Date: Feb 3, 2013 4:56 PM
Author: Graham Cooper
Subject: Re: This is False. 0/0 {x | x ~e x} e {x | x ~e x} A single Principle<br> to Resolve Several Paradoxes
On Feb 4, 7:18 am, Charlie-Boo <shymath...@gmail.com> wrote:

> On Feb 3, 4:03 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

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> > On Feb 4, 3:01 am, Charlie-Boo <shymath...@gmail.com> wrote:

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> > > On Feb 1, 3:35 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

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> > > > On Feb 2, 4:09 am, Charlie-Boo <shymath...@gmail.com> wrote:

>

> > > > > There is a peculiar parallel between Semantic Paradoxes, Set Theory

> > > > > Paradoxes and ordinary formal Arithmetic.

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> > > > > Consider the following 3 pairs of expressions in English, Set Theory

> > > > > and Mathematics:

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> > > > > A

> > > > > This is false.

> > > > > This is true.

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> > > > > B

> > > > > 1/0

> > > > > 0/0

>

> > > > > C

> > > > > {x | x ~e x} e {x | x ~e x}

> > > > > {x | x e x} e {x | x ~e x}

> > > > > {x | x ~e x} e {x | x e x}

> > > > > {x | x e x} e {x | x e x}

>

> > > > > A is the Liar Paradox, B is simple Arithmetic, and C is Russell?s

> > > > > Paradox.

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> > > > This is Russells Paradox

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> > > > {x | x ~e x} e {x | x ~e x}

> > > > <->

> > > > {x | x ~e x} ~e {x | x ~e x}

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> > > > To make a consistent set theory the formula { x | x ~e x }

> > > > must be flagged somehow.

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> > > How do you define a wff - precisely? That is the problem. Frege was

> > > right, Russell was wrong, and all you need is an exact (formal)

> > > definition of wff.

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> > > C-B

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> > in the usual manner by Syntactic construction.

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> > IF X is a WFF

> > THEN ALL(Y) X is a WFF

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> > and so on.

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> The problem isn't with the connectives. What can X be for starters -

> the most primitive wffs from which we build others?

>

> C-B

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>

http://en.wikipedia.org/wiki/First_order_logic#Formation_rules

In PROLOG we use lowercase words for TERMS

and uppercase words for VARIABLES

ATOMIC PREDICATE

p( a1, a2, a3, ... an)

where ak is either a term or a variable.

p is also a term.

The connectives are superfluous, just use

if( X, Y )

not( X )

and( X, Y)

which are special predicates in that their arguments are predicates

themselves.

-----

For Quantifiers, all solved variables EXIST()

and I need a routine for SUBSET( var, set1, set2 )

which can do quantifier ALL(var).

A(x):D P(x)

<=>

{ x | x e D } C { x | P(x) }

Now all Predicate Calculus can be expressed in Atomic Predicates.

p(a,b,c)

Herc