User:Lord Farin/Sandbox/Equivalence of Definitions of Consistent

Theorem
Let $\mathcal L$ be a logical language.

Let $\mathscr P$ be a proof system for $\mathcal L$.

The following definitions of consistent are equivalent:

Proof
Let $\mathcal F$ be a set of logical formulas in $\mathcal L$.

Definition 1 implies Definition 2
Let $\mathcal F$ be consistent by definition 1.

Suppose $P$ derivable from $\mathcal F$.

Hence, by definition, $P$ is a tautology.

Then by Tautology is Negation of Contradiction, $\neg P$ is a contradiction.

So, by definition, $\neg P$ is not derivable from $\mathcal F$.

Similarly, suppose $\neg P$ is derivable from $\mathcal F$.

Hence, by definition, $\neg P$ is a tautology.

Then by Contradiction is Negation of Tautology, $P$ is a contradiction.

So, by definition, $P$ is not derivable from $\mathcal F$.

Thus either $P$ or $\neg P$ is not derivable from $\mathcal F$.

Hence $\mathcal F$ is consistent by definition 2.

Definition 2 implies Definition 3
Let $\mathcal F$ be consistent by definition 2.

Let $P$ be derivable from $\mathcal F$.

Then, by definition, $\neg P$ is not a derivable formula.

That is, there exists a formula that is not derivable from $\mathcal F$.

Hence $\mathcal F$ is consistent by definition 3.

Definition 3 implies Definition 2
Let $\mathcal F$ be consistent by definition 3.

that there exists $P$ such that both $P$ and $\neg P$ are derivable from $\mathcal F$.

Let $Q$ be a logical formula.

From the Rule of Explosion:
 * $\left({P \land \neg P}\right) \implies Q$

and thus it follows that $Q$ is a derivable formula.

Thus, whatever $Q$ is, it is derivable from $\mathcal F$.

That is, $\mathcal F$ is not consistent by definition 3.

This contradicts the hypothesis.

Hence $\mathcal F$ must be consistent by definition 2.