Révision 212 CSL17/preliminaries.tex
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%\nb{TODO: extend with $\mu$s.} |
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\anupam{Should add $+(;x,y)$ as a basic function (check satisfies polychecking lemma!), since otherwise not definable with safe input. Then define unary successor $\succ{} (;x)$ as $+(;x,\succ 1 0)$. } |
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%\anupam{Should add $+(;x,y)$ as a basic function (check satisfies polychecking lemma!), since otherwise not definable with safe input. Then define unary successor $\succ{} (;x)$ as $+(;x,\succ 1 0)$. }
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\begin{example} |
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[Some simple functions] |
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\end{itemize} |
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\end{definition} |
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One then obtains: |
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\begin{theorem}[\cite{BellantoniThesis, Bellantoni95}] |
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\begin{theorem}[\cite{BellantoniThesis, Bellantoni95}]\label{thm:mubc}
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The class of functions representable by $\mubc$ programs is $\fph$, and for each $i\geq 1$ the class representable by $\mubc^{i-1}$ programs is $\fphi{i}$. |
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\end{theorem} |
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We will also need an intermediary lemma, also proved in \cite{BellantoniThesis}. |
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\begin{lemma}[Polychecking Lemma, \cite{BellantoniThesis}] |
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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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\end{lemma} |
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The following property follows from \cite{BellantoniThesis, Bellantoni95}] |
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\begin{proposition} |
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If $\Phi$ is a class of polymax bounded polynomial checking functions, then $\mubc(\Phi)$ satisfies the properties of Thm \ref{thm:mubc}. |
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\end{proposition} |
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Consider the addition function $+$ with two safe arguments, so denoted as $+(;x,y)$. It is clear that if $+(;x,y)=z$, then we have $|z| \leq \max(|x|,|y|)+1$, so $+$ is a polymax bounded function. Moreover for any $n$, the $n$th least significant bits of $z$ only depend on the $n$th least significant bits of $x$ and $y$, so $+(;x,y)$ is a polynomial checking function (on no normal argument), with threshold $q(x,y)=0$. |
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In the following we will consider $\mubc(\Phi)$ with $\Phi$ containing the function $+(;x,y)$, but write it simply as $\mubc$. Note that addition could be defined in $\bc$ (hence in $\mubc$), but not with two safe arguments because the definition would use safe recursion. We will also use the unary successor $\succ{} (;x)$, which is defined thanks to addition as as $\succ{} (;x)=+(;x,\succ 1 0)$. |
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