Article View: comp.lang.prolog
Article #992Find the bug
From: Sanjai Narain
Date: Wed, 18 Jul 1990 20:54
Date: Wed, 18 Jul 1990 20:54
35 lines
1303 bytes
1303 bytes
What is the bug in the following reasoning?
Claim: The following set of clauses, S, is inconsistent:
{p(x)->x=a, x=a->p(x)} U EQ_0
where EQ_0 is a set of clauses containing ->z=z, z a variable, and for
each pair of distinct 0-ary function symbols c and d, the clause c=d->
(i.e. ~(c=d)).
Proof: We first show that S->exists(u).p(u) is valid. We do this by
assuming ~exists(u).p(u), adding it to S, and showing the result to be
inconsistent. We obtain:
S, (u).~p(u)
In clausal form we obtain:
S, p(u)->
Now, p(u)-> resolves with x=a->p(x) to yield x=a->. This resolves
with ->z=z in EQ_0 to yield ->. Contradiction.
We now show that S->~exists(u).p(u) is valid. We add exists(u).p(u) to S.
Skolemizing, we obtain:
S, ->p(c) where c is a new 0-ary constant.
->p(c) resolves with p(x)->x=a to yield ->c=a. This resolves with c=a->
in EQ_0 to yield ->. Contradiction.
Thus, S is inconsistent. QED.
Sanjai Narain
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Path: arc1-ca.pugleaf.net!utzoo!attcan!uunet!seismo!dimacs.rutgers.edu!bcm!rice!cs.utexas.edu!usc!randvax!narain From: narain@randvax.UUCP (Sanjai Narain) Newsgroups: comp.lang.prolog Subject: Find the bug Keywords: skolemization Message-ID: <2635@randvax.UUCP> Date: 18 Jul 90 20:54:46 GMT Organization: Rand Corp., Santa Monica, Ca. Lines: 35