Equivalence of inconsistency definitions Let T be inconsistent if it exists sentences $\phi_1,...\phi_n$ such that $T\vdash \neg(\phi_1\wedge...\wedge\phi_n)$ can be deduced from T. 
Prove that the following statement are equivalent :
a) T is inconsistent
b) It exists a L-Sentence $\phi$ such that $T\vdash\phi$ & $T\vdash\neg\phi$
The path from b) to a) seems pretty straightforward using the Theorem of Deduction and Reduction to the Absurd but I have a hard time with the converse.
 A: If in this definition of inconsistency the $\phi_i's$ are just sentences and not theorems then I do not believe that this theorem of yours is true. Consider propositional calculus $L$. Then for some proposition $p$ let $\phi_1=p$ and $\phi_2=\lnot p$. Then as $\lnot p \land p$ is a contradiction it follows that $\lnot (p \land \lnot p)$ is a tautology, hence
$$L\vdash \lnot(\phi_1\land \phi_2).$$
However we know that $L$ is consistent in the traditional sense so there can't exist any $\phi$ such that
$$L\vdash \phi ~ \text{and} ~ L\vdash \lnot \phi.$$
So let us assume that the $\phi_i's$ are indeed theorems in $T$. So assume that for some natural $n$ there exists a set of theorems $\{\phi_i\}_{i\leq n}$ such that $T\vdash \lnot(\phi_i\land\dots\land\phi_n)$. Then using modus ponens and De
Morgan's law
$$\vdash \lnot \phi_1 \lor \dots \lor \lnot \phi_n.$$
From the definition of $\lor$ there must exist some $i, 1\leq i \leq n$ such that $T\vdash\lnot \phi_i$. This gives the result. Note I was not quite as formal as I should have been, but this gives the general form of the solution, which can be adapted to whatever level of formality you desire.
