Gödel's Proof book - circular reasoning? 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?
Thanks!
 A: The truth of a formula in propositional calculus is defined through the semantics for the language, based on valuations.
A formula of propositional logic is called a tautology when it is true for every valuation, like : $p \lor \lnot p$.
A simple way to check if a formula of propositional logic is a tautology or not is to use the truth-table method.
For propositional logic, we can prove two meta-theorems :

(i) Soundness : if a formula $\varphi$ is provable in the propositional calculus, it is a tautology (i.e. if $\vdash \varphi$, then $\vDash \varphi$).
(ii) A logical calculus is consistent iff there is at least one formula $\varphi$ not provable in it (i.e. $\nvdash \varphi$).

Now, we can cook them together in the following way : $p \lor q$ is not a tautology, because with the valuation $v$ such that $v(p)=v(q)=$f we have that $v(p \lor q)=$f. Thus, having found a valuation that does not satisfy it, we conclude that the formula is not true for all valuations, and thus it is not a tautology.
Thus, by (i) (soundness), it is not provable in the calculus.
Having found a formula not provable in the propositional calculus, we conclude by (ii) that the calculus is consistent.
