Download a theorem on well-finite standard consequence operations

Bulletin of the Section of Logic
Volume 2/3 (1973), pp. 159–163
reedition 2014 [original edition, pp. 159–165]
Stephen L. Bloom
A THEOREM ON WELL-FINITE STANDARD
CONSEQUENCE OPERATIONS
In this note we obtain a theorem characterizing those finite consequence
operations on a sentential language that can be defined in a proof-theoretic
way from a finite number of axiom schemes and/or structural (polynomial)
rules. This theorem can be regarded as a strengthening of a theorem of
Tarski ((T), theorem 27) on finite axiomatizability.
Let L be an absolutely free algebra of a finite similarity type having a
countably infinite number of generators. We will refer to the members of
L as polynomials. P (L) is the collection of all subsets of L.
1. Definition. A consequence operation on L is a function C : P (L) →
P (L) such that
1.1. X ⊆ C(X) = C(C(X)), all X ⊆ L.
1.2. If X ⊆ Y , then C(X) ⊆ C(Y ), all X, Y ⊆ L.
C is a finite consequence operation if
[
C(X) = (C(Y ) : Y ⊆ X, Y finite).
C is a structural consequence operation if
for every endomorphism h of L,
h(C(X)) ⊆ C(h(X)).
C is standard if C is both finite and structural.
2. Definition. An n-ary rule r is a subset of Ln ; if (π, δ1 , . . . , δn ),
n ≥ 0, is a sequence of members of L, the polynomial rule r(π1 δ1 ...δn ) is the
(n + 1)-ary rule defined as follows:
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Stephen L. Bloom
(a, b1 , . . . , bn ) ∈ r(π1 δ1 ...δn ) iff for some endomorphism h of L, h(π) = a
and h(δi ) = bi , i = 1, . . . , n.
Let Q be any set of rules.
3. Definition. CQ : P (L) → P (L) is the function defined as follows:
a ∈ CQ (X) iff there is a finite sequence x1 . . . xn such that xn = a and
for each i, 1 ≤ i ≤ n, xi ∈ X or there is some (k + 1)-ary rule r ∈ Q,
and j1 , . . . , jk < i such that (xi , xj1 , . . . , xjk ) ∈ r. (In this case, we call the
sequence x1 , . . . , xn a CQ -proof of a from X.)
The following theorem is due to Roman Suszko and has been quoted
in (RW).
4. Theorem. C is a standard consequence operation on L iff there is a
finite or countable set Q of polynomial rules such that C = CQ .
5. Definition. A consequence operation C is well-finite if C = CQ for
some finite set of polynomial rules.
6. Example. The classical two-valued propositional calculus can be given
by 3 unary polynomial rules (axiom schemes) and one 3-ary polynomial
rule (modus ponens, considered as the rule r(p,q,(q→p)) (where p, q are free
generators of L.)
7. Definition. Let C1 , C2 be consequence operations on L. C1 ≤ C2 iff
every X ⊆ L, C1 (X) ⊆ C2 (X).
It is well-known that with this ordering the set of all consequence operations on L forms a complete lattice.
8. Lemma. Suppose C1 ≤ C2 ≤ C3 ≤ . . . is an infinite chain of finite
consequence operations. Let C = sup{Cn : n = 1, 2, . . .}. Then, for every
X ⊆ L,
[
C(X) =
Cn (X).
n
S
Proof. Let D(X) = Cn (X). It is easy to check that D satisfies 1.1
and 1.2. In order to see that D(D(X)) ⊆ D(X), assume a ∈ D(D(X)).
Then for some n, a ∈ Cn (D(X)). Since Cn is a finite consequence operation, there are b1 , . . . , bk ∈ D(X) such that a ∈ Cn ({b1 , . . . , bk }). Since the
Cn ’s are nested, there is an m such that {b1 , . . . , bk } ⊆ Cm (X). Thus
A Theorem on Well-Finite Standard Consequence Operations
161
a ∈ Cn (Cm (X)). Now Cm (X) ⊆ Cm+n (X), since Cm ≤ Cm+n ; also
Cn (Cm+n (X)) ⊆ Cn+m (Cn+m (X)). Hence a ∈ Cn+m (Cm+n (X)) =
Cm+n (X). Thus a ∈ D(X), and we have shown D is a consequence operation.
S
Clearly Cn ≤ D, so C ≤ D. Since Cn ≤ C for all n, we see Cn (X) ⊆
C(X), so that D ≤ C. Thus D = C, completing the proof.
9. Lemma. Suppose
Ci = CQi , i ∈ I, where for each i, Qi is a set of
S
rules. Let Q = (Qi : i ∈ I). Then CQ = sup{Ci : i ∈ I}.
Proof. Clearly Ci ≤ CQ , so that sup{Ci : i ∈ I} < CQ . Let D be any
consequence operation such that Ci ≤ D, all i ∈ I. Then, by induction
on the length of CQ -proofs, it is easy to see that CQ ≤ D, so that CQ =
sup{Ci : i ∈ I}.
10. Lemma. Let C be a standard consequence operation. If C is not
well-finite, there is an infinite sequence C1 ≤ C2 ≤ . . . of standard (and
well-finite) consequence operations such that C = sup{Cn : n = 1, 2, . . .}
and C 6= Cn , any n.
Proof. Let Q be an infinite set of polynomial rules such that C = CQ .
Say Q = {r0 , r1 , r2 , . . .}. Let Qn = {r0 , . . . , rn−1 }, n = 1, 2, . . . and set
Cn = CQn .
From 9 it follows that C = sup{Cn }, and since C is not well-finite,
C 6= Cn , any n.
We may now prove the main theorem.
11. Theorem. A finite structural consequence C on L is not well-finite iff
there is a sequence C1 ≤ C2 ≤ C3 ≤ . . . of standard consequence operations
such that C = sup{Cn }, C 6= Cn , any n.
Proof. Lemma 10 showed that the condition is necessary. Now suppose
C is the sup of the Cn as above. By 4 there are sets S and Qn of polynomial
rules such that C = CS and Cn = CQn , n = 1, 2, . . .. For each (1 + m)ary rule r(π,δ1 ,...,δm ) , m ≥ 0, in S, we have π ∈ CS (δ1 , . . . , δm ) (if m = 0,
π ∈ CS (∅).). Then π ∈ CQn (δ1 , . . . , δm ) for some n. Let A(π,δ1 ,...,δm ) be the
finite subset of Qn used in the CQn -proof of π from δ1 , . . . , δn . Let A be the
union of all sets A(π,δ1 ,...,δn ) . In order to obtain a contradiction, suppose S
is a finite set of rules. Then A is finite, and we can find an integer n0 such
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Stephen L. Bloom
that if r(π,δ1 ,...,δm ) is a rule in S and A(π,δ1 ,...,δm ) ∈ Qn then n ≤ n0 .
We will show C = CS ≤ CA ≤ Cn0 . Indeed, suppose a ∈ CS (X). Let
x1 , . . . , xt be a CS -proof of a from X. We show that for each i, 1 ≤ i ≤ t,
xi ∈ CA (X).
It i = 1, then either x1 ∈ X or x1 follows from a unary rule rπ in S.
In the first case, clearly x1 ∈ CA (X). In the second case, x1 = h(π) for
some endomorphism h of L. But by construction, π ∈ CAπ (∅), and hence
π ∈ CA (∅). By 4 then x1 = f (π) ∈ CA (∅) ⊆ CA (X).
Now assume for j < i, xj ∈ CA (X). If (xi , xj1 , . . . , xjm ) belongs to the
rule r(π,δ1 ,...,δm ) in S, then, for some endomorphism h of L, xi = h(π) and
xjk = h(δk ), k = 1, . . . , m. By construction, π ∈ CA(π,δ1 ,...,δm ) (δ1 , . . . , δm ) ⊆
CA (δ1 , . . . , δm ). Since all rules in A are polynomial rules,
h(π) ∈ CA (h(δ1 ), . . . , h(δm )), by 4; i.e.
xi ∈ CA (xj1 , . . . , xjm ).
But by the induction assumption, {xj1 , . . . , xjm } ⊆ CA (X). Thus xi ∈
CA (CA (X)) = CA (X), completing the proof that CS ≤ CA .
Since A is a finite set of rules, each one of which belongs to some Qn ,
n < n0 , it follows from 8 that CA ≤ sup{Cn : n < n0 } ≤ Cn0 .
Thus on the assumption that S is finite, we have shown C = CS ≤ Cn
for some n0 . But this contradicts the assumption that Cn < C, all n, and
completes the proof of the theorem.
12. Example. Let L be the sentential language having only one unary
connective, say F . Let q be one of the free generators of L, and let F n x
be an abbreviation for F F . n. . F x. If 2 = p0 , 3 = p1 , . . . are the primes in
increasing order, let Rn be the binary polynomial rule determined by the
sequence (F pn q, q). Finally, put Q = {R0 , R1 , . . .}. CQ is not well-finite.
Indeed, if so, CQ ≤ CQn for some n, where Qn = {R0 , . . . , Rn }. But for
example
F pn+1 q ∈ CQ (q) but
F pn+1 q 6∈ CQn (q),
as can be easily proved by induction on the length of CQn -proofs.
The author is pleased to acknowledge several helpful conversations with
Roman Suszko on the topic of this paper.
A Theorem on Well-Finite Standard Consequence Operations
163
References
[T] A. Tarski, Fundamental Concepts of the Methodology of the Deductive Sciences, Reprinted in Logic, Semantics and Metamathematics,
Oxford (1956).
[RW] R. Wójcicki, Investigations into methodology of sentential calculi
(I), Institute of Philosophy and Sociology, Polish Academy of Sciences,
Warsaw (1971).
Stevens Institute of Technology
Hoboken, N.J. 07030