# Carl Mummert

Marshall University

science.marshall.edu/mummertc

Age: 37

I work in mathematical logic. My main areas of interest are arithmetic, reverse mathematics, computability, and proof theory.
 52m revised For a compact logic, strong completeness follows from weak completenessadded 60 characters in body 1h comment For a compact logic, strong completeness follows from weak completenessI added a counterexample showing that the notion of compactness from the question is not sufficient. 1h revised For a compact logic, strong completeness follows from weak completenessadded 2191 characters in body 2h revised For a compact logic, strong completeness follows from weak completenessedited body 2h revised For a compact logic, strong completeness follows from weak completenessedited body 2h comment For a compact logic, strong completeness follows from weak completeness@mrp: the problem is that if we do not have $\emptyset \vdash \phi$ then the axiom just stated has no bearing on $\phi$. In general, we can't expect that every finite subset of will prove $\phi$. Put another way, if every finite subset of $\Phi$ proves $\phi$ then $\emptyset$ proves $\phi$, so $\Phi$ proves $\phi$ by monotonicity. So, unless every monotonic logic is compact we need a different form of compactness. 2h answered For a compact logic, strong completeness follows from weak completeness 2h comment For a compact logic, strong completeness follows from weak completenessI think that the statement of Definition 1 is not quite what you meant. Did you mean that, for all $\Phi \subseteq L$, if $\Phi' \vDash \phi$ for some finite $\Phi' \subseteq \Phi$ then $\Phi \vDash \phi$? I would imagine that compactness might be more useful if stated differently: if $\Phi \vDash \phi$ then there is some finite $\Phi' \subseteq \Phi$ with $\Phi' \vDash \phi$. 22h comment Hilbert style proof for $\left( \left( A\rightarrow \left( A\wedge \neg A\right) \right) \rightarrow \left( A\rightarrow A \right) \right)$@Doug Spoonwood: Thanks for the examples, but I am a little slow this morning - could you let me know what they are showing? Since you pinged me I think it must be something I wrote, but I can't remember the relationship. 1d comment Is it true that hausdorff and continuous lead to first separation axiomThe question as posed is indistinguishable from homework. Please edit the post to improve it by including more context: where did you encounter the problem? What have you tried? 1d comment Is there a bijection between $(0,1)$ and $(0,infinity)$This seems to be a duplicate of math.stackexchange.com/questions/160738/… 1d reviewed Approve suggested edit on Is there a bijection between $(0,1)$ and $(0,infinity)$ 1d answered Is there a bijection between $(0,1)$ and $(0,infinity)$ 1d comment Henkin semantics for Second-order Logic@Andrej I am not sure exactly what you are asking - is it about whether a model of $\mathsf{ACA}_0$ is a model of second-order PA? 1d revised Henkin semantics for Second-order Logicadded 7 characters in body 1d answered Henkin semantics for Second-order Logic 1d comment a problem about truth in first order logicWhat have you tried already? Where did the question arise? Please see the advice at meta.math.stackexchange.com/questions/9959/… and edit the post to make it better. 2d awarded computer-science 2d revised About the Words "recursion" and "recursive"edited tags 2d revised Is $(\neg q \rightarrow \neg p) \rightarrow (p \rightarrow q)$ equivalent to $p \vee \neg p$ in intuitionistic logic?added 88 characters in body