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5 Craig Interpolation Theorem
5 Craig Interpolation Theorem
THEOREM 4 (CRAIG INTERPOLATION THEOREM FOR Lm1,). Let cp, I)be sentences of L,,, such that != cp + I). Then there is a sentence 8 of L,,, such that t cp + 8, t 8 + I),and every relation, function, or constant symbol of L which occurs in 8 occurs both in cp and in Ic/. (The above result is due to LOPEZ-ESCOBAR [1965], who gave a prooftheoretic argument using cut elimination. The result generalizes a theorem of Craig for first order logic. The proof below is due to Makkai, and is an adaptation to L,,, of a proof of Craig’s Theorem for L by Henkin.) PROOF.Let X , be the set of all sentences cp‘ of M,,, such that every relation, constant, or function symbol of L which occurs in cp’ occurs in cp, and also only finitely many c E C occur in cp’. Define X, similarly. Let S be the set of all finite sets s of sentences of M,,, which can be expressed as a union s = s1 u s2 such that: (1)
81
= X,,
s2
= X,.
(2) If 8,, O2 E Xp n X , and k A s1 -+ consistent.
e l , c A s2
-+
02, then 8, A O2 i s
We claim that S is a consistency property. Let us show, for example, that (C5) holds for S. Let s = s1 u s2 E S and V Z E s. Assume first that V Z E S ~Then . V Z E X, and C c X,. We shall show that for some a E Z, s’ = s u {a} E S with s; = s, u {a}, si = s2. Suppose not. Then for all a E Z there exist el,, OZg E X, n X, such that 19
20
and
CRAIG INTERPOLATION THEOREM
el, A 8,,
k
AS, A C
-+
el,,
k AS, -, e,,,
is inconsistent. But then
c A S,
+
v el,,
1 S,
4
aEz
A
e,,.
ad
Thus if we put = VasXOla,8, = /\,EZ8Z,, we see that (1) and (2 are contradicted. This shows that S has the property (C5). The case that V Z E s, is similar. The other parts of the proof that : is a consistency property are left to the reader. We remark that (C6) de pends on the fact that only finitely many c E C occur in each s E S. Fo (C7) we note that we may assume that each relation constant, or functioi symbol of L occurs in cp or $. Recalling that the terms in (C7) are basil terms, we see that all the equations c = t belong to either X , or X , . Since S i s a consistency property, by the Model Existence Theorem eacl s E S has a model. But != cp -, $, so the set {cp, 1$} has no model Therefore {cp, 1$} # S. Also, cp E X,, $ E X , This means that we cai find 8,, 8, EX, n X, such that
*
k cp -, 81, I=1
and
.
-+
02,
8, A 8, is inconsistent. It follows that k 8, .+ -, 8,, k- -, 8,
4
$, whence k
-+
$.
Let us write dl = Bl(cl . . . c,) where cl, . . ., c, are all the c E C occur ring in el. Choose new variables y l , . . .y,,, and let 8 = Vyl. . .ynOl(y . .y,). Then 0 is a sentence of L,,, which satisfies the desired conclusion
.
..
COROLLARY. Let cp(x, . x,), $(xl . . . x,) be formulas of L,,,, sucl that k cp -+ $. Then there is a formula 8 ( x , . . . x,) of L,,, such tha k cp + 8, k 8 4 $, and every relation, function, or constant symbol whicl. occurs in 8 occurs in both cp and $.
PROOF.Replace x l , . . ., x, by new constants e l , . . ., en and apply tht interpolation theorem.4 REMARK: In the Craig Interpolation Theorem we cannot always finc 8 SO that:
51
21
CRAIG INTERPOLATION THEOREM
If the equality symbol For example, consider
occurs in 8, then it occurs in both cp and $.
=
b (Vx, Y ) ( X = Y )
+
(Vx, Y ) ( W
-
NY)).
THEOREM 5 (BETH'STHEOREM FOR L,,,). (LOPEZ-ESCOBAR [1965]). Let P and Q be two new relation symbols with n places. Let q (P ) be a sentence of the language (L u {P}),,,, and let q ( Q ) be formed by replacing each occurrence of P by Q. Suppose that (i)
CP(P)A d Q ) (P(x1 * +
*
-
xn)
-
Q(xl** * xn)),
i.e. cp(P) defines P implicitly. Then there is a formula O(xl L,,, such that (ii) k cp(P) + (P(x,
. . . x,,)
i.e. q(P) defines Pexplicitly.
-
8(x1 .
. . .x,,)
of
. . x,,)),
PROOF.From (i) we have
I= ~P(P)AP(xI * * xn)
+
(q(Q) Q(x1 +
.
-
*
xn)).
By the above corollary there is a formula O(x, . .x,,) of L,,, (1) t cp(P>AP(X,. . x,,) -, q x l . . . x,,), XI . xn) (v(Q) Q(x1 * * * X n ) ) . (2) From the second line (2) it follows that, by replacing Q by P, (3) b e(x, . . . x,,) -+ ( c p ( ~-,) ~ ( x .,. . x,,)). Then by (1) and (3) we conclude that (ii) ho1ds.i
- -
.
+
such that
+
In first-order logic L it is easy to see, using the compactness theorem, that Craig's Interpolation Theorem is equivalent to A. Robinson's Consistency Theorem (which was proved at about the same time, perhaps a bit earlier):
THEOREM. Let L',L" be two expansions of the ROBINSON CONSISTENCY language L such that L' n L" = L (i.e. every symbol common to L' and L" belongs to L). Let T be a complete theory in L and let cp be a sentence of L' and $ a sentence of L". If T u {cp} is consistent and T u {$} is consistent, then T u {cp, $} is consistent.
22
CRAIG INTERPOLATION THEOREM
This theorem fails for the logic L,,,; its proof does not go throu; because the compactness theorem is no longer available. The followi set of exercises explains the situation in more detail. A set T of sentenc of L,,, is said to be a complete theory iff T has a model and for any t\ models %,23 of T we have % = 23(L,,,). PROBLEMS 1. (DOWNWARD LOWENHEIM-SKOLEM THEOREM). If CJ is a countable s of sentences of L,,, and @ has a model, then @ has a finite or countal: model. HINT:Use the Model Existence Theorem. 2. If all countable models of a sentence cp of L,,, are isomorphic ai cp has a countable model, then {cp} is a complete theory. (LO$-VAUG. TEST). 3. (A WEAK ROBINSON CONSISTENCY THEOREM). Let L‘, L” be expansio of L with L‘ n L” = L. Let T be a countable complete theory of L,, Let.cp beasentence of L:,, and $ one of Ud,,. If T u {cp} and T u { each has a model, then T u (cp, $} has a model. HINT:Use Scott’s Theorem.
4. Show that problem 1 fails for some uncountable set CJ. 5. Show that problem 2 fails with ‘model of power o’in place of ‘cou able model’.
6. Show that problem 3 fails for uncountable T. 7. (THEMODELEXISTENCE THEOREM WITHOUT EQUALITY). Assume tl L has no function or constant symbols. Let S be a set of countable s of sentences of M,,, such that each of the rules (Cl)-(C6) of Consistc cy Properties except for the equality rule holds. Assume further tl the equality symbol does not occur in any s E S. Then every s E S ha: model %, and the set of elements of 3 is C.
8. (ALTERNATE DEFINITION OF CONSISTENCY PROPERTY). Let X be a to1 logical space. By a topological interpretation of the language M,,, mean a mapping f on the set of all sentences of M,,, into the set of
51
23
CRAIG INTERPOLATION THEOREM
a) Letf be a topological interpretation of M,,,. Let Scf) be the set of all finite or countable sets s of sentences of Ma,,such that
nfm z 0.
(P-
Prove that S ( f ) is a consistency property. b) Let S be a consistency property. Let X be the topological space where S is the set of all points of X,and a set U c X is open 'iff s E U and s c s1 E S implies s1 E U. Letfs be the function
f&)
= (s E
s:cp E s>.
Prove that fsis a topological interpretation of M,,,
.
c) If S is a consistency property, then S ( f s ) is the consistency property S ( j s ) = {sl: s1 c s j o r some s E S ] .
THEOREM 4 (CRAIG INTERPOLATION THEOREM FOR Lm1,). Let cp, I)be sentences of L,,, such that != cp + I). Then there is a sentence 8 of L,,, such that t cp + 8, t 8 + I),and every relation, function, or constant symbol of L which occurs in 8 occurs both in cp and in Ic/. (The above result is due to LOPEZ-ESCOBAR [1965], who gave a prooftheoretic argument using cut elimination. The result generalizes a theorem of Craig for first order logic. The proof below is due to Makkai, and is an adaptation to L,,, of a proof of Craig’s Theorem for L by Henkin.) PROOF.Let X , be the set of all sentences cp‘ of M,,, such that every relation, constant, or function symbol of L which occurs in cp’ occurs in cp, and also only finitely many c E C occur in cp’. Define X, similarly. Let S be the set of all finite sets s of sentences of M,,, which can be expressed as a union s = s1 u s2 such that: (1)
81
= X,,
s2
= X,.
(2) If 8,, O2 E Xp n X , and k A s1 -+ consistent.
e l , c A s2
-+
02, then 8, A O2 i s
We claim that S is a consistency property. Let us show, for example, that (C5) holds for S. Let s = s1 u s2 E S and V Z E s. Assume first that V Z E S ~Then . V Z E X, and C c X,. We shall show that for some a E Z, s’ = s u {a} E S with s; = s, u {a}, si = s2. Suppose not. Then for all a E Z there exist el,, OZg E X, n X, such that 19
20
and
CRAIG INTERPOLATION THEOREM
el, A 8,,
k
AS, A C
-+
el,,
k AS, -, e,,,
is inconsistent. But then
c A S,
+
v el,,
1 S,
4
aEz
A
e,,.
ad
Thus if we put = VasXOla,8, = /\,EZ8Z,, we see that (1) and (2 are contradicted. This shows that S has the property (C5). The case that V Z E s, is similar. The other parts of the proof that : is a consistency property are left to the reader. We remark that (C6) de pends on the fact that only finitely many c E C occur in each s E S. Fo (C7) we note that we may assume that each relation constant, or functioi symbol of L occurs in cp or $. Recalling that the terms in (C7) are basil terms, we see that all the equations c = t belong to either X , or X , . Since S i s a consistency property, by the Model Existence Theorem eacl s E S has a model. But != cp -, $, so the set {cp, 1$} has no model Therefore {cp, 1$} # S. Also, cp E X,, $ E X , This means that we cai find 8,, 8, EX, n X, such that
*
k cp -, 81, I=1
and
.
-+
02,
8, A 8, is inconsistent. It follows that k 8, .+ -, 8,, k- -, 8,
4
$, whence k
-+
$.
Let us write dl = Bl(cl . . . c,) where cl, . . ., c, are all the c E C occur ring in el. Choose new variables y l , . . .y,,, and let 8 = Vyl. . .ynOl(y . .y,). Then 0 is a sentence of L,,, which satisfies the desired conclusion
.
..
COROLLARY. Let cp(x, . x,), $(xl . . . x,) be formulas of L,,,, sucl that k cp -+ $. Then there is a formula 8 ( x , . . . x,) of L,,, such tha k cp + 8, k 8 4 $, and every relation, function, or constant symbol whicl. occurs in 8 occurs in both cp and $.
PROOF.Replace x l , . . ., x, by new constants e l , . . ., en and apply tht interpolation theorem.4 REMARK: In the Craig Interpolation Theorem we cannot always finc 8 SO that:
51
21
CRAIG INTERPOLATION THEOREM
If the equality symbol For example, consider
occurs in 8, then it occurs in both cp and $.
=
b (Vx, Y ) ( X = Y )
+
(Vx, Y ) ( W
-
NY)).
THEOREM 5 (BETH'STHEOREM FOR L,,,). (LOPEZ-ESCOBAR [1965]). Let P and Q be two new relation symbols with n places. Let q (P ) be a sentence of the language (L u {P}),,,, and let q ( Q ) be formed by replacing each occurrence of P by Q. Suppose that (i)
CP(P)A d Q ) (P(x1 * +
*
-
xn)
-
Q(xl** * xn)),
i.e. cp(P) defines P implicitly. Then there is a formula O(xl L,,, such that (ii) k cp(P) + (P(x,
. . . x,,)
i.e. q(P) defines Pexplicitly.
-
8(x1 .
. . .x,,)
of
. . x,,)),
PROOF.From (i) we have
I= ~P(P)AP(xI * * xn)
+
(q(Q) Q(x1 +
.
-
*
xn)).
By the above corollary there is a formula O(x, . .x,,) of L,,, (1) t cp(P>AP(X,. . x,,) -, q x l . . . x,,), XI . xn) (v(Q) Q(x1 * * * X n ) ) . (2) From the second line (2) it follows that, by replacing Q by P, (3) b e(x, . . . x,,) -+ ( c p ( ~-,) ~ ( x .,. . x,,)). Then by (1) and (3) we conclude that (ii) ho1ds.i
- -
.
+
such that
+
In first-order logic L it is easy to see, using the compactness theorem, that Craig's Interpolation Theorem is equivalent to A. Robinson's Consistency Theorem (which was proved at about the same time, perhaps a bit earlier):
THEOREM. Let L',L" be two expansions of the ROBINSON CONSISTENCY language L such that L' n L" = L (i.e. every symbol common to L' and L" belongs to L). Let T be a complete theory in L and let cp be a sentence of L' and $ a sentence of L". If T u {cp} is consistent and T u {$} is consistent, then T u {cp, $} is consistent.
22
CRAIG INTERPOLATION THEOREM
This theorem fails for the logic L,,,; its proof does not go throu; because the compactness theorem is no longer available. The followi set of exercises explains the situation in more detail. A set T of sentenc of L,,, is said to be a complete theory iff T has a model and for any t\ models %,23 of T we have % = 23(L,,,). PROBLEMS 1. (DOWNWARD LOWENHEIM-SKOLEM THEOREM). If CJ is a countable s of sentences of L,,, and @ has a model, then @ has a finite or countal: model. HINT:Use the Model Existence Theorem. 2. If all countable models of a sentence cp of L,,, are isomorphic ai cp has a countable model, then {cp} is a complete theory. (LO$-VAUG. TEST). 3. (A WEAK ROBINSON CONSISTENCY THEOREM). Let L‘, L” be expansio of L with L‘ n L” = L. Let T be a countable complete theory of L,, Let.cp beasentence of L:,, and $ one of Ud,,. If T u {cp} and T u { each has a model, then T u (cp, $} has a model. HINT:Use Scott’s Theorem.
4. Show that problem 1 fails for some uncountable set CJ. 5. Show that problem 2 fails with ‘model of power o’in place of ‘cou able model’.
6. Show that problem 3 fails for uncountable T. 7. (THEMODELEXISTENCE THEOREM WITHOUT EQUALITY). Assume tl L has no function or constant symbols. Let S be a set of countable s of sentences of M,,, such that each of the rules (Cl)-(C6) of Consistc cy Properties except for the equality rule holds. Assume further tl the equality symbol does not occur in any s E S. Then every s E S ha: model %, and the set of elements of 3 is C.
8. (ALTERNATE DEFINITION OF CONSISTENCY PROPERTY). Let X be a to1 logical space. By a topological interpretation of the language M,,, mean a mapping f on the set of all sentences of M,,, into the set of
51
23
CRAIG INTERPOLATION THEOREM
a) Letf be a topological interpretation of M,,,. Let Scf) be the set of all finite or countable sets s of sentences of Ma,,such that
nfm z 0.
(P-
Prove that S ( f ) is a consistency property. b) Let S be a consistency property. Let X be the topological space where S is the set of all points of X,and a set U c X is open 'iff s E U and s c s1 E S implies s1 E U. Letfs be the function
f&)
= (s E
s:cp E s>.
Prove that fsis a topological interpretation of M,,,
.
c) If S is a consistency property, then S ( f s ) is the consistency property S ( j s ) = {sl: s1 c s j o r some s E S ] .