On the complexity of finding falsifying assignments for Herbrand disjunctions

Result code in 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>
Result on the web
<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>

Result language
angličtina
Original language name
On the complexity of finding falsifying assignments for Herbrand disjunctions
Original language description
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 $Phi$.
Czech name
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Czech description
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Type
J<sub>x</sub> - Unclassified - Peer-reviewed scientific article (Jimp, Jsc and Jost)
CEP classification
BA - General mathematics
OECD FORD branch
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Project
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Continuities
I - Institucionalni podpora na dlouhodoby koncepcni rozvoj vyzkumne organizace

Publication year
2015
Confidentiality
S - Úplné a pravdivé údaje o projektu nepodléhají ochraně podle zvláštních právních předpisů

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