Re: Question: What conditions are sufficient to prove a subset does not exist?

On Jul 31, 12:06 pm, "Jesse F. Hughes" <je...@xxxxxxxxxxxxx> wrote:
Scott <ToaTe...@xxxxxxxxx> writes:
On Jul 31, 9:28 am, LauLuna <laureanol...@xxxxxxxx> wrote:
This is not hard. You have:

1. A exists <-> f exists

2. A exists -> contradiction

Hence:

3. A does not exist
4. f does not exist

In particular any attempt to show that the diagonal argument fails
because A is ill defined is debunked by the biconditional in 1.

Sorry for my slowness, but I still don't understand this particular
point you're trying to make. Does not 1 & 3 mean there is an
inconsistency since 1 & 3 -> A exists <-> A does not exist?

No. You have P <-> Q and ~P. That does not imply P <-> ~P.

We can easily derive ~Q from P <-> Q and ~P.

We can also derive P -> ~P, but that is fairly irrelevant and follows
from ~P without any need for (1). We cannot derive P <-> ~P from
P <-> Q and ~P.

Okay. But the following would still be true?

A -> ~f (2,4)
~f -> ~A (1)
A & ~A (Conjunction)
A <-> ~A (Rewrite)

likewise

A -> ~f (2,4)
A -> f (1)
f & ~f (Conjunction)
f <-> ~f (Rewrite)

.

Loading