The Inhabitation Problem for Rank Two Intersection Types

0.2 MB·English

by  Dariusz Kusmierek

The Inhabitation Problem for⋆Rank Two
Interse
tion Types
Dariusz Ku±mierek
Warsaw University,Instituteof Informati
s
7
Bana
ha 2, 02-097 Warsaw, Poland
0
dakumimuw.edu.pl
0
2

n
a
Abstra
t. We prove that the inhabitation problem for rank two inter-
J
se
tion types is de
idable, but (
ontrary to a
ommon belief)
5
EXPTIME-hard. The exponential time hardness is shown by redu
tion

from thein-pla
e a

eptan
e problem for alternating Turing ma
hines.
]
O
L
Key words: lambda
al
ulus, interse
tion types, type inhabitation problem,
.
s alternating Turing ma
hine
c
[

1 Introdu
tion
v
9 Typeinhabitationproblemisusuallyde(cid:28)nedasfollows:(cid:16)doesthereexista
losed
T τ
2 term of a given type (in an empty environment)(cid:17). The answer to this ques-
0 tion depends on the system and its type inferen
e rules. By the Curry-Howard
1

orresponden
e, the inhabitation problem for a given type system is equivalent
0
to the de
idability of the
orresponding logi
.
7
0 In the simply typed system the inhabitation problem is PSPACE-
omplete
/ (see [7℄).
s
c The interse
tion types system studied in the
urrent paper allows types of
α∩β α∩β
: the form . Intuitively, a term
an be assigned the type if and only if
v α β
i it
an be assigned the type and also the type . This system
orresponds to
X
the logi
of (cid:16)strong
onjun
tion(cid:17) (see [1, 4, 5℄).
r Unde
idabilityofthegeneralinhabitationproblemforinterse
tiontypeswas
a
shown by P. Urzy
zyn in [8℄.
Several weakened systems were studied, and proved to be de
idable. T. Ku-
rataandM.Takahashiin[2℄provedthede
idabilityoftheprobleminthesystem
λ(E∩,≤) (I∩)
whi
h does not use the rule .
Leivantin[3℄de(cid:28)nestherankofaninterse
tiontype.Thenotionofrankturns
out to be veryuseful, sin
e it provides meansfor
lassi(cid:28)
ationand ameasureof

omplexity of the interse
tion types.
One
an noti
e, that
onstru
tion in [8℄ uses only types of rank four. The
inhabitation for rank three is still a well-known open problem. The problem for
rank two was so far believed to be de
idable in polynomial spa
e (see [8℄).
⋆
Partly suported byKBN grant 3 T11C 002 27
2 Dariusz Ku±mierek
Ourresult
ontradi
tsthisbelief.Weprovetheinhabitationproblemforrank
twotobeEXPTIME-hardbyaredu
tionfromthehaltingproblemforAlternat-
ingLinearBoundedAutomata(ALBA in short).The idea ofthe redu
tionisas
n
follows. For a given ALBA and a given word of length we
onstru
t a type of
α1∩...∩αn∩αn+1∩αn+2
the form . The intended meaningof the
omponents
α1 ... αn αi i
, , is that represents the behaviour of the -th
ell of the tape, the
αn+1
represents
hanges in the position of the head of ma
hine, and the last
αn+2 ∩
part simulates
hanges of the ma
hine state. The operator is used here
to holdandpro
essinformationabout thewhole
on(cid:28)gurationoftheautomata.
The fa
t that the problem for rank two is EXPTIME-hard only highlights
how di(cid:30)
ult the still open problem for rank three
an be.
1 Basi
s
1.1 Interse
tion Types
We
onsider a lambda
al
ulus with types de(cid:28)ned by the following indu
tion:
(cid:21) Type variables are types
α β α→β α∩β
(cid:21) If and are types, then and are also types
∩
We assume that the operator is asso
iative,
ommutative and idempotent.
α1∩...∩αm
That is, the meaning of the interse
tiontype of the form does not
α
i
depend on the order or number of o

uren
es of ea
h of the types .
The type inferen
e rules for our system are as follows:
Γ ⊢x:σ (x:σ)∈Γ
(VAR) if
Γ ⊢M:α→β Γ ⊢N:α
(E →)
Γ ⊢(MN):β
Γ,(x:α)⊢M:β
(I →)
Γ ⊢λx.M:α→β
Γ ⊢M:α∩β Γ ⊢M:α∩β
(E∩)
Γ ⊢M:α Γ ⊢M:β
Γ ⊢M:α Γ ⊢M:β
(I∩)
Γ ⊢M:α∩β
By the Curry-Howard
orresponden
e the
al
ulus de(cid:28)ned above
orresponds
to the logi
of (cid:16)strong
onjun
tion(cid:17) of Mints and Lopez-Es
obar (see [1, 4, 5℄).
(I∩)
While
onsidering the rule one
an easilynoti
e the spe
i(cid:28)
binding of the
TheInhabitation Problem for RankTwo Interse
tion Types 3
α β M α ∩β α
types and by the term . A term has a type if it has a type
β
and at the same time it has a type . A

ordingly in the
orresponding logi
,
α ∩β α β
one
an prove a formula if one
an prove and and both formulae
have at least one
ommon proof (strong
onjun
tion). It follows that
he
king
the
orre
tnessof formulae in the logi
of (cid:16)strong
onjun
tion(cid:17) may prove to be

on
eptually di(cid:30)
ult and usually it is more
onvenient to regard the equivalent
problem of inhabitation for the interse
tion type systems.
1.2 Interse
tion Types Classi(cid:28)
ation
τ
De(cid:28)nition 1 Following Leivant ([3℄) we de(cid:28)ne the rank of a type , denoted
rank(τ)
by :
rank(τ)=0 τ ∩
, if is a simple type (without (cid:16) (cid:17));
rank(τ ∩σ)=max(1,rank(τ),rank(σ))
;
rank(τ →σ)=max(1+rank(τ),rank(σ)) rank(τ)>0 rank(σ)>0
, when or .
2 De
idability of the Inhabitation Problem
2.1 The Algorithm
x k Γ Γ
De(cid:28)nition 2 A variable is -ary in an environment , if in
ludes a de
la-
(x:α)
ration , su
h that
α=β1 →···→βk →τ
or
α=γ∩(β1 →···→βk →τ)
k k
The variable is -ary, if one
an apply it to some arguments of
ertain types.
The Algorithm. The task
onsidered by the algorithm in ea
h step is a set of
problems:
[Γ1 ⊢M:τ1 ... Γn ⊢M:τn]
, ,
n n
( environments, types, but only one term to be found)
Rem
Ouralgorithmusesthe(cid:16)interse
tionremoval(cid:17) operation de(cid:28)nedasfollows:
Rem(Γ ⊢M:τ)={Γ ⊢M:τ} τ
if is not an interse
tion
Rem(Γ ⊢M:τ1∩τ2)=Rem(Γ ⊢M:τ1)∪Rem(Γ ⊢M:τ2)
Rem ∩
The purpose for the operation is to eliminate (cid:16) (cid:17) and to
onvert a judge-
Γ ⊢M:τ τ
ment with beingpossiblyaninterse
tiontypeintoasetofjudgements
where the types on the right side are not interse
tions.
τ
For a given type the (cid:28)rst task is:
Z0 =Rem(∅⊢M:τ)
4 Dariusz Ku±mierek
It is worth noting that in the tasks pro
essed by our pro
edure types on the
right side will never be interse
tions.
Let the
urrent task be:
Z =[Γ1 ⊢M:τ1 ... Γn ⊢M:τn]
, ,
τ α →β
i i i
1. Ifea
htype isoftheform ,thenthenexttaskpro
essedre
ursively
by the algorithm will be:
Z′ =Rem(Γ1∪{x:α1}⊢M′:β1)∪...∪Rem(Γn∪{x:αn}⊢M′:βn)
,
x Γ
i
where is a fresh variable not used in any of the .
Z′ M′ M = λx.M′
If the re
ursive
all for returns , then , if on the other
handthe re
ursive
allgivesan answer(cid:16)empty type(cid:17), weshallgivethe same
answer.
τ M
i
2. If at least one of the is a type variable, then the term
annot be a an
abstra
tion, but must be an appli
ation or a variable. Suppose that there
x k i
existsavariable whi
his -aryin ea
hoftheenvironments,andforea
h
it holds that:
Γi ⊢x:βi1 →···→βik →τi
(if there is morethan onesu
h variable, we pi
k nondeterministi
allyone of
them). Then:
k =0 M =x
(cid:21) If , then ,
k > 0 M = xM1...Mk Mi k
(cid:21) If , then , where are solutions for the inde-
pendent problems:
Z1 =[Γ1 ⊢M1:β11 ... Γn ⊢M1:βn1]
, , ,
...
Zk =[Γ1 ⊢Mk:β1k ... Γn ⊢Mk:βnk]
, ,
k
Ifanyofthe re
ursive
alsgivestheanswer(cid:16)emptytype(cid:17),weshallgive
the same answer.
3. Otherwise we give the answer (cid:16)empty type(cid:17).
M τ ⊢M:τ
Fa
t 3 If the above algorithm (cid:28)nds a term for an input type , then .
Proof. Straightforward.
2.2 Soundness
The algorithmdes
ribed above is not
apable of (cid:28)nding all the terms of a given
type.Hen
e,toprovethe
ore
tnessoftheproposedpro
edure,we(cid:28)rstde(cid:28)nethe
notion of a long solution, then we show that every task, whi
h has a solution,
has also a long solution. Finally we
omplete the proof of the soundness of
the algorithm by proving that every long solution
an be found by the given
pro
edure.
TheInhabitation Problem for RankTwo Interse
tion Types 5
M Z =[Γ1 ⊢M:τ1,...,Γn ⊢M:τn]
De(cid:28)nition 4 isalong solutionofthetask ,
when one of the following holds:
τ α →β M =λx.M′ M′
i i i
(cid:21) All types areoftheform and , where isalongso-
Z′ = Rem([Γ1,(x:α1) ⊢ M:β1,...,Γn,(x:αn) ⊢ M:βn])
lution of the task ,
or
τi M = xM1...Mk i = 1...n
(cid:21) Some is a type variable and , where for
Γi ⊢ x:αi1 → ··· → αik → τi M1,...,Mk
and are long solutions of tasks
Z1,...,Zk Zj =[Γ1 ⊢Mj:α1j,...,Γn ⊢Mj:αnj] j =1...k
, where for .
Z
Lemma 5 If there exists a solution of a task , then there exists a long one.
M Z = [Γ1 ⊢ M:τ1,...,Γn ⊢ M:τn]
Proof. Assume that is a solution of .
A(M,Z)
We
onstru
t a long solution in the following way:
τ M
i
(cid:21) If there is a whi
h is a type variable, then is not an abstra
tion and:
• M =x A(M,Z)=M M
If , then , be
ausein this
ase isalongsolution,
• M =xM1...Mk Γi ⊢x:αi1 →···→αik → τi
If ,thenitmustholdthat
i = 1 ... n A(M,Z) = xA(M1,Z1)...A(Mk,Zk)
for , so , where
Zj =[Γ1 ⊢Mj:α1j,...,Γn ⊢Mj:αnj] j =1...k
for .
τ α →β
i i i
(cid:21) Otherwise (if all have the form of ):
• M =xM1...Mk k =0 A(M,Z)=λz.A(Mz,Z′)
If (possibly for ), then ,
Z′ =[Γ1,(z:α1)⊢ Mz:β1,...,Γn,(z:αn)⊢Mz:βn] Mz
where (sin
e is
Z′ λz.Mz Z
a solution of the task , and (cid:21) of the task ).
• M = λx.M′ A(M,Z) = λx.A(M′,Z′)
If , then , where
Z′ =Rem([Γ1,(x:α1)⊢M′:β1,...,Γn,(x:αn)⊢M′:βn])
.
M Z =[Γ1 ⊢M:τ1,...,Γn ⊢M:τn]
Lemma 6 Every long solution of the task

an be found by the above alternating pro
edure.
M
Proof. By indu
tion on the stru
ture of .
M =x M τ
i
(cid:21) .Sin
e islong,atleastoneofthe mustbeatypevariable.Hen
e
Z Γ
i
the algorithm working on the task will sear
h in the environments for
a variable of the right type (
ase 2 of the algorithm). One of the variables
x
that the algorithm
hooses from is of
ourse .
M =xM1...Mk
(cid:21) . Likebefore we
an reasonthat the algorithmshall
hoose
the
ase2, and in oneofits possibleruns thealgorithmwill
hoosethe vari-
x x
able .After is
hosen,thepro
edureshallsear
hforsolutionsofthetasks
Z1,...,Zk M1,...,Mk
.Bythede(cid:28)nitionofalongsolution,wehavethat are
Z1,...,Zk
longsolutionsofthetasks ,andbytheindu
tionhypothesis,these
solutions
anbefoundbythere
ursiverunsofourpro
edure.Itfollowsthat
M
also
an be found.
6 Dariusz Ku±mierek
M =λx.M′ τ α →β
i i i
(cid:21) . Then of
ourse all the types haveto be of the form .
Z
Hen
eforthetask thepro
edureshall
hoose
ase1,andsear
hforsolution
Z′ = Rem([Γ1,(x:α1) ⊢ M′:β1,...,Γn,(x:αn) ⊢ M′:βn])
of the task . By
M′ Z′
the indu
tion hypothesis, a long solution for
an be found by the
algorithm.
Corollary 7 Our algorithm (cid:28)nds an inhabitant for every non-empty type, for
whi
h it terminates.
Proof. A dire
t
on
lusion of Lemmas 5 and 6.
3 The Termination of the Algorithm
τ
Let us
onsider the work of the algorithm for a type of rank two.
Fa
t 8 Types of variables put in the environments during the work of algorithm
are of the rank at most one.
Proof. The environmentsaremodi(cid:28)ed only in
ase1. Ifany ofthe variablesput
τ
in the enviromentswasofranktwo,thanthetype musthavebeen ofthe rank
three.
|τ|
Fa
t 9 In every re
ursive run there is no task with more than simultaneous
problems to solve.
Rem ∩
Proof. When a new problem is generated by the operation, one (cid:16) (cid:17) is re-
τ
moved from the type and the type is split between the problems. There
an
|τ|
neverbe morethan problems.The re
ursive
alls in
ase 2do not
reate any
new parallel problems, be
ause of the way they are
reated. Namely, in these
problems the pro
edure sear
hes for terms whi
h
an serve as arguments for
a variable taken from the environment. As we noti
ed before, in environments
there are only variables with types of rank zero and one, and su
h variables

an only be given arguments with types of rank zero. And these types are sim-
Rem
ple (without interse
tions), so they do not generate new problems by the
operation.
3.1 The De
idability
Theorem 10 The inhabitation problem for ranktwo interse
tion types is de
id-
able.
TheInhabitation Problem for RankTwo Interse
tion Types 7
Proof. First noti
e that the environments
annot grow bigger in(cid:28)nitely during
the work of the algorithm. Variables are added to the environments only when
τ α →β
i i i
all
urrentlyexaminedtypes areoftheform .Theneveryenvironment
Γ α O(|τ|)
i i
isexpanded by anew variableof thetype . Note that thereareonly
types that
an be assigned to a variable in one environment. Sin
e we do not
needtokeepseveralvariablesofthesametype(meaningofthesametypeinea
h
of the environments) it follows that there is (cid:28)nite number of possible distin
t
environmentsthatmayo

urduringtheworkofthealgorithm.Alsothenumber
⊢ O(|τ|)
of the types that may o

ur on the right hand side of ea
h is . Hen
e
ea
h bran
h of the alternating pro
edure must (cid:28)nish or repeat a
on(cid:28)guration
in a (cid:28)nite (although possibly exponential) number of steps.
4 The Lower Bound
4.1 Terms of Exponential Size
First we shall
onsider an instru
tive example. We propose a s
hema for
reat-
ing instan
es of the inhabiation problem for whi
h the above algorithm has to
perform an exponentialnumber of steps before (cid:28)nding the only inhabitant. The
sizeoftheinhabitantwillalsobeexponentialinthesizeofthetype.Ourexample
demonstrates a te
hnique used in the
onstru
tion to follow.
T(n)=τ1∩...∩τn
Let , where
τ =α→Ψ →···→Ψ →(α→β)→(β →α)···→(β →α)→β
i
, and
i−1 n−i
| {z }Ψ =(α→α)∩|(β →β) {z }
.
T(3)=
For instan
e
(α→(α→β)→(β →α)→(β →α)→β)∩
(α→ Ψ →(α→β)→(β →α)→β)∩
(α→ Ψ → Ψ →(α→β)→β)
One
an noti
e that a
onstru
tion of an inhabitant for this type is similar to
βββ ααα
the rewriting pro
ess|Tfr(onm)|=theOw(no2r)d to the word , atnd it is aTl(ent)ter by
l|et|tt=erOr(e2wnr)itng. For , there isαonly one term of type T(,3a)nd
. Forinstan
e,theonly(modulo (cid:21)equivalen
e)termoftype is:
λx1x2x3x4.x2(x3(x2(x4(x2(x3(x2x1))))))
.
T(4)
While for it is:
λx1x2x3x4x5.x2(x3(x2(x4(x2(x3(x2(x5(x2(x3(x2(x4(x2(x3(x2x1))))))))))))))
.
Inwhatfollows,whileprovingEXPTIME-hardnessoftheinhabitationproblem,
T(n)
we shall generate types of a similar form to . For this reason it is worth to
T(n)
use for introdu
ing notions and notations, whi
h we shall use later on.
∩ →
Be
ause of the di(cid:27)erent role played by the (cid:16) (cid:17) and (cid:16) (cid:17) it is
onvenient to
8 Dariusz Ku±mierek

onsider the stru
ture of the type in terms of
olumns and rows. The rows are
∩ → T(n) n

onne
ted with (cid:16) (cid:17), and
olumns with (cid:16) (cid:17) (in the
ase of there are
τ1 ... τn T(3)
rows: , , ). A

ording to this terminology has three rows and (cid:28)ve

olumns. One row represents operations available for a given obje
t and the
α β
initial and (cid:28)nal state of the obje
t (here states are variables and ). One

olumn represents a
ertain operation (that is a step of a
ertain automaton).
T(3) i
In type there are three available operations. The -th operation
hanges
i β α α β
the -th sign from to , and all earlier signs from to . More pre
isely ea
h
(α→β) T(n) β α
in the type represents the
hange from to , and an o

uren
e
Ψ β α
of represents no
hange of sign (
hanges and to themselves).
4.2 EXPTIME-hardness
We shall show the lower bound for the
omplexity of the inhabitation problem
byaredu
tionfromtheEXPTIME-
ompleteproblemofthein-pla
ea

eptan
e
for alternating Turing ma
hines.
De(cid:28)nition 11 An alternating Turing ma
hine is a quintuple:
M =(Q,Γ,δ,q0,g)
, where
Q
(cid:21) is a non-empty, (cid:28)nite set of states.
Γ Γ ={0,1}
(cid:21) is a non-empty, (cid:28)nite set of symbols. We shall assume that .
δ ⊆(Q×Γ)×(Q×Γ ×{L,R})
(cid:21) - is a non-empty, (cid:28)nite transition relation.
q0 ∈Q
(cid:21) is the initial state.
g :Q→{∧,∨,accept}
(cid:21) is a fun
tion whi
h assigns a kind of every state.
De(cid:28)nition 12 A
on(cid:28)guration of an alternating Turing ma
hine ma
hine is
a triple:
C =(q,t,n)
, where
q ∈Q
(cid:21) is a state
t∈Γ∗
(cid:21) is a tape
ontent
n∈N
(cid:21) is a position of the head
De(cid:28)nition 13 AnAlternating Linear Bounded Automaton (ALBA)isanalter-
nating Turing ma
hine whose tape head never leaves the input word.
p=((q1,s1),(q2,s2,k))
De(cid:28)nition 14 We shall say that atransition is
onsis-
δ C1 = (q1,t,n)
tent with the relation in a
on(cid:28)guration , when the following

onditions hold:
p∈δ
(cid:21) ;
t(n)=s1
(cid:21) ;
k =L n>1 k =R n<|t|
(cid:21) ( and ) or ( and ).
TheInhabitation Problem for RankTwo Interse
tion Types 9
p C1
We shall say then that transforms a
on(cid:28)guration to a
on(cid:28)guration
C2 = (q2,t2,n2)
, where
s2 m=n
t2(m)= if
(cid:21) (cid:26)t(m)
otherwise
n+1 k =R
n2 = if
(cid:21) (cid:26)n−1
otherwise
De(cid:28)nition 15 An alternating Turing ma
hine a

epts in
on(cid:28)guration
C = (q,t,n)
, if
g(q)=accept |t|=n
(cid:21) and , or
g(q) = ∨ δ
(cid:21) and there exists a transition
onsistent with , whi
h transforms
C
the
on(cid:28)guration to a
on(cid:28)gurationin whi
h the automaton a

epts, or
g(q)=∧ δ C
(cid:21) and every transition
onsistent with , transforms to a
on(cid:28)gu-
ration in whi
h the automaton a

epts.
It is worth noting, that in
on(cid:28)gurations in whi
h the head s
ans the (cid:28)rst
symbolofthetape,theonlyavailabletransitionsarethese,whi
hmovethehead
to the right, and when the head rea
hes the end of the word, the only a
tive
transitions will move it to the left.
De(cid:28)nition 16 The problem of in-pla
e a

eptan
e is de(cid:28)ned as follows: does
t
a given ALBA a

ept a given word (meaning it a

epts the
on(cid:28)guration
C0 = (q0,t,1)
).
It is known that APSPACE = EXPTIME (see Corollary2 to Theorem 16.5
and Corollary3 to Theorem 20.2 in [6℄).
Lemma 17 The problem of in-pla
e a

eptan
e for ALBA is EXPTIME-
om-
plete (APSPACE-
omplete).
Proof. A simple modi(cid:28)
ation of the proof of Theorem 19.9 in [6℄. First we note
that the in-pla
e a

eptan
e is in APSPACE. Consider a ma
hine
M = (Q,Γ,δ,q0,g) M
. Keeping the
ounter of steps, we simulate the run of
t M
o|tn||Qth||eΓ|i|nt|put word . We reje
t if reje
ts, or if ma
hine makes more than
steps, be
ause after so many steps ma
hine has to repeat a
on(cid:28)gu-
ratioLn. nk M
Let be a Mlanguage in APSPACE a

epted in spa
e by a ma
hine .nIkt
means that does not use in any of its parallel
omputations more than
n

ells of the tape (where is the length of the input word). Let us denote the
⊥ M′
blank symbol by . Let us
onsider a modi(cid:28)ed ma
hine , whi
h during its
M M
work performMs′the samenkm−ovnes as , but when rea
hes an a

epting state,
the head of makeMs stteps to the right and also eMnt′ers an a
t
⊥epntki−nng
state. It is
learthat a

epts if and only if the ma
hine a

epts
M
withouteverleavingthisword(notethata

ordingtothede(cid:28)nition,ma
hine
a

epts with the headat rightmostsymbol of the input word) (cid:22) blanksymbols
⊥
Mat the very end of the wonrdkdo not
hange the bethaviour of thLe ma
hine, and
M′does nottu⊥snekm−nore than
ells of the tape. So belongs to if and only if
a

epts in-pla
e.
10 Dariusz Ku±mierek
Theorem 18 The inhabitation problem for rank two interse
tion types is
EXPTIME-hard.
t = t1t2...tn−1tn
Proof. Let us
onsider the input word . We
onstru
t a type
n+2
with rows and some number of
olumns (a

ording to the terminology
n n
introdu
edin 4.1).The (cid:28)rst rowsshallrepresentthe stateof tape
ells.The
1...n
next row shall represent the position of the head (values ). The last row
q
acc
shall stand for the state of the ma
hine. Let be a new type variable. We
shall begin our
onstru
tion with these two
olumns:
(...→ 2 → t1 )∩
...
(...→ 2 →t )∩
n
(...→ 0 → 1 )∩
(...→qacc →q0 )
q ∈/ Q
acc
where . The last
olumnin thetyperepresentsthe initial
on(cid:28)guration:
t t1...tn
in the
ellstherearesymbolsfromtheinput word , thevariables repre-
q0
sentthe input word,the head isat (cid:28)rst position, and the ma
hineisin state .
The se
ond last
olumn represents the (cid:28)nal state.
In the further
onstru
tion we shall add new
olumns on the left side.
A

epting States: If the ma
hine has any a

epting states, we add a
olumn
responsible for a transition from the a

epting states to our additional state
qacc q1...qr
. Let the be all the a

epting states of the ma
hine. The additional

olumn will be:
S →
... n
S →
K →

Q→
where
S =(2→0)∩(2→1)
,
K =(0→n)
,
Q=(qacc →q1)∩...∩(qacc →qr)
.
Ea
h
olumnofthetype(ex
eptthelastone)will beassignedto onevariablein
aterm.The
omponentsofa
olumnarejustdi(cid:27)erenttypesthatareassignedto
n+2
the samevariablein di(cid:27)erent environments. The variable, whi
h will have
assignedtypesbeingpartsofthis
olumn,isresponsiblefortransitionfromea
h
a

epting state of the ma
hine (for ea
h tape
ontent and for head of ma
hine
q
acc
being at last sign of the word) to the state .