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Mar
3
asked Natural Decision Problem not in PTIME
Feb
13
comment Selecting a unique pair satisfying a condition $\varphi$ with an ordering
Okay, you're right. I tried to make this problem as general as possible. But it back-fired. $A$ is finite. Furthermore, we assume that there is always a pair which satisfies $\varphi$.
Feb
13
revised Selecting a unique pair satisfying a condition $\varphi$ with an ordering
Specialization
Feb
12
asked Selecting a unique pair satisfying a condition $\varphi$ with an ordering
Feb
7
accepted Transforming Nested Fixed-Point Formulas into Infinitary Logic Formulas with Finitely many Variables
Feb
5
answered Transforming Nested Fixed-Point Formulas into Infinitary Logic Formulas with Finitely many Variables
Feb
5
revised Transforming Nested Fixed-Point Formulas into Infinitary Logic Formulas with Finitely many Variables
correct the set
Feb
3
revised Transforming Nested Fixed-Point Formulas into Infinitary Logic Formulas with Finitely many Variables
correct
Feb
3
revised Transforming Nested Fixed-Point Formulas into Infinitary Logic Formulas with Finitely many Variables
correct formula
Feb
2
asked Transforming Nested Fixed-Point Formulas into Infinitary Logic Formulas with Finitely many Variables
Dec
10
awarded  Popular Question
Dec
10
comment Computable Criteria to check whether a given basis is a Gröbner Basis
I think what your describing is the cancellation criterion of Buchberger's algorithm. I was trying to avoid that computation and was looking for some kind of short cut to proof that the computation can be stopped at this point. I am starting to think that the answer is that there is no such short-cut.
Dec
3
comment How much topology for graph theory?
Exactly what I was looking for. Thanks!
Dec
3
accepted How much topology for graph theory?
Nov
27
comment How much topology for graph theory?
Where exactly are the notes?
Nov
27
asked How much topology for graph theory?
Nov
22
accepted Generated equivalence relations in logics
Nov
21
revised $\exists x Px \land \exists x Qx$ does not imply $\exists x (P x \land Q x)$
More explenation
Nov
21
answered $\exists x Px \land \exists x Qx$ does not imply $\exists x (P x \land Q x)$
Nov
21
revised $\exists x Px \land \exists x Qx$ does not imply $\exists x (P x \land Q x)$
spelling prettierer relations