sci.logic
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Re: how to define Thm in PRA

Subject: Re: how to define Thm in PRA
From: "george"
Date: 28 Oct 2006 11:42:17 -0700
Newsgroups: sci.logic
Daryl McCullough wrote:
> george says...
>
> >Peter_Smith wrote:
> >
> >> Take a language lacking quantifiers and variables,
> >
> >If only you knew how.
> >
> >> 0 =/= Sm
> >> Sm = Sn -> m = n
> >> m + 0 = m
> >> m + Sn = S(m + n)
> >> m x 0 = 0
> >> m x Sn = m x n + n
> >> m^0 = S0
> >> m^Sn = m^n x n
> >>
> >> (keep on going in the obvious way ...).   Here the m and n are
> >> placeholders for standard numerals, not variables:
> >
> >JEEzus; just go   yourself.
> >Q: How many legs does a dog have, if you call its tail a leg?
> >A: Four.  Calling a tail a leg doesn't make it one.
> >And calling m and n not-variables here doesn't make them not
> >variables.
>
> They are not variables.

YES THEY ARE, dumbass.
YOU AND PETER YOURSELVES HAVE BOTH ALREADY CONCEDED
that m and n above are universally quantified bound variables.
What you were falsely accuing ME of NOT understanding was that
they were schematic variables in the meta-language as opposed to
variables in the object language.

> Is 0 a variable? Is S(0) a variable?
> Is S(S(0)) a variable?

That was NEVER even being DISCUSSED.
My reply was that *m* was a variable, YOU FLAMING SHITHEEL.

> Peter is saying that the (infinitely many) axioms of his
> theory are
>
>     0 =/= S(0)
>     0 =/= S(S(0))
>     0 =/= S(S(S(0)))
>     etc.
> which he summarizes in the metalanguage (the language he
> is using to *describe* the language) as
>
>     0 =/= S(m)

His decision to do that MAKES m a VARIABLE, DUMBASS.


> Why is it that when you don't understand something,
> you start using profanity and insults?

I did NOT START by using profanity and insults.
I STARTED by replying "m is a variable".

So why are YOU such A DAMN LIAR?


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