What is necessary to show that a theory $T$ is sound?

If I have some first-order language $L$ with just one relation symbol $\Re$, given an interpretation $\Im$ of the semantics $l$, how can I show that a theory $T$ with an axiom is sound?

More concretely, how do I show that a given theory $T$ with an axiom, say

$$\forall x \forall y(\Re xy \rightarrow \exists z(\Re xz \wedge \Re yz))$$ is sound for some interpretation $\Im$?

That is to say, I want to show that any theorem $\varphi$ of $T$ is true. So this requires that my axiom is true, and that my proof system is truth preserving.

Main Question

How can I show, given an interpretation $\Im$ of the semantics $\ell$ of a first-order language $L$, that a theory $T$ is sound?

It is a homework question for me to come up with an interpretation of the axiom I mentioned and then proof that it is sound, but I do not want an answer to my homework question, I want to solve that for my self. I just need some advice on how to go about proving that $T$ is sound by either showing the axiom is true and the proof system is truth preserving, or something else entirely.

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I'm confused about your confusion. Surely all you have to do here is check that your axiom is indeed validated by the interpretation? There's no need to check that the proof system is truth-preserving because you haven't introduced any new rules of inference. – Zhen Lin Jan 12 '12 at 23:56
So, you're telling me that all I need to do is show that for a given interpretation $\Im$ that my axiom is true? Then $T$ is sound? – Samuel Reid Jan 13 '12 at 0:26
Philosophically and intuitively, there is a difference between logical axioms and non-logical axioms (or postulates). The axioms of deductive systems are usually logical axioms: these are formulae that are (usually) sound for obvious reasons and cannot be falsified under any reasonable interpretation. For example, $$p \to p$$ is a logical axiom expressing the tautology ‘$p$ implies $p$’. In order to falsify it one must dream up an interpretation using a non-standard implication operator $\to$. On the other hand, axioms like the kind you give in your question are non-logical and are merely to narrow the scope of the discussion to systems of interest. That said, I suppose there is no real mathematical difference between logical and non-logical axioms...