0

If I have a set of axioms A = {A1, ..., An} and if I create a set of axioms B = {A1, ..., An, Con(A)}, would it be true to say that Con(A) iff Con(B)? Is there a simple counter-example to this?

More generally, if something X is true of A and we construct C = {A1, ..., An, X}, does Con(A) bi-imply Con(C)? Thanks.

  • What is $\mathrm{Con}(A)$? (Are you assuming your axioms interpret arithmetic?) – Andrés E. Caicedo Mar 05 '15 at 22:11
  • Anyway, this is false. Take $A$ to be $\mathsf{GB}+\lnot\mathrm{Con}(\mathsf{GB})$. Then $A$ is consistent but $B$ is not. – Andrés E. Caicedo Mar 05 '15 at 22:12
  • By Con(A) I mean "A is consistent." The axioms needn't be concerning arithmetic, though they could be. –  Mar 05 '15 at 22:17
  • Also, what is meant by GB vs. what is meant by B? –  Mar 05 '15 at 22:24
  • Without an assumption such as the possibility of interpreting some basic recursive processes, the question makes no sense, as $\mathrm{Con}(A)$ may not be expressible, so you need to be more precise on what your assumptions on $A$ are. – Andrés E. Caicedo Mar 05 '15 at 22:51
  • $\mathsf{GB}$ is Gödel-Bernays set theory. $B$ is as in your post. – Andrés E. Caicedo Mar 05 '15 at 22:51
  • I'm new to this stuff, so I'm not sure what kinds of assumptions you're talking about. I was thinking in terms of relations, constants, and FOL statements. Also, if A has the axioms of GB and as an additional axiom says that GB is inconsistent, doesn't that make A inconsistent? –  Mar 05 '15 at 23:48

0 Answers0