Propositional calculus, Post-completeness and beyond For a homework, our professor gave us the following problem: Prove that propositional calculus is Post-complete (term is not in English I am literally translating it), which means that if we add anything that isn't a theorem to propositional calculus as a scheme (or I think the phrase I need is axiom of propositional calculus) the system becomes inconsistent.
Now, this is my first post here, but I have been using this site quite a lot. My English isn't the greatest out there, but I hope that my question is understandable. Ofc, if prove requires use of natural deduction or Hilberts system, I think I am capable of following in both systems.
To be fair, actually I dont have a clue where to start, since I did have a whole lot of pause when it comes to logic, and Set Theory is what I find more appealing. :D 
 A: Let $U$ an unprovable schema of propositional calculus with $p_1, \ldots, p_k$ propositional variables.
By (semantic) completeness of the calculus, being unprovable implies that $U$ is not a tautology.
Thus, there is a valuation $v$ such that $v(U)=\bot$.
Add the schema $U$ as axiom, and consider its istance $B$ obtained replacing in $U$ each $p_i$ with $p \to p$, if $p_i$ is evaluated to $\top$ by the valuation $v$ above, and with $\lnot (p \to p)$ if $p_i$ is evaluated to $\bot$.
Obviously, the resulting formula $B$ is a theorem, being a substitution instance of an axiom of the "enlarged" system.
But $B$ is always false; therefore, $\lnot B$ is a tautology.
By completeness $\lnot B$ is provable in the enlarged system, and thus both $B$ and $\lnot B$ are theorems.
Conclusion: the enlarged system is inconsistent.

Added Feb,14
A different proof, derived from  David Hilbert and Wilhelm Ackermann's Grundzüge der theoretischen Logik, 1st German ed., 1928 [see the English translation of the 2nd ed, 1937, page 43].
Let $A$ be any formula which is not provable from the axioms. Let $B$ be the 
corresponding formula expressed in conjunctive normal form. Since $A$
is unprovable, also $B$ is.
Being unprovable, in $B$ we have at least a conjunct $C$ that is a disjunction containing no mutually contradictory literals.
If in $C$ we put $p$ for every un-negated sentential symbol and $\lnot p$ for every negated sentential symbol, we obtain a disjunction of the form $p \lor p \lor \ldots \lor p$, which is equivalent to $p$.
If now we add $A$ as a new axiom, we have that $B$ will be provable in the enlarged system, as well as $C$ and finally also $p$.
But then we could also substitute $\lnot p$ for $p$, obtaining a contradiction. 
