How can we use a false statement to disprove itself? Suppose, we have statement A which satisfies: if A then -C, where -C is the negation of C. In addition, we have: if A and B, then C. Then, by using the truth table, I found that the combination of:if (if A then -C) and (if A and B, then C), then the negation of (A and B) is always true, I mean it is tautology. Now my question, how do I used a false statement (A and B), which I already prove it is wrong, in the logical induction to disprove itself? 
 A: Note the following theorem of classical logic:
$(A\to B)\wedge (A\to\neg B)\to \neg A$. It says that if from a proposition $A$ I can infer a contradiction (both $B$ and $\neg B$) then necessarily $A$ is not the case. Which is a standard method of proving negation of same claim in science. 
Using this theorem in your case you can prove that from (1) $A\to \neg C$ (which imply $(A\wedge B)\to \neg C$) and (2) ($A\wedge B)\to C$ the desired $\neg (A\wedge B)$ follows. Which does only say that whenever (1) and (2) are true $A\wedge B$ is false.
A: "(if A and B, then C)" does not at all imply "A and B", and so you did not use "A and B" to prove anything. If A and B, then C and not C. But actually (as you have proven) not ( A and B ). If you're wondering about the logical validity of your argument, it is simply because the deductive rules are truth-preserving, meaning that if you start with only true sentences then you deduce only true sentences in their respective contexts. Thus if you can ever deduce both a sentence and its negation in the same context (in this case you deduced "C and not C" under the assumption "A and B"), then the only possible situation is that the context never occurs (in this case you therefore can conclude "not ( A and B )").
I think your problem is that you don't understand contexts. I'll show you the proof of your example in a clear and intuitive style.

$A \to \neg C$.   [Premise 1; assumed true]
$A \land B \to C$   [Premise 2; assumed true]
If $A \land B$:   [This specifies a context, not a statement!]
  $C$.   [by Premise 2 and the context]
  $A$.   [by the context]
  $\neg C$.   [by Premise 1 and the previous line]
  $C \land \neg C$.   [This is a true statement in this context.]
$\neg( A \land B )$.   [The context where "$A \land B$" is true can never occur.]

Note that everything you deduce in any context is true in any subcontext (that is why we could 'drag' in the two premises into the context where "$A \land B$" is true, but not the other way around. In particular, all the things we deduced in the above proof in the context where "$A \land B$" is true may not be true in other contexts!
Also go through every step of the deduction and convince yourself that it is truth-preserving. For instance if we have deduced up to line 5, then "$A$" is true in the context where "$A \land B$" is true, and hence $\neg C$ must also be true in that context because of Premise 1. Therefore line 6 is valid.
Thus when we manage to prove "$C \land \neg C$" in the context where "$A \land B$" is true, we know by truth-preservation that "$C \land \neg C$" is a true sentence in the context where "$A \land B$" is true. Since we also know that "$C \land \neg C$" is always false in any context, we can conclude that "$A \land B$" can never be true. This also means that everything we proved under "If $A \land B$" is useless.
If you know programming this concept is trivial to understand, because when you have the code if(P){ ... } and P evaluates to false then the code inside the if-structure is never executed.
By the way, the style of proof shown above is a slight variant of Fitch-style natural deduction.
A: I am not sure if I have understood your question. However this is a possible explanation.
We have that
$$\mbox{$(A\wedge B)\Rightarrow A\Rightarrow \neg C \quad\mbox{and}\quad (A\wedge B)\Rightarrow C$},$$ 
which implies that $(A\wedge B )\Rightarrow (\neg C  \wedge  C)=$False, and therefore $(A\wedge B)$ is  False.
As a matter of fact, once we know that $(A\wedge B)$ is  False, we can also say that $(A\wedge B)\Rightarrow (\mbox{whatever you want})$!! This not a paradox: if the hypothesis is always false you can not deduce anything from it, and so you can put whatever you want in the thesis. 
