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Re: A formal counter-example of Ax Ey P(x,y) -> Ey Ax P(x,y)
Posted:
Dec 10, 2012 11:54 PM
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On Dec 10, 2:31 pm, fom <fomJ...@nyms.net> wrote: > On 12/8/2012 11:13 PM, Dan Christensen wrote: > > > On Dec 8, 7:05 pm, fom <fomJ...@nyms.net> wrote: > >> On 12/8/2012 4:58 PM, Dan Christensen wrote: > > <snip> > > > I am proving that Ey Ax P(x,y) does not follow from Ax Ey P(x,y). > > | AxAyAz((x=y /\ y=z) -> x=z) > || AxAy(x=y -> y=x) > ||| -(d=c) > |||| AxEy(x=y) -> EyAx(x=y) > |||| AxEy(x=y) > |||| Ey(p=y) > ||||| p=c > |||||| EyAx(x=y) > |||||| Ax(x=d) > |||||| p=d > |||||| Ay(p=y -> y=p) > |||||| p=d -> d=p > |||||| d=p > |||||| AyAz((d=y /\ y=z) -> d=z) > |||||| Az((d=p /\ p=c) -> d=c) > |||||| (d=p /\ p=c) -> d=c > |||||| d=p /\ p=c > |||||| d=c > |||||| d=c /\ -(d=c) > > That is what the proof of your revised statement > looks like. And, contrary to your fast and impatient > response, it requires a formal assumption of > distinctness as George brought to your attention > elsewhere. > > Your model is a counter-example to any purported > proof of the original statement. With a model you > are proving that EyAx(x=y) is not a logical consequence > of AxEy(x=y). Logical consequence is a semantic > notion. A counter-example destroys claims of correct > derivation by virtue of the soundness theorem for > the deductive calculus of first-order logic.
OK, going off on a tangent here, what is the domain of this inference formula?
~p->f & ~p->~f ---------------- p
f,~f |- W
From a contradiction anything follows, but treats ~p
~p -> f
in the LHS as an *ASSUMPTION*.
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I'm trying to get it work similarly to MODUS PONENS
l & l->r -------- r
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Here is the problem:
Although:
~p->f & ~p->~f ---------------- p
it would be a FLUKE to directly match any implied formula f.
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would
(m&~p)->f & (m&~p)->~f ---------------- p
prove p for ALL-CASES IFF a contradiction arrives from ~p ?
M is unbound so it would match to any subset of theorems... should be possible to code in PROLOG.
Herc
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