Questions about mathematical arguments I have a few questions about mathematical arguments
(1) Suppose that I want to prove that if the statements $A, B, C$ hold true, then $Z$ holds. To prove this, I would assume $A,B,C$, which then implies $D,E,F,G,H$, which in turn implies $I,J,K$ which in turn implies $\dots$, and so on. And I eventually get $Z$. But then, how do I know that $Z$ is actually true? That is, how do I know that some combinations of the statements, say $J,K,F$ which are obtained along the way,  do not contradict $Z$?
(2) I know that if assuming A yields a contradiction, then A is false. But, what if assuming A does not contradict anything? Can we conclude that $A$ is true? or what can we say about $A$?
 A: *

*If you assume $A,B,C$ and are able to use them to not only deduce $Z$ but also a contradiction to $Z$, then the assumptions you began with are inconsistent, and then you really have a problem, because literally anything can be deduced from a set of inconsistent hypotheses.  So what you seem to be asking is:  How do you know if a set of assumptions is consistent?  In general, this can be quite difficult; any proof of consistency must necessarily rest on some other set of assumptions, and you then would have to worry about whether those assumptions are consistent.  At the most fundamental level, the problem is fundamentally unsolvable; you might want to read up on Gödel's Incompleteness Theorems (the Wikipedia article is a good place to start).

*This question is in a sense the complement of the previous one.  If a set of assumptions does not lead to any contradictions, then the set is consistent.  That does not necessarily mean that it is true -- to answer that question you would have to decide what "true" means.  A good example of how this has historically worked is the development of non-Euclidean geometry. (Again, the Wikipedia article is a good place to begin.)

A: (1) Suppose that I want to prove that if the statements A, B,C hold true, then Z holds. 
So you want to prove this:
(A and B and C) ==>  z
Here is the truth table: 
(A and B and C) |  Z      | (A and B and C) ==>  Z
----------------------------------------------------
True            |  True   |  True
True            |  False  |  False
False           |  True   |  True
False           |  False  |  True

Notice that when (A and B and C) is False the implication is true either way. 
To prove this, I would assume A,B,C, which then implies D,E,F,G,H, which in turn implies I,J,K which in turn implies …, and so on. And I eventually get Z. 
So your proof shows that whenever (A and B and C) is true Z is true. It does not show "Z is true."
But then, how do I know that Z is actually true? 
You don't. Z could be false if (A and B and C) is false. But you still proved that (A and B and C) ==>  Z.
That is, how do I know that some combinations of the statements say J,K,F which are obtained along the way, do not contradict Z?
This is a different question. Since you are using those statements to show that Z is true, provided that your axioms or "first principles" are consistent you will not have a contradiction. 
(2) I know that if assuming A yields a contradiction, then A is false. But, what if assuming A does not contradict anything?
A must be either true or false. But, there is no guarantee that we can prove that A is true or false if our axioms are consistent. The ambiguity in this question comes from the word "anything" -do you mean anything we can write with the given axioms and symbols? 
If that is the case I don't know how you could prove that "A does not contradict anything" except by proving that it is true since (unless we are in a very boring simple system) we know there are some true statements with no proof within the system. In most example we don't know what these statements are, so how do we check if there is a contradiction.
If we instead show that A contradicts none of the axioms of the system that shows A is true, so ... 
Can we conclude that A is true? or what can we say about A?
I will say yes though this 2nd question was a bit deeper than I thought at first.
