Skip to content
Projects
Groups
Snippets
Help
Loading...
Help
Support
Keyboard shortcuts
?
Submit feedback
Contribute to GitLab
Sign in
Toggle navigation
B
BW_VersionPositive
Project overview
Project overview
Details
Activity
Releases
Repository
Repository
Files
Commits
Branches
Tags
Contributors
Graph
Compare
Issues
0
Issues
0
List
Boards
Labels
Milestones
Merge Requests
0
Merge Requests
0
CI / CD
CI / CD
Pipelines
Jobs
Schedules
Analytics
Analytics
CI / CD
Repository
Value Stream
Wiki
Wiki
Snippets
Snippets
Members
Members
Collapse sidebar
Close sidebar
Activity
Graph
Create a new issue
Jobs
Commits
Issue Boards
Open sidebar
Guillaume GENESTIER
BW_VersionPositive
Commits
9d1d3259
Commit
9d1d3259
authored
Apr 02, 2019
by
Guillaume GENESTIER
Browse files
Options
Browse Files
Download
Email Patches
Plain Diff
k
parent
ec4d67e4
Changes
2
Hide whitespace changes
Inline
Side-by-side
Showing
2 changed files
with
43 additions
and
33 deletions
+43
-33
dp.tex
dp.tex
+26
-16
enTeteArticle.tex
enTeteArticle.tex
+17
-17
No files found.
dp.tex
View file @
9d1d3259
...
...
@@ -23,7 +23,7 @@
\begin{cond}
{
cond-symb-order-wf
}
$
\sqsubset
$
is well-founded.
\todo
{
Note that it is free if
$
\Func
$
is finite.
}
\end{cond}
...
...
@@ -40,7 +40,7 @@
\AxiomC
{$
g
\sqsubset
f
$}
\BinaryInfC
{$
\G\vdash
_{
\sqsubset
f
}
g:
\tau
(
g
)
$}
\end{prooftree}
Note that it introduces predicates well-formed
${}_{
\sqsubset
f
}$
.
\begin{center}
\begin{tabular}
{
rcl
}
...
...
@@ -168,14 +168,18 @@
we have that
$
\rew
_{
arg
}
\cup
{
\call
}$
terminates on
$
\mbb
{
U
}$
.
We now prove that,
$
f
\,\bar
t
\in\interp
{
T'
\pi
}$
by induction on
${
\rew
_{
arg
}}
\cup
{
\call
}$
.
Because of the assumptions we have on rules,
$
f
\,\bar
t
$
is neutral.
Hence, it suffices to prove that,
for all
$
u
$
such that
$
f
\,\bar
t
\rew
u
$
, we have
$
u
\in\interp
{
T'
\pi
}$
.
There are two cases:
\begin{itemize}
\item
$
u
=
f
\bar
u
$
with
$
\bar
t
\rew\bar
u
$
.
\begin{itemize}
\item
If
$
f
\in\FObj\setminus\CObj
$
,
$
f
\,\bar
t
$
is neutral.
Hence, it suffices to prove that,
for all
$
u
$
such that
$
f
\,\bar
t
\rew
u
$
, we have
$
u
\in\interp
{
T'
\pi
}$
.
There are two cases:
\begin{itemize}
\item
$
u
=
f
\,\bar
u
$
with
$
f
\,\bar
t
\rew
_{
arg
}
f
\,\bar
u
$
.
Then, we can conclude by induction hypothesis.
\item
There are
$
fl
_
1
\dots
l
_
k
\rul
r
\in\mcal
{
R
}$
and
$
\sg
$
such that
\item
There are
$
fl
_
1
\dots
l
_
k
\rul
r
\in\mcal
{
R
}$
and
$
\sg
$
such that
$
u
=(
r
\sg
)
\,
t
_{
k
+
1
}
\dots
t
_
n
$
and,
for all
$
i
\in\{
1
,
\dots
,k
\}
$
,
$
t
_
i
=
l
_
i
\sg
$
.
Then,
$
\crochet
{
\overline
{
\sur
{
l
}{
x
}}}
\sg\vDash\G
$
.
...
...
@@ -183,11 +187,17 @@
We now prove that, for all
$
\G
$
,
$
u
$
and
$
U
$
, if
$
\G\vdash
_{
f
\bar
l
}
u:U
$
and
$
\sg\vDash\G
$
, then
$
u
\sg\in\interp
{
U
\sg
}$
.
The proof is the same as for
\indthm
{
TypImpliInterp
}
except for (fun) replaced by (fun').
In this case, for all
$
i
$
, we have
$
\G\vdash
_{
f
\bar
l
}
u
_
i:U
_
i
\g
$
.
By induction hypothesis,
$
u
_
i
\sg\in\interp
{
U
_
i
\g\sg
}$
.
So,
$
\g\sg\vDash\Sg
$
and
$
g
\,\bar
u
\sg\in\mbb
{
U
}$
.
Now,
$
g
\,\bar
u
\sg\tilde
{
<
}
f
\,\bar
l
\sg
$
since
$
g
\,\bar
u<f
\,\bar
l
$
.
Therefore, by induction hypothesis,
$
g
\,\bar
u
\sg\in\interp
{
V
\g\sg
}$
.
\end{itemize}
The proof is the same as for
\indthm
{
thm-adequacy
}
except for (fun) replaced by (dp).
\begin{description}
\item
[(dp)]
In this case, for all
$
i
$
, we have
$
\G\vdash
_{
f
\bar
l
}
u
_
i:U
_
i
\g
$
.
By induction hypothesis,
$
u
_
i
\sg\in\interp
{
U
_
i
\g\sg
}$
.
So,
$
\g\sg\vDash\Sg
$
and
$
g
\,\bar
u
\sg\in\mbb
{
U
}$
.
Now,
$
g
\,\bar
u
\sg\tilde
{
<
}
f
\,\bar
l
\sg
$
since
$
g
\,\bar
u<f
\,\bar
l
$
.
Therefore, by induction hypothesis,
$
g
\,\bar
u
\sg\in\interp
{
V
\g\sg
}$
.
\end{description}
\end{itemize}
\item
If
$
f
\in\CObj
$
, then
$
T'
=
C
\,\bar
u
$
with
$
C
\in\CTyp
$
.
\end{itemize}
\end{proof}
enTeteArticle.tex
View file @
9d1d3259
\usepackage
[T1]
{
fontenc
}
%Permet d'imprimer les caractères spéciaux
\usepackage
[utf8]
{
inputenc
}
%Permet de taper directement les caractères spéciaux dans le fichier Tex
\usepackage
[
francais
]
{
babel
}
%Écrit Démonstration et non Proof
\frenchbsetup
{
StandardLists=true
}
%Met des puces rodes plutôt que des - pour les listes
\usepackage
[
english
]
{
babel
}
%Écrit Démonstration et non Proof
%
\frenchbsetup{StandardLists=true} %Met des puces rodes plutôt que des - pour les listes
\usepackage
{
amsthm
}
%Permet de définir les différents types de théorèmes, remarques...
\usepackage
{
amssymb
}
%Ajoute \mathbb
...
...
@@ -23,7 +23,7 @@
\usepackage
[toc,page]
{
appendix
}
% Pour gérer les annexes
\bibliographystyle
{
apalike
}
% Type de bibliographie
\usepackage
{
multirow
}
\usepackage
{
imakeidx
}
%
\usepackage{imakeidx}
\usepackage
[refpage]
{
nomencl
}
\usepackage
{
listings
}
\usepackage
{
enumitem
}
...
...
@@ -34,24 +34,24 @@
\usepackage
{
diagbox
}
\theoremstyle
{
plain
}
\newtheorem
{
thm
}{
Th
\'
{
e
}
or
\`
{
e
}
me
}
[section]
\newtheorem
{
thm
}{
Th
eorem
}
[section]
\newtheorem
{
propo
}
[thm]
{
Proposition
}
\newtheorem
{
coro
}
[thm]
{
Corolla
ire
}
\newtheorem
{
lem
}
[thm]
{
Lemm
e
}
\newtheorem
{
propri
}
[thm]
{
Prop
ri
\'
et
\'
e
}
\newtheorem
{
propris
}
[thm]
{
Prop
ri
\'
et
\'
es
}
\newtheorem
{
defi
}
[thm]
{
D
\'
efinition
}
\newtheorem
{
algo
}
[thm]
{
Algorithm
e
}
\newtheorem
{
coro
}
[thm]
{
Corolla
ry
}
\newtheorem
{
lem
}
[thm]
{
Lemm
a
}
\newtheorem
{
propri
}
[thm]
{
Prop
erty
}
\newtheorem
{
propris
}
[thm]
{
Prop
erti
es
}
\newtheorem
{
defi
}
[thm]
{
Definition
}
\newtheorem
{
algo
}
[thm]
{
Algorithm
}
\theoremstyle
{
definition
}
\newtheorem
{
exo
}
[thm]
{
Exerci
c
e
}
\newtheorem
{
exo
}
[thm]
{
Exerci
s
e
}
\theoremstyle
{
remark
}
\newtheorem*
{
nots
}{
Notations
}
\newtheorem*
{
rmq
}{
Remar
que
}
\newtheorem*
{
expl
}{
Ex
e
mple
}
\newtheorem*
{
expls
}{
Ex
e
mples
}
\newtheorem
{
sub
}{
But loc
al
}
\newtheorem
{
rest
}{
But restant
}
\newtheorem*
{
locProof
}{
Démonstration
}
\newtheorem*
{
rmq
}{
Remar
k
}
\newtheorem*
{
expl
}{
Ex
a
mple
}
\newtheorem*
{
expls
}{
Ex
a
mples
}
\newtheorem
{
sub
}{
Local go
al
}
\newtheorem
{
rest
}{
Remaining goal
}
\newtheorem*
{
locProof
}{
Proof
}
\newcommand
{
\sur
}
[2]
{
\raisebox
{
0.4ex
}{$
#
1
$}
/
\raisebox
{
-0.7ex
}{$
#
2
$}}
...
...
Write
Preview
Markdown
is supported
0%
Try again
or
attach a new file
Attach a file
Cancel
You are about to add
0
people
to the discussion. Proceed with caution.
Finish editing this message first!
Cancel
Please
register
or
sign in
to comment