On the complexity of finding falsifying assignments for Herbrand disjunctions

Kód výsledku v IS VaVaI

<a href="https://www.isvavai.cz/riv?ss=detail&h=RIV%2F67985840%3A_____%2F15%3A00449171" target="_blank" >RIV/67985840:_____/15:00449171 - isvavai.cz</a>

Výsledek na webu

<a href="http://dx.doi.org/10.1007/s00153-015-0439-6" target="_blank" >http://dx.doi.org/10.1007/s00153-015-0439-6</a>

DOI - Digital Object Identifier

<a href="http://dx.doi.org/10.1007/s00153-015-0439-6" target="_blank" >10.1007/s00153-015-0439-6</a>

Jazyk výsledku

angličtina

Název v původním jazyce

On the complexity of finding falsifying assignments for Herbrand disjunctions

Popis výsledku v původním jazyce

Suppose that $Phi$ is a consistent sentence. Then there is no Herbrand proof of $neg Phi$, which means that any Herbrand disjunction made from the prenex form of $neg Phi$ is falsifiable. We show that the problem of finding such a falsifying assignment is hard in the following sense. For every total polynomial search problem $R$, there exists a consistent $Phi$ such that finding solutions to $R$ can be reduced to finding a falsifying assignment to an Herbrand disjunction made from $neg Phi$. It has beenconjectured that there are no complete total polynomial search problems. If this conjecture is true, then for every consistent sentence $Phi$, there exists a consistent sentence $Psi$, such that the search problem associated with $Psi$ cannot be reducedto the search problem associated with $Phi$.

Název v anglickém jazyce

On the complexity of finding falsifying assignments for Herbrand disjunctions

Popis výsledku anglicky

Suppose that $Phi$ is a consistent sentence. Then there is no Herbrand proof of $neg Phi$, which means that any Herbrand disjunction made from the prenex form of $neg Phi$ is falsifiable. We show that the problem of finding such a falsifying assignment is hard in the following sense. For every total polynomial search problem $R$, there exists a consistent $Phi$ such that finding solutions to $R$ can be reduced to finding a falsifying assignment to an Herbrand disjunction made from $neg Phi$. It has beenconjectured that there are no complete total polynomial search problems. If this conjecture is true, then for every consistent sentence $Phi$, there exists a consistent sentence $Psi$, such that the search problem associated with $Psi$ cannot be reducedto the search problem associated with $Phi$.

Druh

J_x - Nezařazeno - Článek v odborném periodiku (Jimp, Jsc a Jost)

CEP obor

BA - Obecná matematika

OECD FORD obor

—

Projekt

—

Návaznosti

I - Institucionalni podpora na dlouhodoby koncepcni rozvoj vyzkumne organizace

Rok uplatnění

2015

Kód důvěrnosti údajů

S - Úplné a pravdivé údaje o projektu nepodléhají ochraně podle zvláštních právních předpisů

Název periodika

Archive for Mathematical Logic

ISSN

0933-5846

e-ISSN

—

Svazek periodika

54

Číslo periodika v rámci svazku

7

Stát vydavatele periodika

DE - Spolková republika Německo

Počet stran výsledku

15

Strana od-do

769-783

Kód UT WoS článku

000363535900002

EID výsledku v databázi Scopus

2-s2.0-84945478316