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Normalization without reducibility
René David
To cite this version:
René David. Normalization without reducibility. Annals of Pure and Applied Logic, Elsevier Masson, 2001, 107, p 121-130. �hal-00384697�
Rene DAVID
Abstract
In[8 ], general results (dueto Coppo,Dezani and Veneri [5 ], [6 ]) re-
lating properties of pure terms and their typability in some systems
withconjunctivetypesDandD areprovedinauniform waybyusing
thereducibilitymethod.Thispapergivesaveryshortproofofthesame
results(actually,oneofthemisabitstronger)usingpurelyarithmetical
methods.
MSC :03B40,03F05
Keywords :-calculus,normalization
1 Introduction
In[8], Gallierpresentsauniformapproachfor provinggeneralresultsrelating
propertiesofpuretermsandtheirtypabilityinsomesystemswithconjunctive
typesDandD,duetoCoppo,DezaniandVenneri([5],[6]). Gallier'sapproach
uses the reducibility method. The results are not new but the accent is put
on the uniformity of the various proofs. Other proofs of similar results can
also be found in [1], [15] or [11]. Bucciarelli& al show in [4] that thestrong
normalization of system D can be easily derived from the one of the simply
typed-calculus.
I givehereanother proof of thesame resultsasin [8] (cf theorem 6). Ac-
tually the point 4 of theorem 6 is stronger (and this result is new) than the
correspondingonein [8]: Foranunsolvableterm, Igiveaprecise relationbe-
tweenthe arityof its type and the number(up to reduction) of its leading
abstractions.
Thisproofdoesnotuse reducibilityandispurelysyntactic.Themain idea
isgivenatthebeginningofsection3.2. Itisalsocompletelyuniform,veryshort
and(atleastinmymind)... elegant. I alsobelievethatthisproofshould help
tobetterunderstandtherelationsbetweenpure-termsandthesystemsDand
D:Notethataveryelementaryandshort(i.e. ahalfpage)proofofthestrong
normalizationofthesystemDcanbe"extracted"fromthispaper.Itusesonly
thepartoflemma18concernedwithSN andthe(trivial)lemma12.
ReneDavid. Laboratoire deMaths. CampusScientique. F-73376LeBourgetduLac.
email[email protected]
beextendedto,forexample,thesystemF. Idonotknowneitherhowtoextend
itto,forexample,Godel'ssystemT norif,forsomereason,noextensionshould
exist.
AcknowledgmentsThe proofof thepart(1) of theorem6 isaverysim-
pliedversionoftheproofgivenbyRMatthesinaninformaltalkin theLogic
Meeting in Oberwolfach (January1998) after which I had helpful discussions
withhim. ThankstoKNourforhelpfulcommentsandtotheanonymousreferee
whodidaverycarefulreadingofthepaperandpointedoutmanyimprecisions.
2 The theorem
Iassumethebasicnotionsonpureand typedcalculusareknown. Theycan
befound inanytextbook onthesubject(forexample: [10], [2], [9]). Forthe
sakeofcompletenessIrecallsomenotationsandthedenitions concerningthe
systemsD andD:
t!t 0
(resptt 0
)meansthattreducestot 0
byonestep(respsomesteps,
possibly0)of reductions.
Every term can be uniquely written as
~
(R ~u)where
~
is a (possibly
empty)sequenceofabstractions,Riseitheraredex(calledtheheadredex)ora
variable(inthiscasethetermissaidtobeinheadnormalformandthevariable
isthehead variable)and~uisa(possiblyempty)sequenceofarguments.
Thehead reduction consists in reducingtheheadredex. The leftreduction
consists in reducing the head redex (if there is one) or(inductively) in doing
theleft reductionof the argumentsof thehead variable. A (nite orinnite)
reductiont
0
!t
1
!::: is aquasi head (respquasi left) reductioniffor every
j the reductiont
i
! t
i+1
is ahead reduction(resp aleft reduction) for some
ij.
cxty(t)representsthecomplexityoft,i.e. thenumberofsymbolsoccurring
int. If~uisasequenceoftermsandN isaset,~u2N meansthateveryelement
ofthesequence~uisinN.
There are twoways ofpresenting theconjunctive types. (Fora historyof
thedierentformulations,see[14])
The rstone (see[3]) is thefollowing: Thetypes areconstructed from
aset ofbasetypesandtheundenedtype!,usingthetypeconstructors
!and\:ThetypingrulesofthesystemDarethefollowing:
(ax) ;x:A`x:Aand `t:!
(!
i
) If ;x:A`t:B then `xt:A!B
(!
e
) If `u:A!B and `v:Athen `(uv):B
(\
i
) If `t:Aand `t:B then `t:A\B
(\
e
) If `t:A\B then `t:Aand `t:B
ThesystemDisobtainedbyrestrictingthesystemDto!-freetypes(i.e.
typeswhere! doesnotoccur) andbydeleting theaxiom `t:!:
B\C)islessconvenientthan(A!B)\(A!C):
Thesecondway(see[5], [6])isthusthefollowing: Werestrictthesetof
typesbyforbidding\after!:Moreprecisely,thesetT oftypesandthe
set S (ofregular types) aredened bythefollowinggrammars(whereV
isthesetofbasetypes) :
S=V j!jT !S andT=S jS\T
The typing rules are the same as in the rst presentation but the types
occurringintherulesmustbeinT,i.e. in therulesA2T andB 2S.
The following result shows that the two presentations are essentially the
same.
Denition1 The translation
fromDintoT isdenedby:
Fora2V [f!g;a
=a.
(A\B)
=A
\B
.
(A!B)
= T
(A
!B
i
)whereB
= T
B
i
and,for every i;B
i 2S :
Proposition 2 1. If `
T
t:A;then `
D t:A:
2. If `
D
t:Athen
`
T t:A
:
Proof. Immediate, byinduction onthelengthofthetypingderivation.
Notations
I will usethesecond presentation which is, formypurpose, moreconve-
nient. Everytypementionnedintherestofthispaperisthusassumedto
beinT. Inparticular, `t:Ameansthat A andthetypesin are in
T andthatt hastypeAin thecontext :
If =fx
1 :A
1
; :::; x
n :A
n
gand=fx
1 :B
1
; :::; x
n :B
n
gthecontext
fx
1 :A
1
\B
1
; :::; x
n :A
n
\B
n
gwillbedenoted by \.
Denition3 1. t2SN it isstronglynormalizing.
2. t2WN it isweakly normalizing.
3. t2HN i tissolvable (i.e. t reducestoahead normalform).
4. Fork1;
(a) t2H
k
ift beginswith kmany :
(b) t2WHN
k
ieither t 2HN ort reducestoaterm inH
k .
Denition4 LetAbe aregulartype.
2. A2HDif itisnon trivial i.e. A=A
1
!::: !A
n
!a for somea2V
andn0.
3. Fork1;A2WHD
k
iA2HD orA hasthe formA
1
!:::!A
k
!
!:
Examples
Let o be a base type. Then : (o ! o) 2 WD:(! ! o) 2 HD WD:
(o!o!!)2WHD
2 HD:
Denition5 1. t 2ST it istypable inD:
2. t2WT i `t:A for someA 2WD and some suchthatthe types
in areconjunctionsoftypesin WD:
3. t 2HT i `t:Afor some 2T andA 2HD:
4. Fork1; t2WHT
k
i `t:A for some 2T andA 2WHD
k :
Examplesand comments
1. It iseasy to checkthat thepreviousdenitions (ST; WT andHT)cor-
respond to the ones in [8]. For example,t 2 HT i t is typablein (the
original)Dwithanontrivialtype(inthesenseof[8]).
2. Notethat,in3and4,thereisnoconditionon :Itiseasyto checkthat
2(resp. 3) meansthat theclosureof t istypable(in theemptycontext)
ofatypein WD(resp.HD).
3. Lett=(( x:y)(Æ Æ))whereÆ= x:(xx). Sincey :o`t:o(where xis
giventhetype!),t2WT.
4. Lett=x:(x(ÆÆ)):Since`t:(!!o)!o, t2HT:
5. Lett=x:(ÆÆ):Since`t:o!!; t2WHT
1 :
Thefollowingtheoremisthemainresultofthepaper.
Theorem6 Let tbe aterm.
1. t2SN it2ST:
2. t 2 WN i t 2WT i the left reduction of t terminates i every quasi
left reduction oft terminates.
3. t 2HN it 2HT i the head reduction of t terminates i every quasi
headreductionof t terminates.
4. Fork1; t2WHN
k
it2WHT
k
i, byhead reduction,t reducestoa
termeitherinheadnormalformorinH
k
i,byanyquasiheadreduction,
t reducestoaterm either inheadnormal formorin H
k .
3.1 The standardisation results
Someimplicationstobeprovedareimmediateconsequencesofthestandardiza-
tiontheorem. I recallhereonlythemaindenition andthetheorem. Elemen-
tary(andveryshort) proofscanbefoundin [7], [12]. Thefollowingdenition
isnot theusual one(forexample, theonein [2]). It canbefound in [12] (or,
implicitly, in [7]). It is,of course,equivalentto theusual oneand theproofof
thisequivalenceisimmediate.
Denition7 The standardreduction
st
isdenedby thefollowing rules:
If u
st u
0
; thenxu
st xu
0
.
If, foralli;u
i
st u
0
i
then(xu
1 :::u
n )
st (xu
0
1 :::u
0
n ).
If (a[x:=b]~c)
st t
0
then(( x:a)b~c)
st t
0
If a
st a
0
;b
st b
0
and for every i, c
i
st c
0
i
then (( x:a) b ~c)
st
(( x:a 0
)b 0
~c 0
).
Lemma8 Assumet
st t
0
:
If t 0
isnormal, thent reduces, byleft reduction, tot 0
:
Ift 0
isinheadnormalform, thentreduces,by head reduction,toaterm
inhead normalform.
If t 0
2H
k
,thent reduces, byhead reduction, toatermin H
k .
Proof. Immediate.
Thefollowingresultisknownasthestandardizationtheorem.
Theorem9 Let tbe aterm. Iftt 0
;thent
st t
0
:
Corollary10 1. t2WN itheleftreductionoftterminatesieveryquasi
left reduction oft terminates.
2. t2HN itheheadreductionoftterminatesieveryquasiheadreduction
of tterminates.
3. Fork1; t 2WHN
k
i, by head reduction,t reducesto aterm either
in head normal formor inH
k
i, byany quasi head reduction, t reduces
toaterm eitherin head normalformor inH
k .
third) property. In each case,(b) )(a) and (c)) (b)are trivial. I onlygive
the proofs of (a) ) (b) and (b) ) (c) in the third case. The other casesare
similar.
(a) ) (b) : It is enoughto prove that, if t reduces to aterm in H
k , then
t reduces,by headreduction,to atermin H
k
:The resultfollowsimmediately
fromtheorem9andlemma8.
(b))(c) : Theresultis provedbyinduction (simultaneouslyfor allk)on
(lg(t);cxty(t))where lg(t)is the lengthof thehead reductionof t to t 0
where
eithert 0
isinheadnormalform ort 0
2H
k .
If t = x u (because of the result is proved simultaneously for all k) or
t =(x
!
u) theresult is clear. Assume t =(( x:a) b
!
c ) doesnot satisfy the
conclusion.Then its innitequasi headreduction is : t (( x:a
1 )b
1
!
c
1 )!
(a
1 [x:=b
1 ]
!
c
1
)::::Thusthereduction(a[x:=b]
!
c )(a
1 [x:=b
1 ]
!
c
1 )
::: alsoisquasi headand(since lg((a[x:=b]
!
c ))<lg(t))this contradicts the
inductionhypothesis.
Anotherconsequenceisthefollowinggrammarcharacterizationoftheclasses
consideredindenition 3.
Theorem11 The classes consideredin denition 3aregiven bythe following
grammars.
SN =(x SN ...SN)j x:SN j (( x:a)b
!
c )whereb;(a[x:=b]
!
c )2SN.
WN =(xWN ...WN)j xWN j(( x:a)b
!
c)where(a[x:=b]
!
c)2 WN.
HN =(x ...) j x HN j((x:a)b
!
c) where (a[x:=b]
!
c )2HN.
WHN
k
= (x ... ) j x:WHN
k 1
(if k > 1) and x: (if k = 1) j
((x:a)b
!
c )where(a[x:=b]
!
c)2WHN
k
Proof. ForSN,theonlynontrivialthingis: Ifb;(a[x:=b]
!
c)2SN then
t =(( x:a) b
!
c) 2 SN: This follows immediately from lemma 12 (1) below.
This(unusual)formulationofthelemmaishelpful forthenextsection.
Theotherresultsareimmediateconsequences ofcorollary10.
Lemma12 1. Assume a;b;
!
c 2 SN and t = (ab
!
c ) 2= SN. Then, for
somea
1,
a xa
1 and(a
1 [x:=b]
!
c )2= SN.
2. xt2SN it2SN:
3. (xt
1 :::t
n
)2SN it
1
;:::;t
n 2SN:
Proof.
(1)Sincea;b;
!
c 2SN,theinnitereductionoftlookslike: t(( xa
1 )b
1
!
c
1
) ! (a
1 [x := b
1 ]
!
c
1
) ::: The result immediately followsfrom the fact
that(a
1 [x:=b]
!
c)(a
1 [x:=b
1 ]
!
c
1 ).
(2)and(3)areimmediate.
Thissectionisthereal novelty.I prove:
Theorem13 1. ST SN:
2. WT WN:
3. HT HN
4. Fork1,WHT
k
WHN
k .
Theideaoftheproofisthefollowing.
Toprovethestrong normalizationin D;I proveasubstitution lemma(see
lemma18): Iftanduaretypedstronglynormalizingterms,thent[x:=u]also
isstronglynormalizing.Thisisprovedbyinduction onatriple: rstthetype
ofu;thenthelengthofthelongestreductionoftandnallythecomplexityof
t:Thetheorem followsimmediately, byinduction on thecomplexity ofterms,
since(uv)=(xv)[x:=u]wherex isafreshvariable.
Toprovetheotherresults(onWN;HN;WHN
k
),IdeneasetN
1
oftriples
( ;t;A)where is a typing context, t isa termand A is atype. This set is,
intuitively, aweakversionoftyped stronglynormalizingterms.Thekeypoint
isanother substitution lemma which is aweakversionof theone for SN and
which is proved in averysimilar way. Theresults easily follow from the fact
that if `t :A; then ( ;t;A)2N
1
andthis is animmediate consequenceof
thesubstitutionlemma. Notethatoneuniquesubstitutionlemmaisenoughto
dealwithallthesesystems.
The following proposition should help to understand the denition of N
1
andtherelationbetweenthetwosubstitutionlemmas.
Proposition 14 SN ischaracterized by the following rules. Lett =
~
(R~c)
whereR iseitheraredex oravariable.
1. If R=(( x:a)b):LetR 0
bethereductof R :
If xappears ina and
~
(R 0
~c)2SN,then t2SN:
Otherwise, ifb2SN and
~
(R 0
~c)2SN,then t2SN
2. If R=x and, for eachi;c
i
2SN,thent2SN:
Proof. Immediate. y
Denition15 ThesetN
1
of triples( ;t;A)(where isatypingcontext, t is
atermandA isatype)isdenedbythe followingrules :
1. If, foreachj; ( ;t;A
j )2N
1 and A
j
2S,then( ;t;
T
A
j )2N
1 :
Inthe other rules, Iassume A=A
1
!:::!A
n
!a (where a is
avariable or !) i.e. A2S andt=x
1 :::x
p (R
!
u) where R iseither a
redexor avariable.
1
Otherwise:
3. If R =x:Assumethat,for 1ik,( ;u
i
;B
i )2N
1 and
`x:B
1
!:::!B
k
!A
p+1
!:::!A
n
!a
`x
j :A
j
for 1jp
then( ;t;A)2N
1 :
4. If R isaredex and( ;t 0
;A)2N
1
(where t 0
isthe headreductof t),then
( ;t;A)2N
1 :
Denition16 1. Fort2SN;l
0
(t) denotes the lengthof the longestreduc-
tionof t:
2. For ( ;t;A) 2 N
1
; l
1
( ;t;A) denotes the number of rules used to prove
(cf. denition 15) that( ;t;A)2N
1 :
Examplesand comments
1. LetI =xx:Then,l=l
1
(;;(II);(o!o)\!)=4:
Byrule1,l=1+l
1
(;;(II);o!o)+l
1
(;;(II);!)
Byrule4,l
1
(;;(I I);o!o)=1+l
1
(;;I;o!o)
Byrule3,l
1
(;;I;o!o)=1
Byrule2,l
1
(;;(I I);!)=1
2. Itcanbeproved(thisissometimescalledthefundamentallemmaofmaxi-
mality)thatl
0
(t)isequaltothenumberofrulesusedtoprove(cf. proposi-
tion14)thatt2SN:Thisobservationbettershowsthesimilaritybetween
thetwocasesoflemma18. SinceIwillnotusethisresultIdon'tproveit.
3. It is clear that, if t reduces to t 0
by left reduction, then l
1 ( ;t
0
;A)
l
1
( ;t;A)andtheunequalityisstrictexceptifthelastruleusedis2.This
willbeusedwithoutmention.
Lemma17 1. If( ;t;A)2N
1
then eithert issolvable orA=A
1
!:::!
A
n
!! andt reduces, byhead reduction,toa termin H
n .
2. ( [fx:Ag;u;B)2N
1
i( ;xu;A!B)2N
1 :
3. Assume `x:A
1
!:::A
k
!B:
(a) If,for alli; ( ;u
i
;A
i )2N
1
,then( ;(xu
1 :::u
k
);B)2N
1 :
(b) If( ;(xu
1 :::u
k
);B)2N
1
andB6=!then( ;u
i
;A
i )2N
1
foralli:
4. Let R be a redex and R 0
be its reduct. If ( ;(R 0
~
u);A) 2 N
1 , then
( ;(R~u);A) 2N
1 .
