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A solution to a problem left open by lemmon regarding the hallden-completeness of the rmlc logical system. The author, dolph ulrich, proves that rmlc is hallden-incomplete by showing that it cannot prove certain formulas that are provable in both rm and lc, two related but distinct logical systems. The document also discusses the relationship between rmlc and other logical systems, such as dummett's lc and the intuitionistic sentential calculus, ic.
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Notre Dame Journal of Formal Logic Volume 22, Number 2, April 1981
A system S is Hallden-incomplete if and only if there are wffs A and B with no variables in common such that ^A v B but neither $A nor ^B, and strongly Hallden-incomplete if, in addition, A and B have but one variable apiece.* Evidently, all strongly Hallden-incomplete systems are Hallden- incomplete; Lemmon [5] poses the converse as an open problem. Consider the system RMLC, with detachment and adjunction as rules and, using standard conventions concerning relative binding strengths of connectives and omission of parentheses, the following axiom schemes: RO A-+(A-+A) Rl A^A R2 (A -> B) -* ((£ -> C) -> (A ~> C)) R3 A-+((A-> B)-» B) R4 (A -+(A-+B))-» (A -+B) R5 A&B-+A R6 A&B-+B R7 (A-+B)&(A-+C)'+(A-+(B& C)) R8 A -> A v B R9 B-^AMB RIO 04 -> C) & (5 -* C) -> ( U v B) -> C) DUMMETT U -* 5 ) v (5 -> A ) R l l i & ( 5 v C ) - > U & 5 ) v C R12 (A-»B)-*(B-+A) _ PRE TRANS U -> (5 -> ^1)) -> (A -• ( I -> 5)) RMLC W->>l)v (B^(C-+B)).
*The author wishes to thank N. D. Belnap, Jr., J. M. Dunn, and the anonymous referee for several suggestions for improving the presentation of this paper.
Received December 27, 1978; revised November 3, 1980
RMLC is clearly a subsystem of Dummett's LC [3], most of the above schemes being among those listed for ZC-duty in [6] (pp. 316-317) and the rest easily derived, e.g., PRE TRANS from the intuitionistiol -> (A -> B) by way of B -* (C->5), and RMLC from the latter by R9. RMLC is also contained in the system /tM(ingle) of [1], for R0-R12 are ^M-axioms (p. 341), DUMMETT is RM64 (p. 397), and PRE TRANS and RMLC are readily established. Indeed, RM and LC may be axiomatized by adding to RMLC (sche- matically) the left disjunct of RMLC for the former and the right for the latter: R0-R12 plus A -> A suffice for RM according to [1] (p. 341), while R2, R4-R10, DUMMETT, R12, PRE TRANS, and B -> (C-»fi) give a set equivalent, with minor adjustments, to one given in [6] (p. 317) for LC. A familiar, Hallden-style argument consequently completes a proof that the theorems of RMLC are precisely the wffs provable in both RM and LC. For assume ^fiC and \jx;C. Then there must be substitution i n s t a n c e s ^ ,.. .,Am ofA-*A and 2?!,.. .,Bn of B -+ (C~+B) such that Ax &... &Am fejxcC and Bt &... & Bn \RMLC C. It follows, by a proof similar to one in [1] (p. 302), that 04! &... & Am) v {Bx &... & Bn) ^MLCC^ whence eventually, after repeated distribution moves licensed by R5-R11 (and the transitivity of \RMLC)> (Ax v Bx) & 04! v B 2 ) &... & {Am v Bn) ^MICC- B^ Y RMLC, however, each A{ v Bj is available in RMLC, so that \RMLCC a s we^» finishing the argument.^1 For a solution to Lemmon's problem, now, let A and B have no variables in common, and just one each, and assume \RMLC^ V^ &- Then \j^A v B also. It is shown in [4] that the extensions (closed under substitution) of LC are linearly ordered, so it follows from Theorem 1 of [5] that LC is Hallden- complete. Thus, \j^A or \j^B. Arbitrarily, say \j^A. Then^l is a tautology of the classical, two-valued truth tables and, since these characterize the one- variable fragment of RM ([ 1 ], p. 413, Corollary 3.1), ^jA as well, whereupon \RMLCA anc* the latter system is thus not strongly Hallden-incomplete. Because A -> A is scarcely in LC, however, and B -> (C -> B) notoriously not in RM, neither disjunct of RMLC can be obtained in RMLC, so that RMLC is Hallden-incomplete..
[1] Anderson, A. R. and N. D. Belnap, Jr., Entailment, Princeton University Press, Prince- ton, New Jersey, 1975.