Trying to make my way through the book Gödel's Proof (Nagel & Newman, edited by Hofstadter).
In chapter V, the authors are showing that the axioms of sentential calculus are not contradictory. In short, they say (or seem to say):
Let's assume that the calculus is contradictory, such that $S$ and its contradiction $\sim S$ are both true. Then the theorem $p \supset ( \sim p \supset q )$ can be substituted with $S$ and $ \sim S$ to force $q$ to become true! Hence, if the calculus is contradictory, then any formula whatsoever is deducible from the axioms.
The way to show that the calculus is not contradictory is to find a formula that's not always true. In the text they use the example $p \vee q$. It's not always true, hence, it is shown that the calculus is consistent.
However... how can you assume that $p \vee q$ cannot be true? If the calculus is inconsistent, why couldn't there be a lengthy transformation that shows that $p \vee q$ is true?
It seems like a somewhat circular argument... what am I missing?