Révision 260 CSL17/tech-report/preliminaries.tex

preliminaries.tex (revision 260)
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$\mubc =\fph$. Furthermore, for $i\geq 1$, $\mubc^{i-1} = \fphi{i}$.
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 \end{theorem}
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In what follows we will recall some of the intermediate results and state a slightly stronger result that directly follows from \cite{bellantoni1995fph}.


  		





  
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In what follows we will recall some of the intermediate results and state a slightly stronger result that directly follows from \cite{Bellantoni95}.


  		





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%\medskip 


  		





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%\noindent


  		





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  If $\Phi$ is a class of functions, we denote by $\mubc(\Phi)$ the class obtained as $\mubc$ but adding $\Phi$ to the set of initial functions.


  		





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 \begin{lemma}[Polychecking Lemma, \cite{BellantoniThesis}]


  		





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 	\label{lem:polychecking}


  		





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 Let $\Phi$ be a class of polymax bounded polynomial checking functions. If $f(\vec u; \vec x)$ is in $\mubc(\Phi)$, then $f$ is  a polymax bounded function  polynomial checking function on $\vec u$.


  		





  
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 Let $\Phi$ be a class of polymax bounded polynomial checking functions. If $f(\vec u; \vec x)$ is in $\mubc(\Phi)$, then $f$ is  a polymax bounded polynomial checking function on $\vec u$.


  		





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 \end{lemma}


  		





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In particular, we also have the following strengthening of Thm.~\ref{thm:mubc},


  		












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