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Re: This is False. 0/0 {x | x ~e x} e {x | x ~e x} A single Principle
to Resolve Several Paradoxes
Posted:
Feb 3, 2013 4:56 PM
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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:
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> > > > > 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
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> > > > > 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}
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> > > > > 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?
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> 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
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