Statistiques
| Révision :

root / CSL17 / tech-report / completeness.tex @ 268

Historique | Voir | Annoter | Télécharger (16,52 ko)

1 251 adas
\section{Completeness}\label{sect:completeness}
2 251 adas
We now turn to proving the converse result to the last section, that our theories $\arith^i$ are strong enough to represent any $\fphi i $ function.
3 251 adas
The main result of this section is the following:
4 251 adas
5 251 adas
\begin{theorem}
6 251 adas
	\label{thm:completeness}
7 251 adas
	For every $\mubci{i-1}$ program $f(\vec u ; \vec x)$ (which is in $\fphi i$), there is a $\Sigma^{\safe}_i$ formula $A_f(\vec u, \vec x)$ such that $\arith^i$ proves $\forall^{\normal} \vec u, \forall^{\safe} \vec x, \exists^{\safe} ! y. A_f(\vec u , \vec x , y )$ and $\Nat \models \forall \vec u , \vec x. A_f(\vec u , \vec x , f(\vec u ; \vec x))$.
8 251 adas
\end{theorem}
9 267 adas
\begin{proof}[Proof sketch]
10 267 adas
 The proof proceeds by induction on the construction of $\mubci{i-1}$ program $f(\vec u ; \vec x)$ to construct a suitable formula $A_f(\vec u, \vec x)$ .  The case of initial functions is easy, by using the formulas we gave in Sect. \ref{sect:graphsbasicfunctions}. The case of safe composition is also simple, and makes use of the $\rais$ rule and simply combines previously defined graphs by \emph{modus ponens} (or, equivalently, $\cut$) and logical steps.
11 267 adas
 Naturally, the interesting cases are predicative recursion and predicative minimisation.
12 267 adas
13 267 adas
 For predicative recursion, the basic idea is to follow the classical simulation of primitive recursion in arithmetic, adapted to recursion on notation, by encoding the sequence of recursive calls as an integer $w$.
14 267 adas
 The main technical difficulty is the representation of (de)coding functions $\beta (k, x)$ (``the $k$th element of the sequence $x$'') and $\langle x,y \rangle$ (``the pair $(x,y)$'').
15 267 adas
 For this we represent instead the length-bounded variants of these functions given towards the end of Sect.~\ref{sect:prelims}, relying on the bounded minimisation and polychecking results, Lemmas~\ref{lem:bounded-minimisation} and \ref{lem:polychecking}, to find appropriate length bounds.
16 267 adas
%  which is sorted in a suitable way so that the formula $A_f$  is in the same class $\Sigma^{\safe}_i$ as $A_g$,  $A_{h_0}$ and $A_{h_1}$.  This can be obtained by defining  the predicate $\beta (k, w, y)$ with $k$ in $\normal$ and $w$, $y$ in $\safe$.
17 251 adas
18 267 adas
19 267 adas
 For predicative minimisation, assume $f$ is defined as  $\succ 1 \mu x .( g(\vec u ; \vec x , x) =_2 0)$ (or $0$ if no such $x$ exists).
20 267 adas
  By induction hypothesis we have a $\Sigma^{\safe}_{i-1}$ formula $A_g (\vec u , \vec x , x , y)$  satisfying the property. We then define $A_f(\vec u ; \vec x , z)$ as the following $\Sigma^{\safe}_{i}$ formula:
21 267 adas
\[
22 267 adas
\begin{array}{rl}
23 267 adas
&\left(
24 267 adas
z=0 \  \cand \ \forall x^\safe , y^\safe . (A_g (\vec u , \vec x , x, y) \cimp y=_2 1)
25 267 adas
\right) \\
26 267 adas
\cor & \left(
27 267 adas
\begin{array}{ll}
28 267 adas
z\neq 0
29 267 adas
& \cand\   \forall y^\safe . (A_g (\vec u , \vec x , p z , y) \cimp y=_2 0 ) \\
30 267 adas
& \cand\ \forall x^\safe < p z . \forall y^\safe . (A_g (\vec u , \vec x , x , y) \cimp y=_2 1)
31 267 adas
\end{array}
32 267 adas
\right)
33 267 adas
\end{array}
34 267 adas
\]
35 267 adas
As for the analogous bounded arithmetic theory $S^i_2$ (\cite{Buss86book}, Thm 20, p. 58),  $\arith^i$ can prove a \emph{minimisation} principle for $\Sigma^\safe_{i-1}$ formulas, allowing us to extract the \emph{least} witness of a safe existential; we prove this in the lemma below. We apply minimisation to $\exists x^\safe , y^\safe .  (A_g (\vec u ,\vec x , x , y) \cand y=_2 0)$. From here we are able to readily prove the totality of $A_f(\vec u ; \vec x , z)$ arguing by cases, notably relying crucially on classical reasoning.
36 267 adas
\end{proof}
37 267 adas
38 257 adas
	\begin{lemma}
39 257 adas
		[Minimisation]
40 267 adas
		For any $\Sigma^\safe_{i-1}$ formula $A$, $\arith^i \proves \exists x^\safe . A(x) \cimp \exists x^\safe . ( A(x) \cand \forall y^\safe \leq x . (A(y) \cimp y=x) ) $.
41 257 adas
	\end{lemma}
42 257 adas
	% see Thm 20 p. 58 in Buss' book.
43 257 adas
	%\begin{proof}
44 257 adas
	%\end{proof}
45 257 adas
	\begin{proof}
46 257 adas
		This Lemma is proved by using the same method as for the proof of the analogous result in the bounded arithmetic $S_2^{i}$ (see \cite{Buss86book}, Thm 20, p. 58).
47 257 adas
48 267 adas
		%Let $A(x)$ be a  $\Sigma^\safe_{i-1}$ formula, with $x$ in $\safe$.
49 267 adas
		We define the formula $B(a,b,c)$ as:
50 257 adas
		$$ \forall x^{\safe}. (x < |a| \moins b \cimp \zerobit(x,c)) \cand \forall y^{\safe}<c. \neg A(y) \cand \exists y^{\safe} < 2^{|a|\moins b}.A(c+y)$$
51 257 adas
		where $a$ is in $\normal$, $b$ in $\normal$ and $c$ in $\safe$.  $B(a,b,c)$ is in $\Sigma^\safe_{i}$.
52 257 adas
53 257 adas
		The intuitive idea of the proof is to observe that, if $A(a)$ is true for $a\neq 0$, then  $\exists x^{\safe}\leq a. B(a,b,x)$ holds for $b=0$, and to try to prove it for $b=|a|$, by using a length induction on $b$.
54 257 adas
55 257 adas
		First, one can prove:
56 257 adas
		$$ (A(a) \cand a \neq 0) \cimp B(a,0,0).$$
57 257 adas
		And thus:
58 257 adas
		$$ (A(a) \cand a \neq 0) \cimp \exists x^{\safe}\leq a .B(a,0,x).$$
59 257 adas
		We then can check that the two following statements are provable:
60 257 adas
		$$
61 257 adas
		\begin{array}{rcl}
62 257 adas
		(b<|a| \cand B(a,b,c)\cand \exists y^{\safe}<2^{|a| \moins (b+1)}.A(c+y)) &\cimp& B(a,\succ{} b,c)\\
63 257 adas
		(b<|a| \cand B(a,b,c)\cand \forall y^{\safe}<2^{|a| \moins(b+1)}.  A(c+y)) &\cimp & B(a,\succ{} b, c+2^{|a| \moins (b+1)})
64 257 adas
		\end{array}
65 257 adas
		$$
66 257 adas
		Moreover we have: $A(a) \cand B(a,b,c) \cimp c\leq a$.
67 257 adas
		From these three statements we deduce:
68 257 adas
		$$(A(a) \cand a \neq 0 \cand b<|a| \cand \exists x^{\safe} \leq a. B(a,b,x)) \cimp \exists x^{\safe } \leq a.B(a,\succ{} b,x).$$
69 257 adas
		Note that here we have used the fact that we are in classical logic.
70 257 adas
71 257 adas
		The formula $\exists x\leq a. B(a,b,c)$ is in $\Sigma^{\safe}_{i}$, so by $\Sigma^{\safe}_{i}$-LIND on the formula $\exists x\leq a. B(a,b,c)$, with the variable $b$ which is in $\normal$,  we obtain:
72 257 adas
		$$(A(a) \cand a \neq 0 ) \cimp \exists x^{\safe } \leq a.B(a,|a|,x).$$
73 257 adas
		But $B(a,|a|,x)$ implies $(\forall y^{\safe}<x. \neg A(y))\cand A(x)$, so we have proven:
74 257 adas
		$$(A(a) \cand a \neq 0 ) \cimp (\exists x^{\safe } \leq a. (\forall y^{\safe}<x. \neg A(y))\cand A(x))$$
75 267 adas
	%	which is the $\Sigma_{i-1}^{\safe}$ axiom for $A$.
76 267 adas
	as required.
77 257 adas
	\end{proof}
78 251 adas
79 267 adas
%\begin{proof}
80 267 adas
%	We proceed by structural induction on a $\mubc^{i-1} $ program, dealing with each case in the proceeding paragraphs.
81 267 adas
%
82 267 adas
%	The property is easily verified for the class of initial functions of  $\mubci{i-1}$: constant, projections, (binary) successors, predecessor, conditional, as already shown in Sect. \ref{sect:graphsbasicfunctions}. Now let us examine the three constructions: predicative minimisation, predicative (safe) recursion and composition.
83 267 adas
%	\paragraph*{Predicative minimisation}
84 267 adas
%	Suppose $f(\vec u ; \vec x)$ is defined by predicative minimisation from $g$ (we then denote $f$ as as $\mu x^{+1} . g(\vec u ; \vec x , x) =_2 0$).
85 267 adas
%	By definition $g$ is in $\mubci{i-2}$, and so by the inductive hypothesis there is a $\Sigma^{\safe}_{i-1}$ formula $A_g (\vec u , \vec x , x , y)$ computing the graph of $g$ such that,
86 267 adas
%	\[
87 267 adas
%	\arith^i \proves \forall \vec u^\normal . \forall \vec x^\safe , x^\safe . \exists ! y^\safe . A_g(\vec u , \vec x , x , y)
88 267 adas
%	\]
89 267 adas
%	Let us define $A_f(\vec u ; \vec x , z)$ as:
90 267 adas
%	\[
91 267 adas
%	\begin{array}{rl}
92 267 adas
%	&\left(
93 267 adas
%	z=0 \  \cand \ \forall x^\safe , y^\safe . (A_g (\vec u , \vec x , x, y) \cimp y=_2 1)
94 267 adas
%	\right) \\
95 267 adas
%	\cor & \left(
96 267 adas
%	\begin{array}{ll}
97 267 adas
%	%z\neq 0
98 267 adas
%	z=\succ{1} p z
99 267 adas
%	& \cand\   \forall y^\safe . (A_g (\vec u , \vec x , p z , y) \cimp y=_2 0 ) \\
100 267 adas
%	& \cand\ \forall x^\safe < p z . \forall y^\safe . (A_g (\vec u , \vec x , x , y) \cimp y=_2 1)
101 267 adas
%	\end{array}
102 267 adas
%	\right)
103 267 adas
%	\end{array}
104 267 adas
%	\]
105 267 adas
%	Notice that $A_f$ is $\Pi^{\safe}_{i-1}$, since $A_g$ is $\Sigma^{\safe}_{i-1}$ and occurs only in negative context above, with additional safe universal quantifiers occurring in positive context.
106 267 adas
%	In particular this means $A_f$ is $\Sigma^{\safe}_i$.
107 267 adas
%
108 267 adas
%	Now, to prove totality of $A_f$, we rely on $\Sigma^\safe_{i-1}$-minimisation, which is a consequence of $\cpind{\Sigma^\safe_i}$:
109 267 adas
%
110 267 adas
%	\begin{lemma}
111 267 adas
%		[Minimisation]
112 267 adas
%		$\arith^i \proves \cmin{\Sigma^\safe_{i-1}}$.
113 267 adas
%	\end{lemma}
114 267 adas
%	% see Thm 20 p. 58 in Buss' book.
115 267 adas
%	%\begin{proof}
116 267 adas
%	%\end{proof}
117 267 adas
%	\begin{proof}
118 267 adas
%		This Lemma is proved by using the same method as for the proof of the analogous result in the bounded arithmetic $S_2^{i}$ (see \cite{Buss86book}, Thm 20, p. 58).
119 267 adas
%
120 267 adas
%		Let $A(x)$ be a  $\Sigma^\safe_{i-1}$ formula, with $x$ in $\safe$. We define the formula $B(a,b,c)$ as:
121 267 adas
%		$$ \forall x^{\safe}. (x < |a| \moins b \cimp \zerobit(x,c)) \cand \forall y^{\safe}<c. \neg A(y) \cand \exists y^{\safe} < 2^{|a|\moins b}.A(c+y)$$
122 267 adas
%		where $a$ is in $\normal$, $b$ in $\normal$ and $c$ in $\safe$.  $B(a,b,c)$ is in $\Sigma^\safe_{i}$.
123 267 adas
%
124 267 adas
%		The intuitive idea of the proof is to observe that, if $A(a)$ is true for $a\neq 0$, then  $\exists x^{\safe}\leq a. B(a,b,x)$ holds for $b=0$, and to try to prove it for $b=|a|$, by using a length induction on $b$.
125 267 adas
%
126 267 adas
%		First, one can prove:
127 267 adas
%		$$ (A(a) \cand a \neq 0) \cimp B(a,0,0).$$
128 267 adas
%		And thus:
129 267 adas
%		$$ (A(a) \cand a \neq 0) \cimp \exists x^{\safe}\leq a .B(a,0,x).$$
130 267 adas
%		We then can check that the two following statements are provable:
131 267 adas
%		$$
132 267 adas
%		\begin{array}{rcl}
133 267 adas
%		(b<|a| \cand B(a,b,c)\cand \exists y^{\safe}<2^{|a| \moins (b+1)}.A(c+y)) &\cimp& B(a,\succ{} b,c)\\
134 267 adas
%		(b<|a| \cand B(a,b,c)\cand \forall y^{\safe}<2^{|a| \moins(b+1)}.  A(c+y)) &\cimp & B(a,\succ{} b, c+2^{|a| \moins (b+1)})
135 267 adas
%		\end{array}
136 267 adas
%		$$
137 267 adas
%		Moreover we have: $A(a) \cand B(a,b,c) \cimp c\leq a$.
138 267 adas
%		From these three statements we deduce:
139 267 adas
%		$$(A(a) \cand a \neq 0 \cand b<|a| \cand \exists x^{\safe} \leq a. B(a,b,x)) \cimp \exists x^{\safe } \leq a.B(a,\succ{} b,x).$$
140 267 adas
%		Note that here we have used the fact that we are in classical logic.
141 267 adas
%
142 267 adas
%		The formula $\exists x\leq a. B(a,b,c)$ is in $\Sigma^{\safe}_{i}$, so by $\Sigma^{\safe}_{i}$-LIND on the formula $\exists x\leq a. B(a,b,c)$, with the variable $b$ which is in $\normal$,  we obtain:
143 267 adas
%		$$(A(a) \cand a \neq 0 ) \cimp \exists x^{\safe } \leq a.B(a,|a|,x).$$
144 267 adas
%		But $B(a,|a|,x)$ implies $(\forall y^{\safe}<x. \neg A(y))\cand A(x)$, so we have proven:
145 267 adas
%		$$(A(a) \cand a \neq 0 ) \cimp (\exists x^{\safe } \leq a. (\forall y^{\safe}<x. \neg A(y))\cand A(x))$$
146 267 adas
%		which is the $\Sigma_{i-1}^{\safe}$ axiom for $A$.
147 267 adas
%	\end{proof}
148 267 adas
%
149 267 adas
%	Now, working in $\arith^i$, let $\vec u \in \normal , \vec x \in \safe$ and let us prove:
150 267 adas
%	\[
151 267 adas
%	\exists !z^\safe  . A_f(\vec u ; \vec x , z)
152 267 adas
%	\]
153 267 adas
%	Suppose that $\exists x^\safe , y^\safe .  (A_g (\vec u ,\vec x , x , y) \cand y=_2 0)$.
154 267 adas
%	We can apply minimisation due to the lemma above to find the least $x\in \safe$ such that $\exists y^\safe .  (A_g (\vec u ,\vec x , x , y) \cand y=_2 0)$, and we set $z = \succ 1 x$. So $x= p z$.
155 267 adas
%	%\todo{verify $z\neq 0$ disjunct.}
156 267 adas
%	%Then $z \neq 0$ holds.
157 267 adas
%	Then $z=\succ{1} p z$ holds.
158 267 adas
%	Moreover we had  $\exists ! y^\safe . A_g(\vec u , \vec x , x , y)$. So we deduce that
159 267 adas
%	$\forall y^\safe . (A_g (\vec u , \vec x , p z , y) \cimp y=_2 0 ) $. Finally, as $p z$ is the least element such that
160 267 adas
%	$\exists y^\safe .  (A_g (\vec u ,\vec x , p z , y) \cand y=_2 0)$, we deduce
161 267 adas
%	$\ \forall x^\safe < p z . \forall y^\safe . (A_g (\vec u , \vec x , x , y) \cimp y=_2 1) $. We conclude that the second member of the disjunction
162 267 adas
%	$A_f(\vec u ; \vec x , z)$ is proven.
163 267 adas
%
164 267 adas
%	Otherwise, we have that $\forall x^\safe , y^\safe . (A_g (\vec u , \vec x , x, y) \cimp y=_2 1)$, so we can set $z=0$ and the first member of the disjunction $A_f(\vec u ; \vec x , z)$ is proven.
165 267 adas
%
166 267 adas
%	So we have proven $\exists z^\safe  . A_f(\vec u ; \vec x , z)$, and unicity can be easily verified.
167 267 adas
%
168 267 adas
%	\paragraph*{Predicative recursion on notation}
169 267 adas
%
170 267 adas
%	\anupam{Assume access to the following predicates (makes completeness easier, soundness will be okay):
171 267 adas
%		\begin{itemize}
172 267 adas
%			%	\item $\pair x y z $ . ``$z$ is the sequence that appends $y$ to the sequence $x$''
173 267 adas
%			\item $\pair x y z$. ``$z$ is the sequence that prepends $x$ to the sequence $y$''
174 267 adas
%			\item $\beta (i; x ,y)$. ``The $i$th element of the sequence $x$ is $y$.''
175 267 adas
%		\end{itemize}
176 267 adas
%	}
177 267 adas
%	\patrick{I also assume access to the following predicates:
178 267 adas
%		\begin{itemize}
179 267 adas
%			%   \item $\zerobit (u,k)$ (resp. $\onebit(u,k)$). " The $k$th bit of $u$ is 0 (resp. 1)"
180 267 adas
%			%   \item $\pref(k,x,y)$. "The prefix of $x$ (as a binary string) of length $k$ is $y$"
181 267 adas
%			\item $\addtosequence(w,y,w')$. "$w'$ represents the sequence obtained by adding $y$ at the end of the sequence represented by $w$". Here we are referring to sequences which can be decoded with predicate $\beta$.
182 267 adas
%		\end{itemize}}
183 267 adas
%		In the following we will use the predicates $\zerobit (u,k)$, $\onebit(u,k)$, $\pref(k,x,y)$ which have been defined in Sect. \ref{sect:graphsbasicfunctions}.
184 267 adas
%
185 267 adas
%		Suppose that $f$ is defined by predicative recursion on notation:
186 267 adas
%		\[
187 267 adas
%		\begin{array}{rcl}
188 267 adas
%		f(0 , \vec u ; \vec x) & \dfn & g(\vec u ; \vec x) \\
189 267 adas
%		f(\succ i u, \vec u ; \vec x) & \dfn & h_i( u , \vec u ; \vec x , f(u , \vec u ; \vec x))
190 267 adas
%		\end{array}
191 267 adas
%		\]
192 267 adas
%
193 267 adas
%		\anupam{using $\beta(i,x,y)$ predicate for sequences: ``$i$th element of $x$ is $y$''. Provably total in $\arith^1$.}
194 267 adas
%
195 267 adas
%		Suppose we have $\Sigma^\safe_i$ formulae $A_g (\vec u ; \vec x,y)$ and $A_{h_i} (u , \vec u ; \vec x , y , z)$ computing the graphs of $g$ and $h_i$ respectively, provably total in $\arith^i$.
196 267 adas
%		We define $A_f (u ,\vec u ; \vec x , y)$ as,
197 267 adas
%		\[
198 267 adas
%		\exists w^\safe . \left(
199 267 adas
%		\begin{array}{ll}
200 267 adas
%		&
201 267 adas
%		%Seq(z) \cand
202 267 adas
%		\exists^{\safe} y_0 . ( A_g (\vec u , \vec x , y_0) \cand \beta(0, w , y_0) ) \cand \beta(|u|, w,y ) \\
203 267 adas
%		%\cand & \forall k < |u| . \exists y_k , y_{k+1} . ( \beta (k, w, y_k) \cand \beta (k+1 ,w, y_{k+1})  \cand A_{h_i} (u , \vec u ; \vec x , y_k , y_{k+1}) )\\
204 267 adas
%		\cand & \forall^{\normal}  k < |u| . \exists^{\safe} y_k , y_{k+1} . ( \beta (k, w, y_k) \cand \beta (k+1 ,w, y_{k+1})  \cand B (u , \vec u ; \vec x , y_k , y_{k+1}) )
205 267 adas
%		\end{array}
206 267 adas
%		\right)
207 267 adas
%		\]
208 267 adas
%		where
209 267 adas
%		\[
210 267 adas
%		B (u , \vec u ; \vec x , y_k , y_{k+1}) = \left(
211 267 adas
%		\begin{array}{ll}
212 267 adas
%		& \zerobit(u,k+1) \cimp  \exists v .(\pref(k,u,v)  \cand A_{h_0}(v,\vec u ; \vec x, y_k, y_{k+1}) )\\
213 267 adas
%		\cand &  \onebit(u,k+1) \cimp  \exists v .(\pref(k,u,v)  \cand A_{h_1}(v,\vec u ; \vec x, y_k, y_{k+1}) )
214 267 adas
%		\end{array}
215 267 adas
%		\right)
216 267 adas
%		\]
217 267 adas
%
218 267 adas
%		%which is $\Sigma^\safe_i$ by inspection, and indeed defines the graph of $f$.
219 267 adas
%
220 267 adas
%		To show totality, let $\vec u \in \normal, \vec x \in \safe$ and proceed by induction on $u \in \normal$.
221 267 adas
%		The base case, when $u=0$, is immediate from the totality of $A_g$, so for the inductive case we need to show:
222 267 adas
%		\[
223 267 adas
%		\exists y^\safe . A_f (u , \vec u ; \vec x , y)
224 267 adas
%		\quad \seqar \quad
225 267 adas
%		\exists z^\safe . A_f (s_i u, \vec u ; \vec x , z)
226 267 adas
%		\]
227 267 adas
%
228 267 adas
%		So let us assume $\exists y^\safe . A_f (u , \vec u ; \vec x , y) $. Then there exists $w$ such that $\safe(w)$ and
229 267 adas
%		$A_f (u , \vec u ; \vec x , w) $.
230 267 adas
%
231 267 adas
%		We know that there exists a $z$ such that $A_{h_i}(u,\vec u ; \vec x, y, z)$. Let then $w'$ be such that
232 267 adas
%		$\addtosequence(w,z,w')$. Observe also that we have $\pref(|u|,s_i u,u)$
233 267 adas
%
234 267 adas
%		So we have, for $k=|u|$:
235 267 adas
%
236 267 adas
%		$$  \beta (k, w', y) \cand \beta (k+1 ,w', z)  \cand B (u , \vec u ; \vec x , y , z).$$
237 267 adas
%
238 267 adas
%		and we can conclude that
239 267 adas
%		\[
240 267 adas
%		\exists w'^\safe . \left(
241 267 adas
%		\begin{array}{ll}
242 267 adas
%		&
243 267 adas
%		%Seq(z) \cand
244 267 adas
%		\exists y_0 . ( A_g (\vec u , \vec x , y_0) \cand \beta(0, w' , y_0) ) \cand \beta(|u|+1, w',z ) \\
245 267 adas
%		\cand & \forall k < |u|+1 . \exists y_k , y_{k+1} . ( \beta (k, w, y_k) \cand \beta (k+1 ,w, y_{k+1})  \cand B (u , \vec u ; \vec x , y_k , y_{k+1}) )
246 267 adas
%		\end{array}
247 267 adas
%		\right)
248 267 adas
%		\]
249 267 adas
%		So $\exists z^{\safe} . A_f (s_i u, \vec u ; \vec x , z)$ has been proven. So by induction we have proven $\forall^{\normal} u,
250 267 adas
%		\forall^{\normal} \vec u, \exists^{\safe} y. A_f (u ,\vec u ; \vec x , y)$. Moreover one can also check the unicity of $y$, and so we have proved the required formula.
251 267 adas
%
252 267 adas
%		\anupam{here need to `add' element to the computation sequence. Need to do this earlier in the paper.}
253 267 adas
%
254 267 adas
%		\anupam{for inductive cases, need $u\neq 0$ for $\succ 0$ case.}
255 267 adas
%
256 267 adas
%		\paragraph*{Safe composition}
257 267 adas
%		Now suppose that $f$ is defined by safe composition:
258 267 adas
%		\[
259 267 adas
%		f(\vec u ; \vec x) \quad \dfn \quad g( \vec h(\vec u;) ; \vec h' (\vec u ; \vec x) )
260 267 adas
%		\]
261 267 adas
%
262 267 adas
%		By the inductive hypothesis, let us suppose that we have $\Sigma^\safe_i $ definitions $A_g , A_{h_i} , A_{h_j'} $ of the graphs of $g , h_i , h_j'$ respectively, which are provably total etc.
263 267 adas
%		In particular, by Raising, we have that $\forall \vec u^\normal . \exists v^\normal . A_{h_i} (\vec u , v)$.
264 267 adas
%
265 267 adas
%		We define $A_f (\vec u , \vec x , z)$ defining the graph of $f$ as follows:
266 267 adas
%		\[
267 267 adas
%		\exists \vec v^\normal . \exists \vec y^\safe .
268 267 adas
%		\left(
269 267 adas
%		\bigwedge\limits_i A_{h_i} (\vec u , v_i)
270 267 adas
%		\wedge
271 267 adas
%		\bigwedge\limits_j A_{h_j'} (\vec u ; \vec x , y_j)
272 267 adas
%		\wedge
273 267 adas
%		A_g ( \vec v , \vec y , z )
274 267 adas
%		\right)
275 267 adas
%		\]
276 267 adas
%		The provable totality of $A_f$ follows from simple first-order reasoning, mostly cuts and basic quantifier manipulations.
277 267 adas
%
278 267 adas
%		\todo{elaborate}
279 267 adas
%
280 267 adas
%		The proof of Thm \ref{thm:completeness} is thus completed.
281 267 adas
%\end{proof}
282 267 adas
%
283 267 adas
%
284 251 adas
285 251 adas
286 251 adas
287 251 adas
288 251 adas
289 251 adas
290 251 adas
291 251 adas
292 251 adas
293 251 adas
294 251 adas
295 251 adas
296 251 adas
297 251 adas
298 251 adas
299 251 adas
300 251 adas
301 251 adas