0
$\begingroup$

Definition: If $A$ is a theory and $B \subseteq A$ then $B$ is a set of axioms for $A$ iff 1) B is recursive and 2) $B \models C$ for all $C \in A$. We say $A$ is axiomatizable iff $A$ has a set of axioms.

How can any theory be axiomatizable with this definition? If both theories prove the same set of sentences then wouldn't those sentences belong in both theories and $B = A$?

Definition: A theory $A$ is consistent iff it's not equal to the set of sentences in the Language of Arithmetic (with no free variables).

I thought that a consistent set is just one that doesn't have any contradictions. Does this definition relate to that? Thanks.

$\endgroup$
0
$\begingroup$

1) Note that a set of sentences is not the same as the set of sentences that can be derived from the given set oif sentences.

2) Once your theory contains any contradiction, you can prove anything (ex falso quodlibet - with mild assumptions about the allowed rules of inference)

$\endgroup$
  • $\begingroup$ Oh ok. I understand the first part, but I still don't understand why we can conclude a theory is consistent as long as it's not equal to the sentences in language of arithmetic? $\endgroup$ – user5306 Apr 21 '14 at 14:26
  • $\begingroup$ @user5306 - clearly, your def is "extrapolated" form some parte regarding first-order theories of arithmetic. As per Hagen's answer, if a theory is incons, then it proves anything. Thus, if there is at least one sentence which is not provable, then the theory is cons. The above def boils down to this fact : a theory (in the language of arithmetic) is cons iff cannot prove all sentences expressible in the language (of arithmetic). $\endgroup$ – Mauro ALLEGRANZA Apr 21 '14 at 16:36
  • $\begingroup$ @MauroALLEGRANZA thanks I understand now $\endgroup$ – user5306 Apr 22 '14 at 2:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.