root / CSL16 / free-cut-elim.tex @ 79
Historique | Voir | Annoter | Télécharger (17,92 ko)
1 |
\section{Free-cut elimination in linear logic} |
---|---|
2 |
|
3 |
|
4 |
\todo{patrick} |
5 |
|
6 |
\anupam{Possibly need to include the cases for affine logic too, e.g.\ if weakening is native, since it does not play nice as an axiom.} |
7 |
|
8 |
\todo{include some bounds on free-cut elimination? Look at paper "Corrected upper bounds for free-cut elimination" by Beckmann and Buss for comparison.} |
9 |
|
10 |
While in plain logical systems such as linear logic cut rules can be eliminated, this is in general not the case anymore when one considers extension with a theory $\mathcal T$ . For this reason we need now to define the kind of cuts that will remain in proofs after reduction. We will call these \textit{anchored cuts}. |
11 |
|
12 |
% \begin{definition}\label{def:anchoredcut} |
13 |
% An instance of cut rule in a $\mathcal T$-proof is an \textit{anchored cut} if its cut-formulas $A$ in the two premises are both principal for their rule, and at least one of these rules is non-logical. |
14 |
% \end{definition} |
15 |
Our first idea would be to consider as anchored a cut whose cut-formulas $A$ in the two premises are both principal for their rule, and at least one of these rules is non-logical. Now, the problem with this tentative definition is that a rule (R) of $\mathcal T$ can contain several principal formulas (in $\Sigma'$, $\Delta'$) and so we would like to allow an anchored cut on each of these principal formulas. See as example Fig. \ref{Fig:twosuccessiveanchorcuts}. Therefore we consider a more general definition: |
16 |
% \begin{figure} |
17 |
% \[ |
18 |
% \vlderivation{ |
19 |
% \vliinf{c1}{!\Gamma , \Gamma_1 \seqar A(t) , ?\Delta }{ |
20 |
% \vlinf{(ind)}{(x \notin \FV(\Gamma, \Delta))}{ !\Gamma , A(\epsilon) \seqar A(t) , ?\Delta }{ !\Gamma , A(x) \seqar A(s x ), ?\Delta }} |
21 |
% {\Gamma_1\seqar A(\epsilon) } |
22 |
% } |
23 |
% \] |
24 |
%% \vlin{\cntr}{}{\Gamma, ?A}{ |
25 |
%% \vlin{\wk}{}{\Gamma, ?A, ?A}{\vlhy{\Gamma, ?A}} |
26 |
%% } |
27 |
% |
28 |
% \caption{Example}\label{Fig:twosuccessiveanchorcuts} |
29 |
% \end{figure} |
30 |
|
31 |
|
32 |
|
33 |
\begin{definition}\label{def:anchoredcut} |
34 |
We define in a mutual inductive way the notion of \textit{hereditarily principal formula} and \textit{anchored cut} in a $\mathcal T$-proof : |
35 |
\begin{itemize} |
36 |
\item A formula $A$ in a sequent $\Gamma \seqar \Delta$ is \textit{hereditarily principal} for a rule (S), if either (i) the sequent is conclusion of the rule (S) and $A$ is principal in it, or |
37 |
(ii) the sequent is conclusion of an anchored cut, the direct ancestor of $A$ in the corresponding premise is hereditarily principal for the rule (S), and the rule (S) is non-logical. |
38 |
\item An instance of cut rule is an \textit{anchored cut} if either its cut-formulas $A$ in the two premises are both hereditarily principal for non-logical rules, or one is hereditarily principal for a non-logical rule and the other one is principal for a logical rule. |
39 |
\end{itemize} |
40 |
\end{definition} |
41 |
% Observe that by this definition if a formula $A$ in a sequent is hereditarily principal for a rule (S) which is a logical rule, then the sequent is conclusion of this rule (S) and $A$ is principal for this rule. The generalization from principal formula to hereditarily principal formula is really made for non-logical rules (R). |
42 |
So as a consequence of this definition, an anchored cut on a formula $A$ is of the following form: |
43 |
\begin{itemize} |
44 |
\item at least one of the two premises of the cut has above it a sub-branch of the proof which starts (top-down) with a non-logical rule (R) with $A$ as one of its principal formulas, and then a sequence of anchored cuts in which $A$ is part of the context; |
45 |
\item the other premise is either of the same form, or a logical rule with principal formula $A$. |
46 |
\end{itemize} |
47 |
|
48 |
% Now, for instance a cut on a (principal) formula $A \lor B$ between a rule $\rigrul{\lor}$ and a rule (R) (where $A \lor B$ occurs in $\Sigma'$) is anchored, while a cut between |
49 |
% a rule $\rigrul{\lor}$ and a rule $\lefrul{\lor}$ is not. |
50 |
|
51 |
A cut which is not anchored will also be called a \textit{free-cut}. |
52 |
|
53 |
Let us first prove a key lemma on hereditarily principal formulas: |
54 |
\begin{lemma}\label{lem:hereditaryprincipalnonlogical} |
55 |
A formula occurrence $A$ on the l.h.s. (resp. r.h.s.) of a sequent and hereditarily principal for a non-logical rule (R) |
56 |
cannot be of the form $A=?A'$ (resp. $A=!A'$). |
57 |
\end{lemma} |
58 |
\begin{proof} |
59 |
Assume that $A$ is an occurrence of formula on the l.h.s. of the sequent and hereditarily principal for a non-logical rule (R), then a direct ancestor of this formula is principal for (R) and thus by condition 4. $A$ cannot be of the form $?A'$. The other case is identical. |
60 |
\end{proof} |
61 |
|
62 |
Now we can state the main result of this section: |
63 |
\begin{theorem} |
64 |
\label{thm:free-cut-elim}[Free-cut elimination] |
65 |
Given a theory $\mathcal{T}$, any $\mathcal T$-proof $\pi$ can be transformed into a $\mathcal T$-proof $\pi'$ with same conclusion sequent and without any free-cut. |
66 |
\end{theorem} |
67 |
The proof will be given below. It will proceed in a way similar to the classical proof of cut elimination for linear logic, but here for eliminating only free-cuts, and we will have to check that all steps of the reasoning are compatible with the fact that the proof here also contains $\mathcal{T}$ rules. Actually the most important part of the proof will be the handling of the commutation steps, since this is where the new non-logical rules could raise some problems. |
68 |
|
69 |
First, observe that the only rules in which there is a condition on the context are the following ones: $(\rigrul{\forall})$, $(\lefrul{\exists})$, $(\rigrul{!})$, $(\lefrul{?})$, $(R)$. These are thus the rules for which the commutation with cut steps are not straightforward. Commutations with rules $(\rigrul{\forall})$, $(\lefrul{\exists})$ are done in the standard way, by using renaming of eigenvariables if necessary. In the following we will have to be careful about rules $(\rigrul{!})$, $(\lefrul{?})$, $(R)$. For that we establish our second key lemma: |
70 |
|
71 |
\begin{lemma}[Key commutations]\label{lem:keycommutations} |
72 |
A cut of the following form, where $?A$ is non principal for $(R)$, can be commuted with the $(R)$ rule: |
73 |
\[ |
74 |
\vliinf{cut}{}{ !\Gamma', \Gamma, \Sigma' \seqar \Delta', ?A, ?\Pi, ?\Pi'} |
75 |
{ \vlinf{(R)}{}{!\Gamma, \Sigma' \seqar \Delta', ?A, ?\Pi}{ \{ !\Gamma, \Sigma_i \seqar \Delta_i, ?A, ?\Pi \}_{i\in I} } } |
76 |
{ \vlinf{}{}{?A, !\Gamma' \seqar ?\Pi'}{} } |
77 |
\] |
78 |
Similarly if $(R)$ is replaced with $(\rigrul{!})$, with $?A$ in its r.h.s. context. Similarly also for the symmetric situations: |
79 |
cut on the l.h.s. of the conclusion of an $(R)$ or a $(\lefrul{?})$ rule on a (non-principal) formula $!A$, with a sequent $!\Gamma' \seqar ?\Pi', !A$. |
80 |
\end{lemma} |
81 |
\begin{proof} |
82 |
The first subderivation can be replaced by: |
83 |
\[ |
84 |
\vlinf{(R)}{}{ !\Gamma', !\Gamma, \Sigma' \seqar \Delta', ?\Pi, ?\Pi'} |
85 |
{ \vliinf{cut}{}{\{!\Gamma', !\Gamma, \Sigma_i \seqar \Delta_i, ?\Pi,?\Pi' \}_{i\in I}} { \vlinf{}{}{ !\Gamma, \Sigma_i \seqar \Delta_i, ?A, ?\Pi}{}} { \vlinf{}{}{?A, !\Gamma' \seqar ?\Pi'}{} } } |
86 |
\] |
87 |
Similarly for the symmetric derivation with a cut on the l.h.s. of the conclusion of an $(R)$ on a formula $!A$. |
88 |
The analogous situations with rules $(\rigrul{!})$ and $(\lefrul{?})$ are handled in the same way, and were already known in linear logic. |
89 |
\end{proof} |
90 |
\begin{proof} (of Thm \ref{thm:free-cut-elim}) |
91 |
Given a cut rule $c$ in a proof $\pi$, we call \emph{degree} $\deg( c)$ the number of logical connectives or quantifiers in its cut-formula. Now the \emph{degree} of the proof $\deg( \pi)$ is the multiset of the degrees of its non-anchored formulas. The degrees will be compared with the multiset ordering. |
92 |
|
93 |
The demonstration will proceed by induction on the degree $\deg( \pi)$. For a given degree we will proceed with a sub-induction on the \textit{height} $\height{\pi}$ of the proof. |
94 |
|
95 |
Consider a proof $\pi$ of non-null degree. We will show how to reduce it to a proof of strictly lower degree. Consider a top-most non-anchored cut $c$ in $\pi$, that is to say such that there is no non-anchored cut above $c$. Let us call $A$ the cut-formula, and $(S_1)$ (resp. $(S_2)$) the rule above the left (resp. right) premise of $c$. |
96 |
\[ |
97 |
\vliinf{c \; \; \cut}{}{\Gamma, \Sigma \seqar \Delta , \Pi}{ \vlinf{S_1}{}{\Gamma \seqar \Delta, A}{} }{\vlinf{S_2}{}{\Sigma, A \seqar \Pi}{}} |
98 |
\] |
99 |
Intuitively we will proceed as follows: if $A$ is not hereditarily principal in one of its premises we will try to commute $c$ with the rule along its left premise $(S_1)$, and if it is not possible commute it with the rule along its right premise $(S_2)$; if $A$ is hereditarily principal in both premises we will proceed with a cut-elimination step. So let us distinguish three cases: |
100 |
|
101 |
\begin{itemize} |
102 |
\item \textbf{First case}: the cut-formula $A$ on the l.h.s. of $c$ is non hereditarily principal. |
103 |
|
104 |
\begin{itemize} |
105 |
\item Consider first the situation where $(S_1)$ is not one of the rules $(\rigrul{!})$, $(\lefrul{?})$, $(R)$. |
106 |
|
107 |
In this case the commutation of $c$ with $(S_1)$ can be done in the usual way. Let us handle as an example the case where $(S_1)=(\rigrul{\laand})$. |
108 |
{\small |
109 |
\[ |
110 |
\vlderivation{ |
111 |
\vliin{c}{}{ \Gamma, \Sigma \seqar B_1\vlan B_2, \Delta, \Pi }{ \vliin{S_1=\rigrul{\vlan}}{}{\Gamma \seqar B_1\vlan B_2, \Delta, A}{ \vlhy{\Gamma \seqar B_1, \Delta, A} }{\vlhy{\Gamma \seqar B_2,\Delta, A}}}{ \vlhy{ \Sigma, A \seqar \Pi} } |
112 |
} |
113 |
\quad\to\quad |
114 |
\vlderivation{ |
115 |
\vliin{\rigrul{\vlan}}{}{ \Gamma, \Sigma \seqar B_1\vlan B_2, \Delta, \Pi }{ |
116 |
\vliin{c_1}{}{\Gamma,\Sigma \seqar B_1, \Delta, \Pi }{ \vlhy{\Gamma \seqar B_1, \Delta, A} }{\vlhy{ \Sigma, A \seqar \Pi} } |
117 |
}{ |
118 |
\vliin{c_2}{}{\Gamma,\Sigma \seqar B_2, \Delta, \Pi }{ \vlhy{\Gamma \seqar B_2, \Delta, A} }{\vlhy{ \Sigma, A \seqar \Pi} } |
119 |
} |
120 |
} |
121 |
\] |
122 |
} |
123 |
%\vlderivation{ |
124 |
%\vliin{\vlan}{}{\Gamma,\Delta,A\vlan B }{ |
125 |
%\vliin{c_1}{}{\Gamma,\Delta,A }{ \vlhy{\Gamma,A,C} }{ \vlhy{ \Delta,C^\bot } } |
126 |
%}{ |
127 |
%\vliin{ c_2 }{}{\Gamma,\Delta,B}{ \vlhy{\Gamma,B,C } }{ \vlhy{\Delta,C^\bot} } |
128 |
%} |
129 |
%} |
130 |
%\] |
131 |
|
132 |
|
133 |
Observe that here $c$ is replaced by two cuts $c_1$ and $c_2$. Call $\pi_i$ the sub-derivation of last rule $c_i$, for $i=1,2$. As for $i=1, 2$ we have |
134 |
$\deg{\pi_i}\leq \deg{\pi}$ and $\height{\pi_i}< \height{\pi}$ we can apply the induction hypothesis, and reduce $\pi_i$ to a proof $\pi'_i$ of same conclusion and with |
135 |
$\deg{\pi'_i} < \deg{\pi_i}$. Therefore by replacing $\pi_i$ by $\pi'_i$ for $i=1, 2$ we obtain a proof $\pi'$ such that $\deg{\pi'}<\deg{\pi}$. |
136 |
|
137 |
The case (S)=($\lefrul{\laor}$) is identical, and the other cases are similar (see the Appendix for more examples). |
138 |
|
139 |
\item Consider now the case where $(S_1)$ is equal to $(\rigrul{!})$, $(\lefrul{?})$ or $(R)$. Let us also assume that the cut-formula is hereditarily principal in its r.h.s. premise, because if this does not hold we can move to the second case below. |
140 |
|
141 |
First consider $(S_1)=(\rigrul{!})$. As $A$ is not principal in the conclusion of $(\rigrul{!})$ it is of the form $A=?A'$. By assumption we know that $A=?A'$ in the conclusion of $(S_2)$ is hereditarily principal on the l.h.s., so by Lemma \ref{lem:hereditaryprincipalnonlogical} it cannot be hereditarily principal for a non-logical rule, so by definition of hereditarily principal we deduce that $(S_2)$ is not an $(R)$ rule. It cannot be an $(\rigrul{!})$ rule either because then $?A'$ could not be a principal formula in its conclusion. Therefore the only possibility is that |
142 |
$(S_2)$ is an $(\lefrul{?})$ rule. So the r.h.s. premise is of the shape $?A',!\Gamma' \seqar ?\Pi'$ and by Lemma \ref{lem:keycommutations} the commutation on the l.h.s. is possible. We can conclude as previously. The case where $(S_1)=(\lefrul{?})$ is the same. |
143 |
|
144 |
Now consider the case where $(S_1)=(R)$. As $A$ is not hereditarily principal in the conclusion of $(R)$, it is a context formula and it is on the r.h.s., so by definition of $(R)$ rules it is the form $A=?A'$. So as before by Lemma \ref{lem:hereditaryprincipalnonlogical} we deduce that $(S_2)=(\lefrul{?})$, and so the r.h.s. premise is of the shape $?A',!\Gamma' \seqar ?\Pi'$. By Lemma \ref{lem:keycommutations} the commutation on the l.h.s. is possible, and so again we conclude as previously. |
145 |
\end{itemize} |
146 |
\item \textbf{Second case}: the cut-formulas on the l.h.s. and r.h.s. of $c$ are both non hereditarily principal. |
147 |
|
148 |
After the first case we are here left with the situation where $(S_1)$ is equal to $(\rigrul{!})$, $(\lefrul{?})$ or $(R)$. |
149 |
\begin{itemize} |
150 |
\item Consider the case where $(S_1)$=$(\rigrul{!})$, $(\lefrul{?})$, so $A$ is of the form $A=?A'$. All cases of commutation of $c$ with $(S_2)$ are as in standard linear logic, except if $(S_2)=(R)$. In this case though we cannot have $A=?A'$ because of the shape of rule $(R)$. So we are done. |
151 |
\item Consider $(S_1)=(R)$. Again as $A$ is not principal in the conclusion of $(S_1)$ and on the r.h.s. of the sequent it is a context formula, and thus of the form $A=?A'$. As $?A'$ is not principal in the conclusion of $(S_2)$, it is thus a context formula on the l.h.s. of sequent, and therefore $(S_2)$ is not a rule $(R)$. So $(S_2)$ is a logical rule. If it is not an $(\rigrul{!})$ or an $(\lefrul{?})$ it admits commutation with the cut, and we are done. If it is equal to $(\rigrul{!})$ or $(\lefrul{?})$ it cannot have $?A'$ as a context formula in the l.h.s. of its conclusion, so these subcases do not occur. |
152 |
\end{itemize} |
153 |
|
154 |
|
155 |
\item \textbf{Third case}: the cut-formulas on the l.h.s. and r.h.s. of $c$ are both hereditarily principal. |
156 |
|
157 |
By assumption $c$ is non anchored, so none of the two cut-formulas is hereditarily principal for a non-logical rule $(R)$. We can deduce from that |
158 |
that the l.h.s. cut-formula is principal for $(S_1)$ and the r.h.s. cut-formula is principal for $(S_2)$. Call $\pi_1$ (resp. $\pi_2$) the subderivation |
159 |
of last rule $(S_1)$ (resp. $(S_2)$). |
160 |
|
161 |
Then we consider the following sub-cases, in order: |
162 |
\begin{itemize} |
163 |
\item \textbf{weakening sub-case}: this is the case when one of the premises of $c$ is a $wk$ rule. W.l.o.g. assume that it is the left premise of $c$ which is conclusion of $\rigrul{\wk}$, with principal formula $A$. We eliminate the cut by keeping only the l.h.s. proof $\pi_1$, removing the last cut $c$ and last $\rigrul{\wk}$ rule on $A$, and by adding enough |
164 |
$\rigrul{\wk}$, $\lefrul{\wk}$ rules to introduce all the new formulas in the final sequent. The degree has decreased. |
165 |
|
166 |
\item \textbf{exponential sub-case}: this is when one of the premises of $c$ is conclusion of a $cntr$, $\rigrul{?}$ or $\lefrul{!}$ rule on a formula $?A$ or $!A$, and the other one is not a conclusion of $\wk$. |
167 |
|
168 |
Assume w.l.o.g. that it is the right premise which is conclusion of $\lefrul{\cntr}$ or $\lefrul{!}$ on $!A$, and thus the only possibility for the left premise is to be conclusion of $\rigrul{!}$. We will use here a global rewriting step. For that consider in $\pi_2$ all the top-most direct ancestors of the cut-formula $!A$, that is to say direct ancestors which do not have any more direct ancestors above. Let us denote them as $!A^{j}$ for $1\leq j \leq k$. Observe that each $!A^{j}$ is principal formula of a rule $\lefrul{!}$ or $\lefrul{wk}$. Denote by $\rho$ the subderivation |
169 |
of $\pi_2$ which has as leaves the sequence premises of these $\lefrul{!}$ or $\lefrul{wk}$ with conclusion containing $!A^{j}$. |
170 |
Let $\rho'$ be a derivation obtained from $\rho$ by renaming if necessary free variables occurring in premises of rules $\lefrul{\exists}$, $\rigrul{\forall}$, $(R)$ so that none of them belongs to $FV(!\Gamma, ?\Delta)$, where $!\Gamma \seqar ?\Delta,!A$ is the l.h.s. premise of the cut $c$. |
171 |
Now, let $\pi'_2$ be the immediate subderivation of $\pi_2$, of conclusion $!\Gamma \seqar ?\Delta,A$. We then define the derivation |
172 |
$\rho''$ obtained from $\rho'$ in the following way: |
173 |
\begin{itemize} |
174 |
\item add a cut $c_j$ with (a copy) of $\pi'_2$ on $A^j$ at each leaf which is premise of a rule $\lefrul{!}$; |
175 |
\item add to each sequent coming from $\rho'$ a context $!\Gamma$ on the l.h.s. and a context $?\Delta$ on the r.h.s., and additional $wk$ rules to introduce these formulas below the $\lefrul{wk}$ rules on formulas $!A^{j}$; |
176 |
\item introduce suitable $\lefrul{cntr}$ and $\rigrul{cntr}$ rules after multiplicative binary rules $\rigrul{\land}$, $\lefrul{\lor}$, $mix$ in such a way to replace $!\Gamma, !\Gamma$ (resp. $?\Delta, ?\Delta$) by $!\Gamma$ (resp. $?\Delta$). |
177 |
\end{itemize} |
178 |
|
179 |
It can be checked that $\rho''$ is a valid derivation, because all the conditions for context-sensitive rules $(\rigrul{\forall})$, $(\lefrul{\exists})$, $(\rigrul{!})$, $(\lefrul{?})$, $(R)$ are satisfied. In particular the rules $(\rigrul{!})$, $(\lefrul{?})$, $(R)$ are satisfied because the contexts have been enlarged with $!$ formulas on the l.h.s. of the sequents ($!\Gamma$) and ? formulas on the r.h.s. of the sequents ($?\Gamma$). |
180 |
|
181 |
Now, let $\pi'$ be the derivation obtained from $\pi$ by removing the cut $c$ and replacing the sub derivation $\rho$ by $\rho''$. The derivation $\pi'$ is a valid one, and with respect to $\pi$ we have replaced one non-anchored cut $c$ with at most $k$ ones $c_j$, but which are of strictly lower degree. So $\deg(\pi')<\deg(\pi)$ and we are done. |
182 |
|
183 |
|
184 |
\item \textbf{logical sub-case}: we are now left with the case where both premises of $c$ are conclusions of rules others than $?$, $!$, $wk$, $cntr$. |
185 |
|
186 |
If one of the premises is an axiom $\lefrul{\bot}$, $\id$ or $\rigrul{\bot}$, then $\pi$ can be rewritten to a suitable proof $\pi'$ by removing $c$ and the axiom rule. Otherwise both premises introduce the same connective, either $\land$, $\lor$, $\laor$, $\laand$, $\forall$ or $\exists$. In each case a specific rewriting rule replaces the cut $c$ with one cut of strictly lower degree. See the Appendix. |
187 |
|
188 |
\end{itemize} |
189 |
|
190 |
|
191 |
\end{itemize} |
192 |
|
193 |
\end{proof} |