Re: bXc = b -> b=0 with regularity but not infinity

["Rupert"]

Rupert wrote:
> MoeBlee wrote:
> > Here's one that has me stumped:
> >
> > In ZR(-I) (that's Z set theory with the axiom of regularity but not the
> > axiom of infinity), prove:
> >
> > bXc = b -> b=0
> >
> > where 'X' stands for the Cartesian product.
> >
> > You may not use the 'infinite sequence of nested members' version of
> > regularity. But the following previous theorems are available (where
> > 'e' stands for 'is a member of') along with the usual "first chapter"
> > theorems of set theory:
> >
> > ~SeS
> >
> > ~SeTeS
> >
> > ~SeTeVeS
> >
> > S subset SXS -> S=0
> >
>
> If bXc=b, then (bXc)Xc=b, so b=bXc=bX(cXc), and ran(b)=c=cXc=0, so b=0.
>
> But I still have no idea how you prove this result with regularity but
> without infinity.
>

I see how to do it now.

> > Source: 'Axiomatic Set Theory' by Suppes, pg. 56, exercise 2 (Dover
> > edition).