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- Newsgroups: sci.logic
- Path: sparky!uunet!destroyer!sol.ctr.columbia.edu!zaphod.mps.ohio-state.edu!magnus.acs.ohio-state.edu!wjcastre
- From: wjcastre@magnus.acs.ohio-state.edu (W.Jose Castrellon G.)
- Subject: Re: Why invalid?
- Message-ID: <1992Nov18.002216.15277@magnus.acs.ohio-state.edu>
- Summary: Solution
- Sender: news@magnus.acs.ohio-state.edu
- Nntp-Posting-Host: bottom.magnus.acs.ohio-state.edu
- Organization: The Ohio State University,Math.Dept.(studnt)
- References: <1992Nov16.091653.1@woods.ulowell.edu> <lgfubaINN7v@peaches.cs.utexas.edu> <1992Nov17.164246.18220@uklirb.informatik.uni-kl.de>
- Date: Wed, 18 Nov 1992 00:22:16 GMT
- Lines: 48
-
- In article <1992Nov17.164246.18220@uklirb.informatik.uni-kl.de> ayala@uklirb.informatik.uni-kl.de (Mauricio Ayala R.) writes:
-
- >Can anyone give me a counterexample of the following:
- >
- >Let T a consistent first order theory with equality which
- >decidable set of quatifier-free logical consequences:
- >
- > QF(T) = { B | t |= B, B quantifier-free }
- >
- >Let A be a T-consistent quantifier-free formula.
- >Then (T,A) is also a consistent first order logic with
- >equality and QF(T,A) is ALSO DECIDABLE.
- >
- >M. Ayala
- >Universitaet Kaiserslautern
- >FRG
- >ayala@informatik.uni-kl.de
- >
-
- Solution follows.
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- Take the theory over the language with one constant 0 , one function f(x) ,
- one unary predicate P(x), and a binary relation R(x,y), as follows:
-
- For all x R(0,f(x))
- R(0,0)
- and for each n in a non-recursive r.e. set, the axioms:
-
- [ There exists an x such that not R(0,x) ] --> [ P(n) ]
-
- (where n is the abbreviation of f(x) applied n times to 0 )
-
- Then one can check that the set of quantifier free sentences that are logical
- consequences of this axioms is recursive.
-
- However if one expands the language with one more constant c , and add the
- sentence not R(0,c), then that set becomes non-recursive. [if you dont allow
- an expansion of the language, then it can be shown that there is no counter-
- example].
-