Re: The MetaMathematical Theorem that Almost Was


Charlie-Boo wrote:
> What does "If the system is inconsistent, we can prove everything."
> mean?

Exactly that.
That's the DEFINITION of "inconsistent system",
for these purposes.

There is a natural/intuitive definition of "inconsistent" that seems,
at first, superficially, to be different from this, namely, that
system is inconsistent if it derives a contradiction, if there is
some P such that the system derives both P and ~P.
But under the logical inference rules normally used,
it follows that from that contradiction, one can, derive/prove
any&every sentence whatsoever ("ex falso quodlibet").
So a system/theory is inconsistent if and only if (BY DEFINITION)
it proves everything.

Through slightly non-standard rules of inference, it is possible
to be able to prove a contradiction but then NOT be able to
prove everything-else-you-can-write as a further consequence
of it. THOSE systems are called "paraconsistent". We are not
bothering with any of them in the current discussion. So it will
HELP you if you migrate AWAY from thinking of "inconsistent"
as meaning "proves a contradiction" TOWARD thinking of "inconsistent"
as meaning "proves everything" (therefore including contradictions).


That accomplished, we must now return to your original question,
which was "how do you prove" (whatever -- in your case it was
"the existence of the empty set" but it doesn't even matter).
The point is that in principle, anything that is not a contradiction
can be proven: YOU JUST MAKE IT AN AXIOM, and voila, there
is the proof, the shortest proof possible: "It's an axiom."

If you're not going to make it an axiom then it might be a validity
or a tautology, in which case you could prove it just from the rules
of inference of your logic. But if it is neither a validity nor a
contradiction
then there is by definition NO way to prove it without assuming SOME
axioms. Only You can decide what axioms You think are relevant or
worthy
for the proof; obviously you always COULD decide to make the statement
itself an axiom.

If your question was specifically, "Does ZF actually NEED an axiom
asserting the existence of an empty set", the answer is, no, it
doesn't.
The standard semantics of classical first-order logic assumes
the domain is non-empty. If there exists a set at all (which there
must,
even without any set-existence axioms) then the existence
of the empty set follows from the application of the axiom of
separation
to that set.

to that set

.

Relevant Pages

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