S-storage operators

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S-storage operators

Karim Nour

To cite this version:

Karim Nour. S-storage operators. MLQ / MLQ - Math Log Quart; MLQ Mathematical Logic Quar-

terly; Math Log Quart, 1998, 44, pp.99-108. �hal-00381607�

S-STORAGE OPERATORS

Karim NOUR

LAMA - Equipe de Logique, Universit´e de Savoie - 73376 Le Bourget du Lac cedex

Abstract In 1990, J.L. Krivine introduced the notion of storage operator to simulate, for Church integers, the “call by value” in a context of a “call by name” strategy. In this present paper, we define, for every λ-term S which realizes the successor function on Church integers, the notion of S-storage operator. We prove that every storage operator is a S-storage operator. But the converse is not always true.

Mathematics Subject Classification : 03B40, 68Q60

Keywords : Church integer ; Storage operator ; Call by value ; Call by name ; Head reduction ; Solvable ; Successor ; S-storage operator.

1 Definitions and notations

• We denote by Λ the set of λ-terms modulo α-equivalence, and by V the set of λ-variables.

• Let t, u, u

, ..., u

be λ-terms, the application of t to u is denoted by (t)u. In the same way we write (t)u

...u

instead of (...((t)u

)...)u

.

• The sequence of λ-terms u

, ..., u

is denoted u.

• If u = u

, ..., u

, we denote by (t)u the λ-term (t)u

...u

.

• The β-equivalence relation is denoted by u ≃

v.

• The notation σ(t) represents the result of the simultaneous substitution σ to the free variables and the constants of t after a suitable renaming of the bounded variables of t.

• Let us recall that a λ-term t either has a head redex [i.e. t = λx

...λx

(λxu)vw, the head redex being (λxu)v ], or is in head normal form [i.e. t = λx

...λx

(x)w].

• The notation u ≻ v means that v is obtained from u by some head reductions.

• If u ≻ v , we denote by h(u, v) the length of the head reduction between u and v.

• A λ-term t is said solvable iff the head reduction of t terminates.

The following results are well known (see [3]):

1We wish to thank Ren´e David for helpful discussions.

2e-mail nour@univ-savoie.fr

–If u is β-equivalent to a solvable λ-term, then t is solvable.

–If u ≻ v, then, for any substitution σ, σ(u) ≻ σ(v), and h(σ(u), σ(v))=h(u,v).

In particular, if for some substitution σ, σ(t) is solvable, then t is solvable.

• We define (u)

v by induction : (u)

v = v and (u)

v = (u)(u)

v.

• For each integer n, we define the Church integer n = λf λx(f )

x.

• A closed λ-term S is called successor iff, for every k ≥ 0, (S)k ≃

k + 1.

Examples Let S

= λnλf λx(f)((n)f )x and S

= λnλf λx((n)f)(f )x.

It is easy to check that S

and S

are successors. 2

2 Introduction

In λ-calculus the left reduction strategy (iteration of the head reduction) has much ad- vantages : it always terminates when applied to a normalizable λ-term and it seems more economic since we compute a λ-term only when we need it. But the major drawback of this strategy is that a function must compute its argument every time it uses it. In 1990 J-L. Krivine introduced the notion of storage operators in order to avoid this problem and to simulate call-by-value when necessary.

Let F be a λ-term (a function), and n a Church integer. During the computation, by left reduction, of (F )θ

(where θ

≃

n), θ

may be computed several times (as many times as F uses it). We would like to transform (F )θ

to (F )τ

where τ

is a fixed closed λ- term β-equivalent to n. We also want this transformation depends only on θ

(and not F ).

Therefore the definition : A closed λ-term T is called storage operator if and only if for every n ≥ 0, there is a closed λ-term τ

≃

n such that for every θ

≃

n, (T )θ

f ≻ (f )τ

(where f is a new variable).

Let’s analyse the head reduction (T )θ

f ≻ (f)τ

, by replacing each λ-term which comes from θ

by a new variable.

If θ

≃

n, then θ

≻ λgλx(g)t

, t

≻ (g)t

1 ≤ k ≤ n − 1, t

≻ x, and t

≃

(g)

x 0 ≤ k ≤ n − 1.

Let x

be a new variable (x

represents θ

). (T )x

f is solvable, and its head normal

form does not begin by λ, therefore it is a variable applied to some arguments. The free

variables of (T )x

f are x

and f, we then have two possibilities for its head normal form

: (f )δ

(in this case we stop) or (x

)a

...a

.

Assume we obtain (x

)a

...a

. The variable x

represents θ

, and θ

≻ λgλx(g)t

, therefore (θ

)a

...a

and ((a

)t

[a

/x, a

/g])a

...a

have the same head normal form.

The λ-term t

[a

/g, a

/x] comes from θ

. Let x

be a new variable (x

represents t

[a

/g, a

/x]). The λ-term ((a

)x

)a

...a

is solvable, and its head normal form does not begin by λ, therefore it is a variable applied to some arguments.

The free variables of ((a

)x

)a

...a

are among x

, x

, and f, we then have three possibilities for its head normal form : (f )δ

(in this case we stop) or (x

)b

...b

or (x

)b

...b

.

Assume we obtain (x

)b

...b

. The variable (x

represents t

[a

/g, a

/x], and t

≻ (g)t

, therefore (t

[a

/g, a

/x])b

...b

and ((a

)t

[a

/g, a

/x])b

...b

have the same head normal form. The λ-term t

[a

/g, a

/x] comes from θ

. Let x

be a new variable (x

represents t

[a

/g, a

/x]). The λ-term ((a

)x

)b

...b

is solvable, and its head normal form does not begin by λ, therefore it is a variable applied to arguments. The free variables of ((a

)x

)b

...b

are among x

, x

, v

, and f , therefore we have four possibilities for its head normal form : (f )δ

(in this case we stop) or (x

)c

...c

or (x

)c

...c

or (x

)c

...c

... and so on...

Assume we obtain (x

)e

...e

during the construction. The variable x

represents t

[d

/g, d

/x], and t

≻ x, therefore (t

[d

/g, d

/x])e

...e

and (d

)e

...e

have the same head normal form ; we then follow the construction with the λ-term (d

)e

...e

. The λ- term (T )θ

f is solvable, and has (f )τ

as head normal form, so this construction always stops on (f )δ

. We can prove by a simple argument that δ

≃

n.

According to the previous construction, the reduction (T )θ

f ≻ (f)τ

can be divided into two parts :

- A reduction that does not depend on n :

(T )x

f ≻ (x

)a

...a

((a

)x

)a

...a

≻ (x

)b

...b

((a

)x

)b

...b

≻ (x

)b

...b

. . .

- A transformation that depends on n (and not on θ

) :

(x

)a

...a

; ((a

)x

)a

...a

(x

)b

...b

; ((a

)x

)c

...c

.

.

.

(x

)e

...e

; (d

)e

..e

We add new constants x

and x

in λ-calculus, and we consider the following set of head reduction rules :

(λxu)vw ≻ (u[v/x])w (x

)abc ≻ ((a)x

)c

(x

)abc ≻ (b)c

(x

)w ≻ ((a)x

)w (x

)w ≻ (b)w

We write t ≻

t

if t

is obtained from t by applying these rules finitely many times.

With this formalisme we have the following result (see [1] and [4]):

A closed λ-term T is a storage operator iff for every n ≥ 0, (T )x

f ≻

(f)τ

and where τ

is a closed λ-term β-equivalent to n.

The constants x

and x

represent intuitively the λ-terms which come from a non calculated Church integer. The uniform shape of Church integers allows to describe the behaviour of these constants when they are in the head position. However, another method to describe a Church integer is simply to say that it is zero or a successor.

Formally, we add new constants X

et X

in λ-calculus, and we consider, for every successor S, the following set of head reduction rules :

(λxu)vw ≻ (u[v/x])w (X

)abc ≻ ((S)X

)abc

(X

)abc ≻ (0)abc

(X

)uvw ≻ ((S)X

)uvw (X

)uvw ≻ (0)uvw

We write t ≻

t

if t

is obtained from t by applying these rules finitely many times.

A S-storage operator is defined as follows :

A closed λ-term T is a S-storage operator iff for every n ≥ 0, (T )X

f ≻

(f )τ

where τ

is a closed λ-term β-equivalent to n.

This paper studies the link betwen the storage operators and the S-storage operators. We

prove that every storage operator is a S-storage operator. But the converse is not always

true.

3 Storage operators and S-storage operators

Definition Let T be a closed λ-term. We say that T is a storage operator iff for every n ≥ 0, there is a closed

λ-term τ

≃

n, such that for every θ

≃

n, (T )θ

f ≻ (f )τ

. Remark Let F be any λ-term (for a function), and θ

a λ-term β-equivalent to n. Dur- ing the computation of (F )θ

, θ

may be computed each time it comes in head position.

Instead of computing (F )θ

, let us look at the head reduction of (T )θ

F . Since it is {(T )θ

f }[F/f ], we shall first reduce (T )θ

f to its head normal form, which is (f )τ

, and then compute (F )τ

. The computation has been decomposed into two parts, the first being independent of F . This first part is essentially a computation of θ

, the result being τ

, which is a kind of normal form of θ

. So, in the computation of (T )θ

F , θ

is computed first, and the result is given to F as an argument, T has stored the result, before giving it, as many times as needed, to any function. 2

Examples Let S be a successor. If we take :

T

= λn((n)G)δ where G = λxλy(x)λz(y)(S)z and δ = λf (f )0 T

= λnλf (((n)F )f )0 where F = λxλy(x)(S)y,

then it is easy to check that (see [1] and [3]):

for every θ

≃

n, (T

)θ

f ≻ (f )(S)

0 (i = 1 or 2).

Therefore T

and T

are storage operators. 2

Let {x

}

be a set of different constants. We define a set of terms (denoted by Λ

) in the following way :

- If x ∈ V

{x

}

, then x ∈ Λ

; - If x ∈ V , and u ∈ Λ

, then λxu ∈ Λ

; - If u ∈ Λ

, and v ∈ Λ

, then (u)v ∈ Λ

; - If n ∈ IN, and a, b, c ∈ Λ

, then x

∈ Λ

.

x

is considered as a constant which does not appear in a, b, c.

The terms of the set Λ

are called λx-terms.

We have the following result (see [1] and [4]) :

A closed λ-term T is a storage operator iff for every n ≥ 0, there is a finite sequence of head reduction {U

≻ V

}

such that :

1) U

and V

are λx-terms ;

3In his definition of storage operator, J.L. Krivine autorizes theτ_n to contain free variables which are replaced by terms depend ofθ_n. The results of this paper remain valid with this definition but the proofs will be too technical.

2) U

= (T )x

f and V

= (f )τ

where τ

is closed λ-term β-equivalent to n ; 3) V

= (x

)abc or V

= (x

)d 0 ≤ l ≤ n − 1;

4) If V

= (x

)abc, then U

= (b)c if n = 0 and U

= ((a)x

)c if n 6= 0 ; 5) If V

= (x

)d 0 ≤ l ≤ n−1, then U

= (b)d if l = 0 and U

= ((a)x

)d if l 6= 0.

Definitions

1) Let {X

}

be a set of different constants. We define a set of terms (denoted by Λ

) in the following way :

- If x ∈ V

{X

}

, then x ∈ Λ

; - If x ∈ V , and u ∈ Λ

, then λxu ∈ Λ

; - If u ∈ Λ

, and v ∈ Λ

, then (u)v ∈ Λ

; - If n ∈ IN, and a, b, c ∈ Λ

, then X

∈ Λ

.

X

is considered as a constant which does not appear in a, b, c.

The terms of the set Λ

are called λX-terms.

2) Let S be a successor. A closed λ-term T is called a S-storage operator iff for every n ≥ 0, there is a finite sequence of head reduction {U

≻ V

}

such that :

1) U

and V

are λX-terms ;

2) U

= (T )X

f and V

= (f )τ

where τ

is closed λ-term β-equivalent to n ; 3) V

= (X

)abc or V

= (X

)uvw 0 ≤ l ≤ n − 1;

4) If V

= (X

)abc, then U

= (0)abc if n = 0 and U

= ((S)X

)abc if n 6= 0 ;

5) If V

= (X

)uvw 0 ≤ l ≤ n − 1, then U

= (0)uvw if l = 0 and U

= ((S)X

)uvw if l 6= 0.

Examples It is easy to check that, for 1 ≤ i, j ≤ 2, the above operator T

is an S

-storage operator. We check here (for example) that T

is an S

-storage operator:

Let n be an integer.

If n = 0, then we check that (T

)X

f ≻ (X

)F f 0 and (0) F f 0 ≻ (f )0.

If n 6= 0, then we check that:

(T )X

f ≻ (X

) F f 0

((S

)X

) F f 0 ≻ (X

) F (F )f 0 .

. .

((S

)X

) F (F )

f 0 ≻ (X

) F (F )

f 0 (0) F (F )

f 0 ≻ (F )

f 0

We prove (by induction on k) that, for every λ-term u, and for every 0 ≤ k ≤ n, we have (F )

f u ≻ (f )(S

)

u.

- For k = 0, it is true.

- Assume that is true for k, and prove it for k + 1.

(F )

f u = (F )(F )

f u ≻ (F )

f (S

)u. By induction hypothesis we have that for every λ-term v , (F )

f v ≻ (f)(S

)

v, then (F )

f u ≻ (f )(S

)(S

)

u = (f)(S

)

u.

In particular, for u = 0 and k = n, we have (F )

f 0 ≻ (f)(S

)

0.

Therefore T

is a S

-storage operator. 2

A question arizes : Is there a link between the storage operators and the S- storage operators ?

4 Link between the storage operators and the S-storage operators

Theorem 1 If T is a storage operator, then, for every successor S, T is a S-storage operator.

Proof Let S be a successor and T a storage operator.

Then for every n ≥ 0, there is a closed λ-term τ

≃

n such that for every θ

≃

n, (T )θ

f ≻ (f )τ

. In particular ((T )(S)

0)f ≻ (f )τ

.

Let σ : Λ

→ Λ the simultaneous substitution defined by : σ(X

) = (S)

0

for every 0 ≤ k ≤ n − 1, σ(X

) = (S)

0 σ(x) = x if x 6= X

, X

For every n ≥ 0, we construct a set of head equation {U

≻ V

}

such that :

1) U

and V

are λX-terms ;

2) V

= (f)δ

;

3) for every 1 ≤ i ≤ r − 1, V

= (X

)abc or V

= (X

)uvw ; 4) σ(V

) is solvable.

Let U

= (T )X

f. We have σ(U

) = ((T )(S)

0)f is solvable, then U

is solvable and U

≻ V

where V

= (f)δ

or V

= (X

)abc. It is clear that σ(V

) is solvable.

Assume that we have the head reduction U

≻ V

and V

6= (f )δ

.

- If V

= (X

)abc, then, by induction hypothesis, σ(V

) = ((S)

)0)σ(a)σ(b)σ(c) is solvable.

- If n = 0, let U

= (0)abc. Then σ(U

) = (0)σ(b)σ(b)σ(c) is solvable.

- If n 6= 0, let U

= ((S)X

)abc. Then σ(U

) = ((S)(S)

)0)σ(a)σ(b)σ(c) = σ(V

) is solvable.

- If V

= (X

)uvw, then, by induction hypothesis, σ(V

) = ((S)

)0)σ(u)σ(v)σ(w) is solvable.

- If l = 0, let U

= (0)uvw. Then σ(U

) = (0)σ(u)σ(v )σ(w) is solvable.

- If l 6= 0, let U

= ((S)X

)uvw. Then σ(U

) = ((S)(S)

)0)σ(u)σ(v)σ(w) = σ(V

) is solvable.

Therefore U

is solvable and U

≻ V

where V

= (f)δ

or V

= (X

)a

b

c

or V

= (X

)a

b

c

. Since σ(U

) is solvable, then σ(V

) is also solvable.

This constraction always terminates (i.e there is a r ≥ 0 such that V

= (f )δ

). Indeed, if not, we check easily that the λ-term ((T )(S)

0)f is not solvable.

Let y be a variable, ˆ S = (λxS)y, and ˆ0 = (λx0)y.

Let ˆ σ : Λ

→ Λ the simultaneous substitution defined by : ˆ

σ(X

) = ( ˆ S)

ˆ0

for every 0 ≤ k ≤ n − 1, ˆ σ(X

) = ( ˆ S)

ˆ0 ˆ

σ(x) = x if x 6= X

, X

Since ( ˆ S)t ≻ (S)t and ˆ0 ≻ 0, we check easily that ((T )( ˆ S)

ˆ0)f ≻ (f)ˆ σ(δ

). But ( ˆ S)

ˆ0 ≃

n, then ((T )( ˆ S)

ˆ0)f ≻ (f)τ

. Therefore ˆ σ(δ

) = τ

. Since τ

is closed, then δ

is also closed and δ

= τ

≃

n.

Therefore T is a S-storage operator. 2

Definition We say that a λX-term U satisfies the property (P ) iff for each constant

X

of U we have :

- a, b, c satisfy (P )

- X

is applied to a and b ;

- a, b do not contain free variables which are bounded in U.

Lemma 1 Let U, V be λX-terms which do not begin by λ. If U satisfies (P ) and U ≻ V , then V satisfies (P ).

Proof It is enough to do the proof for one step of head reduction. We have U = (λxu)vw and V = (u[v/x])w. Since U satisfies (P ), then u, v, w satisfy (P ) and x is not free in a, b if the constant X

appears in u. Therefore u[v/x], u

, ..., u

satisfy (P ) and V satisfies (P ). 2

Let △ : Λ

→ Λ

the simultaneous substitution defined by :

△(x

) = X

for every 0 ≤ k ≤ n − 1, △(x

) = (X

)△(a)△(b) σ(x) = x if x 6= x

, x

Lemma 2 If U is a λX-term satisfies (P ), then there is a λx-term U

such that △(U

) = U .

Proof By induction on U.

- For U = x, it is true.

- If U = λxV , then V satisfies (P ), and, by induction hypothesis, there is a λx-term V such that △(V

) = V . We put U

= λxV

. We have △(U

) = U .

- If U = (U

)U

(where U

does not begin by a constant), then U

, U

satisfy (P ), and, by induction hypothesis, there are λx-terms U

, U

such that △(U

) = U

and

△(U

) = U

. We put U

= (U

)U

. We have △(U

) = U .

- If U = (X

)abV , then a, b, c, V satisfy (P ), and, by induction hypothesis, there are λx-terms a

, b

, c

, V

such that △(a

) = a, △(b

) = b, △(c

) = c, and △(V

) = V . We put U

= (x

)V

. We have △(U

) = U. 2

Theorem 2 T is a S

-storage operator iff T is a storage operator.

Proof Let n ≥ 0. If T is a S

-storage operator, then there is a finite sequence of head reduction {U

≻ V

}

such that :

1) U

and V

are λX-terms ;

2) U

= (T )X

f and V

= (f )τ

where τ

is closed λ-term β-equivalent to n ;

3) V

= (X

)abc or V

= (X

)uvw 0 ≤ l ≤ n − 1 ;

4) If V

= (X

)abc, then U

= (0)abc if n = 0 and U

= ((S

)X

)abc if n 6= 0 ;

5) If V

= (X

)uvw 0 ≤ l ≤ n − 1, then U

= (0)uvw if l = 0 and U

= ((S

)X

)uvw if l 6= 0.

We prove (by induction on i) that, for every 1 ≤ i ≤ r, V

satisfies (P ).

- For i = 1, it is true.

- Assume that is true for i, and prove it for i + 1.

If V

= (X

)abc, we have two cases :

- if n = 0, then U

= (0)abc. By induction hypothesis V

satisfies (P ), then a, b, c satisfy (P ), therefore U

and V

satisfy (P ).

- if n 6= 0, then U

= ((S)X

)abc. By induction hypothesis V

satisfies (P ), then a, b, c satisfy (P ). Since U

≻ ((a)(X

)ab)c, then V

satisfies (P ).

If V

= (X

)uvw 0 ≤ l ≤ n − 1, then u = a, v = b, and w = c since, by induction hypothesis, V

satisfies (P ). We have two cases :

- if n = 0, then U

= (0)abc. By induction hypothesis V

satisfies (P ), then a, b, c satisfy (P ), therefore U

and V

satisfy (P ).

- if n 6= 0, then U

= ((S)X

)abc. By induction hypothesis V

satisfies (P ), then a, b, c satisfy (P ). Since U

≻ ((a)(X

)ab)c, then V

satisfies (P ).

Therefore there is a finite sequence of head reduction {M

≻ N

}

such that : 1) M

and N

are λX-terms ;

2) M

= (T )X

f and N

= (f )τ

where τ

is closed λ-term β-equivalent to n ; 3) N

= (X

)abc or N

= (X

)abd 0 ≤ l ≤ n − 1;

4) If N

= (X

)abc, then M

= (b)c if n = 0 and M

= ((a)(X

)ab)c if n 6= 0 ;

5) If N

= (X

)abd 0 ≤ l ≤ n − 1, then M

= (b)d if l = 0 and M

= ((a)(X

)ab)d if l 6= 0.

Since, for every 1 ≤ i ≤ r, M

and N

satisfy (P ), let M

and N

the λx-terms such that : △(M

) = M

and △(N

) = N

. We have :

1) M

and N

are λx-terms ;

2) M

= (T )x

f and N

= (f)τ

where τ

is closed λ-term β-equivalent to n ; 3) N

= (x

)a

b

c

or N

= (x

)d

0 ≤ l ≤ n − 1;

4) If N

= (x

)a

b

c

, then M

= (b

)c

if n = 0 and M

= ((a

)x

)c

if n 6= 0 ;

5) If N

= (x

)d

0 ≤ l ≤ n − 1, then M

= (b

)d

if l = 0 and M

= ((a

)x

)d

if l 6= 0.

Therefore T is a storage operator. 2

Theorem 3 There is a S

-storage operator which is a no storage operator.

Proof Let T = λx(x) a b 0 S where a = λxλyλz((x)(z)(x)IIλx0)λx(S)(z)x, b = λxλyλz(z)x,

and S a successor.

Let n be an integer.

If n = 0, then we check that :

(T ) X

f ≻ (X

) a b 0 S f (0) a b 0 S f ≻ (f )0

If n 6= 0, then we check that :

(T ) X

f ≻ (X

) a b 0 S f

((S

)X

) a b 0 S f ≻ (X

) a (a)b 0 S f .

. .

((S

)X

) a (a)

b 0 S f ≻ (X

) a (a)

b 0 S f (0) a (a)

b 0 S f ≻ (a)

b 0 S f

We define two sequences of λ-terms (P

)

and (Q

)

by :

Q

= S, and, for every 0 ≤ k ≤ n − 1, we put Q

= λx(S)(Q

)x P

= 0, and, for every 0 ≤ k ≤ n − 1, we put P

= (Q

)((a)

b)IIλx0 It is easy to check that, for every 1 ≤ k ≤ n, Q

≃

λx(S)

x.

We prove (by induction on k) that, for every 0 ≤ k ≤ n, we have (a)

b 0 S f ≻ (a)

b P

Q

f.

- For k = 0, it is true.

- Assume that is true for k, and prove it for k + 1.

(a)

b P

Q

f = (a) (a)

b P

Q

f ≻

((a)

b) (Q

)((a)

b)IIλx0 λx(S)(Q

)x f = (a)

b P

Q

f.

In particular, for k = n, we have (a)

b 0 S f ≻ (b) P

Q

f ≻ (f )P

.

P

= (Q

)(b)IIλx0 ≃

(λx(S)

x)(b)IIλx0 ≃

(S)

(λx0)I ≃

(S)

0 ≃

n.

Therefore T is a S

-storage operator.

We define a sequence of λ-terms (P

)

by :

P

= 0, and for every 0 ≤ k ≤ n − 1, we put P

= (Q

)(x

)IIJ We check (as before) that :

(T ) x

f ≻ (x

) a b P

Q

f

(a)x

P

Q

f ≻ (x

) P

Q

f .

. .

(a)x

) P

Q

f ≻ (x

) P

Q

f (b) P

Q

f ≻ (f )P

But P

= (Q

)(x

−1,Qn−1,f

)IIλx0 is not closed.

Note that P

≃

(S)

(X

−1,Qn−1,f

)IIλx0 6≃

n. Indeed, if (S)

(X

−1,Qn−1,f

)IIλx0 ≃

n, then (S)

(λx

λx

λx

(S)0)IIλx0 ≃

n, therefore n + 1 ≃

n. A contradiction.

Therefore T is a no storage operator. 2

References

[1] R. David and K. Nour Storage operators and directed λ-calculus Journal of symbolic logic vol. 60, num. 4, pp. 1054-1086, 1995 [2] J.L. Krivine Lambda-calcul, types et mod`eles

Masson, Paris 1990

[3] J.L. Krivine Op´erateurs de mise en m´emoire et traduction de G˝odel Archiv for Mathematical Logic 30, 1990, pp. 241-267

[4] K. Nour Op´erateurs de mise en m´emoire en lambda-calcul pur et typ´e

Th`ese de Doctorat, Universit´e de Chamb´ery, 1993