The Journal of Symbolic Logic
Volume 71, Number 3, Sept. 2006
A POLYNOMIAL TRANSLATION OF S4 INTO INTUITIONISTIC LOGIC
DAVID FERNANDEZ
§1. Introduction. It is known that both S4 and the Intuitionistic propositional
calculus Int are P-SPACE complete. This guarantees that there is a polynomial
translation from each system into the other.
However, no sound and faithful polynomial translation from S4 into Int is commonly known. The problem of finding one was suggested by Dana Scott during
a very informal gathering of logicians in February 2005 at UCLA. Grigori Mints
then brought it to my attention, and in this paper I present a solution. It is based
on Kripke semantics and describes model-checking for S4 using formulas of Int.
A simple translation from Int into S4, the Gödel-Tarski translation, is wellknown; given a formula ϕ of Int, one obtains ϕ 2 by prefixing 2 to every subformula.
For example,
(p ∨ ¬p)2 = 2(2p ∨ 2¬2p).
That the translation is sound and faithful can be seen by considering topological
semantics, which assign open sets both to 2-formulas of S4 and arbitrary formulas
of Int; the interpretation of ϕ and ϕ 2 turn out to be identical. See Tarski’s paper
[6] for details. Gödel’s original paper can be found in [3].
In [2], Friedman and Flagg present a kind of inverse to Gödel-Tarski. Given a
formula ϕ of S4 and a finite set of formulas Γ of Int, for each ε ∈ Γ one gets an
intuitionistic formula ϕΓ(ε) given recursively by
αΓ(ε) = (α → ε) → ε
(α ∨
â)(ε)
Γ
=
(αΓ(ε)
∨
âΓ(ε)
for α atomic;
→ ε) → ε;
(ε)
(ε)
(α ∧ â)(ε)
Γ = αΓ ∧ âΓ ;
(ε)
(ε)
(α → â)(ε)
Γ = αΓ → âΓ ;
^ (ã)
(2α)(ε)
αΓ → ε → ε.
Γ =
ã∈Γ
Usually ϕ = ø 2 for some ø and Γ = sub(ø). This translation is sound independently of the choice of ε and Γ, but is not faithful in general. For example, if
ε ≡ ⊥ → ⊥ and p is a propositional variable, then
pΓ(ε) = (p → ε) → ε,
which is clearly provable.
Received November 1, 2005.
c 2006, Association for Symbolic Logic
0022-4812/06/7103-0015/$2.30
989
990
DAVID FERNANDEZ
For this paper we will use the following notation: if ϕ is a formula of S4, m(ϕ)
denotes its modal depth and N (ϕ) the number of 2 symbols appearing in ϕ.
The set of subformulas of ϕ will be denoted sub(ϕ), while sub2 (ϕ) is the set of
subformulas α of ϕ such that 2α ∈ sub(ϕ). Var(ϕ) represents the propositional
variables appearing in ϕ, and PV is the set of all propositional variables. Other
notation for Modal and Intuitionistic logic mainly follows that used in [1] and [5],
respectively. Any required background knowledge can also be found in these two
books.
S4 and Int have semantics introduced in [4] by Kripke. Models for both systems
are triplets hW, R, V i where W is a set of worlds, R is a transitive and reflexive
relation on W and V is a truth valuation. This makes the two seem very similar.
However, for Int the valuation V must be R-monotone; that is, if x, y ∈ W and
p ∈ PV, then Rxy implies V (p, x) ≤ V (p, y). Models for S4 in general contain
clusters (a cluster is a subset C ⊂ W such that x, y ∈ C =⇒ Rxy). For example,
the formula 23p ∧ 23¬p cannot be satisfied in a finite model unless it contains a
cluster. For Int we only need to consider partially ordered frames. These differences
are the main obstacle in finding a natural translation from S4 into Int.
Suppose M = hW, R, V i is a model for S4. Assume there is a root world, 0,
which sees the entire model (i.e., R0, x for all x ∈ W ). In such cases we may say M
is a pointed model and write M = hW, R, V, 0i, identifying M |= ø with 0 |=M ø.
To simulate M in Int we use a “layering” scheme. Take the quotient W/R, with
induced accessibility relation R̄. This gives us a new, partially ordered Kripke frame.
If x ∈ W , we refer to its R-equivalence class by x̄. A cluster x̄ will be in layer L(x̄)
if that is its distance from 0; in other words, if the maximum length of any strictly
increasing R̄-chain from 0 to x is L(x̄) + 1. In particular, L(0̄) = 0, since the only
such chain is the singleton {0̄}, which has length 1. Note that the model may branch
and we might have many different clusters at any given layer.
For any S4 formula ϕ, every propositional variable p appearing in ϕ will be
n
represented in our translation by a sequence of propositional variables pm
. The
index n indicates the layer where the variable is evaluated. For a cluster x̄, variables
L(x̄)
of the form pm
represent the values of p in the original S4 model. In this way,
monotonicity conditions for Int can be satisfied. The index m indicates the specific
world we are referring to within the cluster. Thus if x̄ = {x0 , . . . , xN −1 }, the variable
L(x̄)
pm
represents p on the world xm .
The second type of propositional variables represent 2-formulas. If 2α ∈
sub(ϕ), we assign to it a sequence of variables bαn . The index n again represents the
layer where the variable is evaluated. We do not need a second index since the value
of 2α is always constant within a given cluster.
Finally, we include a series of variables l n , which function as level indicators.
n
These are needed to “turn on” those propositional variables pm
and bαn which
correspond to our current level. The translation will have the form
ϕ Int = P(ϕ) → ϕ00 ,
where P(ϕ) is a conjunction of assumptions designed to mimic truth-definitions in
S4 and ϕ00 is essentially a copy of ϕ, corresponding to the root 0.
Section 2 describes our translation formally. Section 3 describes the properties
that hold in models satisfying P(ϕ) and which are used throughout the paper.
A POLYNOMIAL TRANSLATION OF S4 INTO INTUITIONISTIC LOGIC
991
Section 4 then shows how to transform models for Int disproving ϕ Int into models
for S4 disproving ϕ. Section 5 shows that the class of Kripke frames of a special
form is complete for S4. Finally, Section 6 establishes a transformation of S4
models into Int models. Thus we use only semantic arguments to establish that
S4 ⊢ ϕ if and only if Int ⊢ ϕ Int .
§2. The translation. This translation will take an S4 formula ϕ and assign to it
a formula ϕ Int in Int which is derivable if and only if ϕ is.
n
To each p ∈ PV and each nonnegative n, m < N (ϕ) assign a variable pm
. To
each α ∈ sub2 (ϕ) we assign N (ϕ) variables bαn . We will also use N (ϕ) variables l n .
We need some preliminary definitions before giving the actual translation.
n
n
). For a formula ø of S4 and n, m ∈ N define øm
recursively as
Definition (øm
follows:
n
;
(p)nm = pm
⊥nm = ⊥;
(ø1 ⊙ ø2 )nm = (ø1 )nm ⊙ (ø2 )nm ,
(2ø1 )nm
=
for ⊙ = ∧, ∨, →;
bøn 1 .
Let N = N (ϕ).
n
in
Next we define a set of formulas that encode instructions for interpreting øm
Int as ø would be in S4. Intuitively, Lev(ϕ) will set up the meaning of the variables
l n as layer indicators, Midn (ϕ) will restore the principle of excluded middle when
we need it (namely, when n = L(x)) and Boxαn (ϕ) tells how to interpret bαn as 2α.
Later we will make these notions precise.
Definition (Lev(ϕ), Midn (ϕ), Boxnα (ϕ), P(ϕ)). Let ϕ be a formula of S4 and
N = N (ϕ). Then,
Lev(ϕ) = l 0 ∧ ¬l N ∧
N^
−1
(l k+1 → l k );
k=0
n
^
n
Mid (ϕ) = l →
^
(bøn ∨ ¬bøn ) ∧
^
l k → l k+1 ∨
n
N^
−1
m=0
k≥n
Boxnα (ϕ)
n+1
k
,
αm
→l
∨ Anα (ϕ) ∧ Anα (ϕ) → l n+1 ∨ bαn ,
^
^
^
P(ϕ) = Lev(ϕ) ∧
Midn (ϕ) ∧
Boxnα (ϕ).
=l →
bαn
n
n
(pm
∨ ¬pm
) ,
p 0≤m≤N −1
ø∈sub2 (ϕ)
Anα (ϕ) =
^
0≤n≤N −1
0≤n≤N −1 α∈sub2 ϕ
Now we can define the translation itself, which we will denote ϕ Int :
Definition (ϕ Int ).
ϕ Int = P(ϕ) → ϕ00 .
The aim of this paper is to prove the following:
992
DAVID FERNANDEZ
Theorem 2.1.
S4 ⊢ ϕ ⇐⇒ Int ⊢ ϕ Int .
§3. Properties of models satisfying P(ϕ). The heart of our translation is that any
Kripke model satisfying P(ϕ) can be interpreted as if it were an S4 Kripke model
for ϕ, after properly decoding the new indexed variables. In this section we explore
the meaning of the clauses appearing in P(ϕ) and draw some parallels to model
checking for S4.
Throughout this section, M = hW, R, V i is an Intuitionistic Kripke model.
Lemma 3.1. Let ϕ be a formula of S4 and assume M |= Lev(ϕ). Let x ∈ W be
any world.
Then there exists a unique natural number L(x) such that
V (l n , x) = 1 ⇐⇒ n ≤ L(x).
Proof. Recall that
Lev(ϕ) = l 0 ∧ ¬l N ∧
N^
−1
(l k+1 → l k ).
k=0
0
0
V (l , x) = 1 (because M |= l ) and V (l , x) = 0 (because M |= ¬l N ).
Now,
M |=
N
N^
−1
(l k+1 → l k ),
k=0
which implies that if V (l n , x) = 1 and k ≤ n, then V (l k , x) = 1 as well; likewise, if
V (l n , x) = 0 and k ≥ n, V (l k , x) = 0. So for some L, the sequence {V (l k , x)}N
k=0
consists of L consecutive ones followed by N − L consecutive zeros; this L is the
alleged L(x).
⊣
From here on, we will refer to L(x) as the level of x.
Next we elucidate the meaning of Midn (ϕ), which essentially allows us to evaluate
n
a formula øm
Classically based only on the truth values of the present world.
Definition (Classical satisfaction in Intuitionistic Kripke frames). Suppose x ∈
W and let M (x) be the singleton submodel h{x}, R |{x} , V |{x} i of M .
A formula ø is satisfied Classically in a world x ∈ W , denoted by x |=C ø, if
x |=M (x) ø.
Lemma 3.2. Let ø be a formula of Int and M = hW, R, V i an Intuitionistic Kripke
model. Assume that for some x ∈ W , x |=M p ∨ ¬p for all p ∈ Var(ø).
Then
x |=M ø ⇐⇒ x |=C ø.
Proof. This can be proven by a standard induction on formulas. The induction
base for propositional variables is evident as is the induction step for conjunction
and disjunction.
For “→” we need to check that the condition Rxy =⇒ V (ø1 , y) ≤ V (ø2 , y)
(i.e., the Intuitionistic condition for implication) is equivalent to the condition
V (ø1 , x) ≤ V (ø2 , x).
A POLYNOMIAL TRANSLATION OF S4 INTO INTUITIONISTIC LOGIC
993
But if x |=M q ∨ ¬q and Rxy, then V (q, x) = V (q, y); indeed, this is the case
unless V (q, x) = 0 and V (q, y) = 1. The first clearly implies x 6|=M q, while the
second implies that x 6|=M ¬q, and therefore x 6|=M q ∨ ¬q.
Since this holds for all q ∈ Var(ø), V (øi , y) = V (øi , x), and we only need to
evaluate the formulas on the current world.
⊣
Corollary 3.1. Let ϕ be a formula of S4. If
M |= Lev(ϕ) ∧
N^
−1
Midn (ϕ),
n=0
and ø ∈ sub(ϕ), then
L(x)
L(x)
x |=M øm
⇐⇒ x |=C øm
.
Proof. Substituting n = L(x) we get M |= MidL(x) (ϕ).
L(x)
But x |=M l L(x) , so x |= q ∨ ¬q for all q ∈ Var(øm
) (note that all variables
L(x)
L(x)
appearing in øm are either of the form pm or of the form bαL(x) ).
Applying Lemma 3.2 concludes the proof.
⊣
Lemma 3.3. Let N = N (ϕ), x ∈ W , α ∈ sub2 (ϕ) and assume
M |= Lev(ϕ) ∧
N^
−1
Midn (ϕ).
n=0
Then M |=
VN −1
Boxnα (ϕ) if and only if
L(y)
x |=M bαL(x) ⇐⇒ ∀m < N, Rxy =⇒ y |=C αm
.
n=0
Proof. We defined
Now
Boxnα (ϕ) = l n → bαn → l n+1 ∨ Anα (ϕ) ∧ Anα (ϕ) → l n+1 ∨ bαn .
M |=
N^
−1
Boxnα (ϕ)
n=0
if and only if for all x ∈ W and α ∈ sub2 (ϕ), x |=M l n implies that
x |=M bαn → l n+1 ∨ Anα (ϕ) ∧ Anα (ϕ) → l n+1 ∨ bαn .
First note that if L(x) < n, x 6|=M l n while if L(x) > n, x |=M l n+1 (because
M |= Lev(ϕ)).
Hence this is equivalent to
x |=M bαL(x) =⇒ x |=M AL(x)
(ϕ)
α
and
x |=M AL(x)
(ϕ) =⇒ x |=M bαL(x) .
α
Now, substituting
AL(x)
(ϕ) =
α
^
k≥L(x)
l k → l k+1 ∨
N^
−1
m=0
k
αm
994
DAVID FERNANDEZ
we expand
x |=M
^
l k → l k+1 ∨
k
αm
m=0
k≥L(x)
to
N^
−1
Rxy =⇒ y |=M l k =⇒ y |=M l k+1 ∨
N^
−1
k
αm
m=0
for all k ≥ L(x).
Again this is trivially true unless k = L(y), and this expression can be replaced
by
N^
−1
Rxy =⇒ y |=M
L(y)
αm
.
m=0
Now by Corollary 3.1 we can replace
N^
−1
y |=M
L(y)
αm
m=0
by
y |=C
N^
−1
L(y)
αm
.
m=0
Plugging this into the equivalence we had earlier for
M |=
N^
−1
Boxnα (ϕ)
n=0
we get
x |=M bαL(x) =⇒ Rxy =⇒ y |=M
N^
−1
m=0
and
Rxy =⇒ y |=M
N^
−1
m=0
L(y)
αm
L(y)
=⇒ x |=M bαL(x) .
αm
This is what we wanted.
Thus bαn variables correspond very naturally to 2-formulas in S4.
⊣
§4. Transformation of Int models into S4 models.
Definition (M S4 ). Let M = hW, R, V, 0i be a model for Int falsifying ϕ Int . Assume without loss of generality that M |= P(ϕ) (that is, in order to falsify ϕ Int we
need to exhibit a world w ∈ W where V (P(ϕ), w) = 1 and V (ϕ00 , w) = 0. We then
take the submodel generated by {w} ).
To each x ∈ W assign a set x̄ = {x0 , . . . , xN −1 } of N = N (ϕ) worlds. Let W S4
be the (disjoint) union of all x̄.
A POLYNOMIAL TRANSLATION OF S4 INTO INTUITIONISTIC LOGIC
995
Define RS4 by
RS4 xm yn ⇐⇒ Rxy.
Set
L(x)
V S4 (p, xm ) = V (pm
, x).
Then
M S4 = hW S4 , RS4 , V S4 , 00 i.
L(x)
Theorem 4.1. If ø ∈ sub(ϕ), V (øm
, x) = V S4 (ø, xm ).
Proof. We will use induction on the depth of ø.
L(x)
is identical to ø except that the variables have been indexed
The formula øm
and any 2-formula 2α has been replaced by bαL(x) .
L(x)
L(x)
By Corollary 3.1, x |=M øm
if and only if x |=C øm
, and we have set
L(x)
S4
V (p, xm ) = V (pm , x).
Therefore we only need to check that V S4 (2α, xm ) = V (bαL(x) , x) to prove the
Theorem.
We will do this by induction on m(α).
By Lemma 3.3, we have that
x |= bαL(x) ⇐⇒ ∀n < N, Rxy =⇒ y |=C αnL(y)
⇐⇒IH ∀n < N, Rxy =⇒ yn |= α.
The reason we can apply IH is that all 2-formulas appearing in α are of lesser
modal depth than 2α itself.
Now pick an index m. Then, Rxm yn if and only if Rxy; so we have that
x |= bαL(x) ⇐⇒ (Rxm yn =⇒ yn |= α),
which is equivalent to xm |= 2α.
Applying the Theorem to x = 0, ø = ϕ and m = 0, we conclude that
⊣
Int 6⊢ ϕ Int =⇒ S4 6⊢ ϕ.
§5. Standard form for Kripke frames. Our objective is to turn a model disproving
ϕ into one disproving ϕ Int . For this we need the models for S4 to have a specific
form. In this section we define this form and construct Kripke frames with the
properties we need.
Let K = hW, Ri be a transitive, reflexive Kripke frame.
If x ∈ W , we will denote by x̄ its R-equivalence class (i.e., the set { y :
Rxy and Ryx }). The quotient W/R forms a partial order with induced accessibility relation R̄x̄ ȳ ⇐⇒ Rxy (this is rather standard and easy to check). We
will refer to the quotient frame hW/R, R̄i by K̄. Abusing notation a bit, we may
also write M̄ to refer to the quotient frame of a model M .
Definition (N -Standard Kripke frame). Let K = hW, Ri be a Kripke frame and
N ≥ 1.
We say K is N -standard if it has the following properties:
(a) Any strictly ascending chain in K̄ has length at most N ;
(b) ∀x ∈ W , x̄ has exactly N elements;
(c) K̄ forms a tree.
996
DAVID FERNANDEZ
Theorem 5.1. Let M = hW, R, V, 0i be any pointed Kripke model and ϕ a formula
of S4.
Then there exists a model M ϕ = hW ϕ , Rϕ , V ϕ , 0ϕ i such that the frame hW ϕ , Rϕ i
is N (ϕ)-standard and
M ≃sub(ϕ) M ϕ .
We need a few preliminaries to prove the Theorem. Throughout this section,
M = hW, R, V, 0i is a pointed Kripke model for S4, ϕ is a fixed formula and
N = N (ϕ).
ϕ
ϕ
ϕ
ϕ
ϕ
Definition (M−2 ). M−2 = hW−2 , R−2 , V−2 , 0i, where
ϕ
W−2
= W;
ϕ
R−2
xy ⇐⇒ V (2ø, x) ≤ V (2ø, y)
ϕ
V−2
(p, x)
for all ø ∈ sub2 (ϕ);
= V (p, x) for all p ∈ PV, x ∈ W .
Note that the only real change is that we are increasing the accessibility relation.
ϕ
The definition of R−2
might seem familiar; it is similar, for example, to the one that
is obtained when taking a sub(ϕ)-filtration of the canonical model.
ϕ
Lemma 5.1. M−2
≃sub(ϕ) M .
ϕ
ϕ
Proof. The relation R−2 is obviously transitive and reflexive, so M−2 is a valid
ϕ
model for S4. Also note that Rxy =⇒ R−2
xy.
Let x ∈ W . We will prove the result by induction on the depth of ø ∈ sub(ϕ).
The base case (when ø is a propositional variable or ⊥) works by the way we
ϕ
defined V−2
, and induction goes through trivially if ø = ø1 ⊙ ø2 , ⊙ = ∨, ∧, →.
Only the case ø = 2α requires any work.
First assume V (2α, x) = 0.
Then there is some world y ∈ W such that Rxy and V (α, y) = 0. Now, by
ϕ
ϕ
induction hypothesis, V−2
(α, y) = 0. Since we have R−2
xy, we conclude that
ϕ
V−2 (2α, x) = 0.
ϕ
ϕ
On the other hand, if V (2α, x) = 1 and R−2
xy, by definition of R−2
, V (2α, y) ≥
V (2α, y) = 1. This implies that V (α, y) = 1. Like before we apply our induction
ϕ
hypothesis to get V−2 (α, y) = V (α, y) = 1. Since y was arbitrary, we conclude
ϕ
that V−2 (2α, x) = 1.
⊣
ϕ
Lemma 5.2. All strictly ascending chains of M̄−2 have length at most N .
Proof. If Rx̄ ȳ and x̄ 6= ȳ, then V (2α, x) < V (2α, y) for some α ∈ sub2 ϕ.
Since we only have N of these formulas, no strictly ascending chain can have more
than N distinct elements.
⊣
ϕ
ϕ
ϕ
ϕ
Lemma 5.3. There exists a model M−1
= hW−1
, R−1
, V−1
, 0i such that
ϕ
ϕ
ϕ
ϕ
hW/R−1 , R̄−1 i ∼
= hW/R−2 , R̄−2 i,
all R-clusters have exactly N worlds and such that
ϕ
ϕ
M−1 ≃sub(ϕ) M−2 .
ϕ
Proof. Let x̄ = {x0 , . . . , xn−1 } be an R-cluster of M−2
.
ϕ
We will replace x̄ by a cluster (x̄) which has N worlds without affecting V (ϕ, 0).
Repeating this for each cluster of W will give us the desired model.
A POLYNOMIAL TRANSLATION OF S4 INTO INTUITIONISTIC LOGIC
997
If n < N , we can simply add copies of one xi and clearly nothing changes.
Otherwise, for each α ∈ sub2 (ϕ) which is disproved in some world of x̄, pick an
index i(α) < n such that V (ϕ, xi(α) ) = 0. Since there are only N formulas we need
to pick at most N such worlds. Then delete all worlds in the cluster that were not
selected. This gives us a submodel M ′ = hW ′ , R′ , V ′ , 0i.
Note: It is important that we not delete the world 0. The only case in which we
would be forced to do this is if the root world is in a cluster and we actually need N
distinct worlds in that very cluster, all different from 0. We will not discuss this case
in detail, but some inspection shows that such a model can be transformed into one
with two clusters of size N , one of which consists of N copies of the root world and the
other, its successor, the cluster of worlds that were selected from x̄. The frame of such
a model is N -standard.
Now take any world y ∈ W ′ . We show by induction that
ϕ
y |=M−2
ø ⇐⇒ y |=M ′ ø,
for ø ∈ sub(ϕ).
If ø ∈ PV or ø = ⊥, the result is trivial. The induction step also goes through
trivially if ø = ø1 ⊙ ø2 , ⊙ a Boolean operator.
ϕ 2α, y |=
If y |=M−2
M ′ 2α as well, because by IH, all successors model α, which
is of lower depth.
ϕ
ϕ such that R
ϕ
Otherwise there exists z in WM−2
−2 y, z and z 6|=M−2 α. By induction
′
on modal depth we know z 6|=W ′ α either if z ∈ W . However, if z was removed and
ϕ
ϕ
is not in W ′ , then z ∈ x̄ and therefore xi(α) ∈ W ′ satisfies R−2
z, xi(α) , R−2
xi(α) , z
ϕ
α.
and xi(α) 6|=M−2
ϕ
Since the model is transitive we know that R−2
y, xi(α) . By induction we know
that xi(ø) 6|=W ′ α.
⊣
So y 6|=W ′ 2α.
Definition (M ϕ ). If S is some partially ordered finite set and s ∈ S, denote by
Chain(S, s) the set of all strictly ascending chains in S beginning on s. Note that
Chain(S, s), ordered by inclusion, naturally forms a tree with root {s}.
For X ∈ Chain(S, s), denote its maximum element by Last(X ).
Then, M ϕ = hW ϕ , Rϕ , V ϕ , 0ϕ i, where
ϕ
ϕ
W ϕ = { (x, X ) : X ∈ Chain(W−1
/R−1
, 0̄), x ∈ Last(X ) };
Rϕ (x, X )(y, Y ) ⇐⇒ X ⊂ Y ;
V ϕ (p, x) = V (p, x) for all p ∈ PV, x ∈ W ;
0ϕ = 0, {0̄} .
Lemma 5.4. M̄ ϕ is a tree and
ϕ
M ϕ ≃sub(ϕ) M−1
.
Proof. Clusters of M ϕ are clearly of the form
{ (x, X ) : x̄ = Last(X ) },
where X is fixed.
ϕ
ϕ
Therefore, M̄ ϕ can be identified with Chain(W−1
/R−1
), which is already a tree.
Now we must check the second claim of the Lemma.
998
DAVID FERNANDEZ
For this we will prove for ø ∈ sub(ϕ) and (x, X ) ∈ W ϕ ,
ϕ
ø.
(x, X ) |=M ϕ ø ⇐⇒ x |=M−1
This is obvious for propositional variables and follows inductively for formulas built
by Boolean connectives from subformulas.
As before, only the step ø = 2α requires any work.
ϕ 2α and let (y, Y ) be such that R ϕ (x, X )(y, Y ).
Assume first x |=M−1
ϕ
ϕ
Then, R̄−1 Last(X ) Last(Y ), which in turn implies R−1 xy, as x ∈ Last(X ) and
y ∈ Last(Y ).
ϕ
We know y |=M−1
α, so by IH we conclude (y, Y ) |=M ϕ α; hence,
(x, X ) |=M ϕ 2α.
ϕ
ϕ
ϕ α.
On the other hand, if x 6|=M−1
ø, pick y ∈ W−1
such that y 6|=W−1
Then consider (y, Y ) = y, X ∪ {ȳ} .
Clearly Rϕ (x, X )(y, Y ) and (y, Y ) 6|=W ϕ α by IH.
So (x, X ) 6|=W ϕ α.
⊣
§6. Transformation of an S4 model into an Intuitionistic model. Throughout this
section, ϕ is a fixed formula of S4, N = N (ϕ) and
M = hW, R, V, 0i
is a Kripke model on an N -standard frame.
We will assume the elements of each x̄ ∈ W/R are numbered, so that
x̄ = {x0 , . . . , xN −1 },
with 0 = (0̄)0 . The immediate predecessor of x̄ will be denoted by x̄−1 .
Denote by L̄(x̄) the distance from 0̄ to x̄; that is, L̄(0̄) = 0, L̄(x̄) = 1 if 0̄ = x̄−1
and, in general, L̄(x̄) + 1 is the length of the longest strictly increasing R̄-sequence
beginning at 0̄ and ending at x̄. This definition is different from that of L given in
Lemma 3.1, but in our model, the two will turn out to be equivalent. Note that
L̄(x̄) < N .
Definition (M Int ). We set M Int = hW/R, R̄, V Int , 0̄i, where V Int is defined by
the following clauses:
(
1 if n ≤ L̄(x̄);
V Int (l n , x̄) =
(1)
0 otherwise.


if L̄(x̄) < n;
0
Int
n
Int
n
(2)
V (pm , x̄) = V (pm , (x̄)−1 ) if L̄(x̄) > n;


V (p, xm )
if L̄(x̄) = n.


if L̄(x̄) > n;
0
V Int (bαn , x̄) = V Int (bαn , (x̄)−1 ) if L̄(x̄) < n;
(3)


V (2α, x0 )
if L̄(x̄) = n.
A POLYNOMIAL TRANSLATION OF S4 INTO INTUITIONISTIC LOGIC
999
Theorem 6.1.
M Int |= P(ϕ)
and
(M Int )S4 ∼
= M.
The latter is a full isomorphism, which is stronger than what we’ve had in previous cases. Once we prove this Theorem we will have established the aim of this
paper, since if M 6|= ϕ, we automatically have (M Int )S4 6|= ϕ. By Theorem 4.1,
V Int (ϕ Int , 0̄) = (V Int )S4 (ϕ, 0).
This Theorem would then imply that
V Int (ϕ Int , 0̄) = V (ϕ, 0)
as well, and thus
S4 6⊢ ϕ =⇒ Int 6⊢ ϕ,
hence concluding the proof of Theorem 2.1.
The remainder of this section will establish partial results leading to the Theorem.
Lemma 6.1. V Int is R̄-monotone with respect to l n variables and
M Int |= Lev(ϕ).
Proof. Assume R̄x̄ ȳ and V Int (l n , x̄) = 1. Then, L̄(x̄) ≤ L̄(ȳ), so by (1),
V (l n , ȳ) = 1. Hence V Int is monotone, at least when restricting to l n variables.
On the other hand, if V Int (l n+1 , x̄) = 1, then L̄(x̄) ≥ n + 1, hence V Int (l n , x̄) = 1.
It is then readily seen that M Int is a valid Intuitionistic model at least when
restricted to the variables l n and that
Int
M Int |=
n−1
^
(l k+1 → l k ).
k=0
It is also easy to see, using (1), that M Int |= l 0 .
Finally, using (1) one sees that if n is greater or equal to the depth of the tree
hW/R, R̄i, V (l n , x̄) = 0 for all x̄ ∈ W/R. Thus M Int |= ¬l N .
⊣
Thanks to the previous lemma, we can speak of the level of x̄ ∈ W/R, L(x̄), as
before. Actually, L(x̄) = L̄(x̄). We will also use the fact that, in this case, L(x̄) is
strictly monotone.
n
Lemma 6.2. V Int is R̄-monotone with respect to the variables pm
and bαn , and for
all n ≤ N ,
M Int |= Midn (ϕ).
n
Proof. For the first part of the proof we only treat pm
variables, as the arguments
n
for bα variables is entirely analogous.
If n > L(x̄), by (2) we know that
n
V Int (pm
, x̄) = 0.
So if R̄x̄ ȳ,
n
n
V Int (pm
, x̄) ≤ V Int (pm
, ȳ).
1000
DAVID FERNANDEZ
Now if x̄ ∈ W/R and n ≤ L(x̄), we claim
n
n
V Int (pm
, x̄) = V Int (pm
, ȳ)
whenever R̄x̄, ȳ.
We proceed by induction on the depth of ȳ (i.e., the maximum length of any chain
beginning at 0 and ending at ȳ ).
First consider ȳ−1 . Certainly R̄x̄, ȳ−1 so by IH,
n
n
V Int (pm
, x̄) = V Int (pm
, ȳ−1 ).
Recall that L(ȳ) has the property that V Int (l k , ȳ) = 1 for k ≤ L(ȳ).
Now, observe that L(x̄) ≤ L(ȳ−1 ) < L(ȳ). Therefore we must have
V Int (l L(x̄)+1 , ȳ) = 1.
But by (2), this implies
n
n
V Int (pm
, ȳ−1 ) = V Int (pm
, ȳ).
Hence,
n
n
, ȳ) = V Int (pm
, x̄).
V Int (pm
With this we can show that
Midn (ϕ) = l n →
^
^
(bαn ∨ ¬bαn ) ∧
^
n
n
(pm
∨ ¬pm
)
p 0≤m≤N −1
α∈sub2 (ϕ)
is valid in M Int .
If V Int (l n , x̄) = 1, L(x̄) ≥ n so by the previous argument,
n
n
V Int (pm
, x̄) = V Int (pm
, ȳ)
whenever R̄x̄ ȳ. But we have seen previously (section 3) that this is equivalent to
n
n
x̄ |=M Int pm
∨ ¬pm
.
n
yields
Repeating the argument with bαn in place of pm
x̄ |=M Int bαn ∨ ¬bαn .
Since m and α ∈ sub(ϕ) are arbitrary,
^
^
V Int
(bαn ∨ ¬bαn ) ∧
α∈sub2 (ϕ)
^
n
n
(pm
∨ ¬pm
) , x̄ = 1.
p 0≤m≤N −1
Since the result holds for all x̄ ∈ W/R, we conclude
M Int |= Midn (ϕ).
Lemma 6.3.
M Int |= Boxnα (ϕ).
Proof. Note that x̄ |= bαL(x) if and only if, for all m < N ,
Rx̄, ȳ =⇒ ym |=M α.
This in turn is equivalent to
L(y)
Rx̄, ȳ =⇒ ȳ |=C αm
.
⊣
A POLYNOMIAL TRANSLATION OF S4 INTO INTUITIONISTIC LOGIC
1001
And we know by Lemma 3.3 that this is equivalent to
M Int |= Boxnα (ϕ).
⊣
Proof of Theorem 6.1. It only remains to prove that
(M Int )S4 ∼
= M.
But an isomorphism f : M → (M Int )S4 is in fact given very naturally by
f(xm ) = (x̄)m .
One can check that the definitions of accessibility functions and valuations of
the translations are then literal inverses of each other and thus f is clearly an
isomorphism.
⊣
REFERENCES
[1] Patrick Blackburn, Maarten de Rijke, and Yde Venema, Modal logic, Cambridge University
Press, 2001.
[2] H. Friedman and R. C. Flagg, Epistemic and intuitionistic formal systems, Annals of Pure and
Applied Logic, vol. 32 (1986), pp. 53–60.
[3] Kurt Gödel, Collected works (Solomon Feferman, ed.), Oxford University Press, 1986.
[4] Saul Kripke, Semantical analysis of intuitionistic logic I, Crossley Formal Systems And Recursive
Functions, (1965), pp. 92–130.
[5] G. E. Mints, A short introduction to intuitionistic logic, Plenum Publishers, 2000.
[6] Alfred Tarski, Der Aussagenkalkül und die Topologie, Fundamenta Mathematica, vol. 31 (1938),
pp. 103–134.
DEPARTMENT OF MATHEMATICS
STANFORD UNIVERSITY
STANFORD, CALIFORNIA 94305, USA
E-mail: [email protected]