=x(x-1)$, it becomes clear that the most natural cases to consider are $x\leq0$, $x\geq1$, and $0<x<1$, and as I said one cannot prove the statement in the final case.
One can prove any statement by considering several cases and proving it separately in each case, provided that they cover all possible cases (in any situation where the hypotheses of the statement are true, at least one of the specified cases holds), and of course that one can actually prove the conclusion in each of the cases considered. If you fail to prove the statement in one (or more) of the cases, this is an indication that the statement might not be true in general; after all, you can use for this case all the original hypotheses plus the assumption that the case in question applies, and the task is therefore not harder than proving what was to be proved originally.
This being said, one would only want to consider cases if this is somehow useful in the argument, since if the proof goes essentially the same way in each case, one has just multiplied the work. So only distinguish cases when there are two separate routes to the desired conclusion, but you cannot know which one applies. For instance $x^2\geq0$ can be proved separately for $x\geq0$ and for $x\leq0$ (which conditions cover all possibilities), but the proofs follow separate trails. This also show that (1) cases may overlap, and (2) one may avoid naming separately cases (here $x=0$) whose proof is subsumed by at least one of the other proofs (here in fact by both).
