Okay, a truth tree tests for the consistency of an argument:

If we want to test for a tautology (something that is true in every case) you:

Take Argument (A)

Negate it ~(A)

Then run a truth tree, if it is closed completely, it is a tautology.

A contradiction is FALSE in every case...How could we use truth trees to prove that something is a contradiction?

My guess is that you first negate the argument at hand -- ~(A), run a truth tree, and if it is open at EVERY end, then it is a contradiction. I haven't given it too much thought, and I would like some help very soon if possible.

Thank You.

I would like to update this:

My new guess is that in order to test for a contradiction you:

Take original argument, (A) , do not negate it, and run a truth tree. If it is completely closed at the bottom, it is a contradiction.

Just need some confirmation on this.

Answering for first-order logic, where you have proof by negation, formally this goes something like this:

Given any sentence P in First Order Logic (atomic or complexily composed) there is another sentence NOT P (or ~P). This sentence is true only and only if P is false.

Once u commit yourself to the truth of ~P this is tantamount to committing yourself to the falsity of P.

Why would you do this? Well sometimes the negation of statements is less complex, yet equivalent so you'd want to use a less complex statement. Also with negation elimination/introduction one can 'undress' the First Order sentence, to it's atomic parts.

'Undressing' the formulas goes a great way in providing a formal proof of the sentence.

For a great introduction into first order logic and a playfull approach to negation, try to get your hands on Tarski's World.

Edit:
A great way to test for contradictions in statements is to prove the statement holds a contradictio in terminis in its parts.
So if the sentence is like: -A && B --> (C || D) && E you try by using logical rules (such as modus ponens) to get something like (~A && A). If (~A && A) can be derived from the First order logic sentence you have proven the statement to hold contradiction.
The tautology rule is an essential rule in first order logic. It simply says for every statement A you are allowed to posit A --> A (and A --> A --> A etc.)
So you are not testing for a tautology, you are even free to introduce them, so you can 'undress' the sentence further. A tautology does not change the truth value.

Also I don't remember using truth trees in first order logic, but for semantics and maybe lambda calculus. Using truth tables will make logical equivalence visable in a flash.

A tautology is an argument that only consists of a conclusion and no premises, that is necessarily true in virtue of logical laws. An argument with premises and conclusion that is necessarily true is just a sound argument.

You can test for a contradiction by making a truth table. It should tell you how to make these in most logic textbooks but I'll see if I can do it here anyway. To test a statement like (A & B) > ¬A, (where > is the material conditional), you do this:

A--B--(A & B)-->--¬A
T--T----T-----F----F
T--F----F-----T----F
F--T----F-----T----T
F--F----F-----T----T

You would fill in the > column last, which is the truth-conditions for (A & B) > ¬A. That sentence is always true except when both A and B are true. For a sentence to be a contradiction it needs to have an F in every row. Likewise to be a tautology it must have a T in every row. If you have 3 or more variables in your sentence, then you need a bigger truth table, with 8 rows in it (e.g. TTT, TTF, TFF etc.)

What is the symbol in front of the last A?

¬ = negation.

also written as ~ or NOT.