A proof by contradiction takes the form of if ~p then c, hence p. Or in other words, assume something is false, and if there is a contradiction that arises from this assumption, it must be true.
Can we also:
A) Assume something is true, and show by contradiction that it must be false
B) Assume something is true with a condition, for example, that a positive function is decreasing when its derivative is negative; then, show by lack of contradiction it is indeed true with said condition, outside of our assumption (a positive function is decreasing when its derivative is negative because the derivative is not positive). This might take the form of if p then ~c, hence p.
The first would be a rather elementary consequence of the second, but I am interested nonetheless. In particular, though, I'm curious as to whether B could be called circular reasoning.
A) Sure. Generally speaking, proof by contradiction can refer to the strategy where you assume something to be false (or true), and then show that we get some contradiction, and therefore that the original assumption had to be true (or false). For example: Suppose you want to prove a statement of the form p. Then you could suppose that p is false, and reach some contradiction c, so you conclude that p must be true. On the other hand, instead of supposing that p is false, you could also suppose that ~p is true, which is equivalent, so it is really just a rephrasing of things.
However, some people take the definition of “proof by contradiction” more strictly, e.g. that you can only start by assuming something is false. In a sense this is fine and equivalent because you can always rephrase the statement “assume that ~q is true” by “assume that q is false”. In other words, you can rephrase the supposition that something is true by the supposition that its negation is false (and note that ~(~p) is equivalent to p). I think the important and practical thing to understand is that it doesn’t really matter what you call it; so long as you know what you are doing and using sound logic, you can do what you like — the remaining details are then merely a matter of clear writing, conciseness, “proper form”, etc (not to say these aren’t important, but I’d say they usually come secondary to being logically sound).
Also, there are different systems of logic (like intuitionistic logic, as was mentioned in another answer), and other sorts of subtleties, but I’m not referring to those, nor am I knowledgeable about them. You can explore an entire world about logic and set theory.
B) It sounds to me that you want to assume what you want to prove; this is circular. Let p be the thing you want to show, and let Cond be the condition you are referring to. Then it seems you are saying the following: to prove that Cond implies p, you want to assume that Cond does indeed imply p, and then “explore the implications” of this assumption, and, if you never reach a contradiction (whatever that means), your assumption must have been justified, in other words, you have “proved” what you wanted. Correct?
Well, consider the following: Let Cond be the statement “$x$ is a non-negative real number”, and let p be the statement “$x^2$ is negative”. I’m guessing that you might reason as follows: suppose that non-negative real numbers have negative squares. I’ll explore the implications of this assumption, and if I never reach a contradiction, then indeed my claim must have been true. But, wait, I can come up with a counterexample: 2 is non-negative, and 2^2 = 4 is positive, not negative. So, I reached a “contradiction”, and so my initial claim was false.
The problem with this line of reasoning is: What if you never came across this “contradiction”? Sometimes the claim at hand is a lot more complicated, and we don’t have any immediate obvious examples that might disprove our claim or we don’t have any known true statements that would conflict with our claim. Then in such a case, you couldn’t find a “contradiction”, but that doesn’t mean that one doesn’t exist. Also, if your initial claim is false, but you suppose it is true, then you can deduce anything. For that matter, why not just say “p is true, therefore p is true”. Or perhaps more subtly, we could have the following scenario: We suppose that p (which is actually false) is true. Then, we are able to deduce that q (which is also false) is true. And we can keep on going. For a silly example, suppose our claim is that $-1 > 0$ (suppose we don't know whether this is true or not). Now assume that $-1 > 0$. Then $-1 + 2 > 0 + 2$, i.e. $1 > 2$. And so on. Now you might object by saying “but I happen to know that $1$ is not greater than $2$, so here is a 'contradiction' ”. But you are taking for granted that, at some point, we will come across known truths that will produce a contradiction with our line of reasoning. Sometimes this is not so easily done. In fact, this is the idea, and the caveat, behind proof by contradiction. If you want to prove p, you can suppose that it is false, and then at some point of our reasoning we come across some known truths that produce a contradiction/conflict with our reasoning, so we can conclude that p cannot be false, i.e. it is true. But if you don’t come across any such contradictions/conflicts, you have not proven p, because there could be other contradictions/conflicts lurking out there that we just haven’t come across yet.
So, I suspect that what you mean to show is that there are no counterexamples to our claim that Cond implies p. This is fine. But that’s the hard part — show that there are no counterexamples. If you start out with assuming that there are no counterexamples, then there is nothing left to prove.
Sometimes it helps to sit down with the logical definitions. The definition of p implies q is $\neg p \vee q$, where “$\vee$” is the “inclusive or”. Therefore the negation of p implies q is $p \wedge \neg q$, where “$\wedge$” is “and”. In other words, to disprove p implies q, you have to show that you can have p true but q false. Equivalently, to prove p implies q, you have to show that you never have p true but q false. To take the above example: to disprove that “$x$ is a non-negative real number” implies “$x^2$ is negative”, we need only produce one counterexample (in this case, let $x = 2$). But if we wanted to show it is true, we can do so by showing that no such counterexamples exist. If we didn’t know that $2^2$ was not negative, or for that matter if we didn’t know anything about whether squares of non-negative real numbers were negative or not, then we might never be able to produce a counterexample to our claim. But this doesn’t prove our claim. Just because you failed to come up with a counterexample to the claim doesn’t necessarily mean you’ve proved the claim, unless you can prove that you’ve considered all cases, and showed that your claim is true under all cases — one might call this proof by exhaustion or proof by cases. This is also valid, since in this case we’ve “considered our entire universe” and showed that our claim is always true.