Re: 1-1/2+1/3-1/4+1/5-1/6+1/7

Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> writes:

Jesse F. Hughes wrote:

Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> writes:

Time to hook in with an improved version of my purported "Proof that ZFC
is inconsistent" (well, it's not the proper thread, but ..). Scrutinize,
please.

Proof that ZFC is inconsistent
------------------------------
You are *so* cute.

Lemma
-----
Let x, y be any sets, then (the _same_ function S as above):

y is in S[x] iff there is some n such that for all m > n
y is in S_m[x]

Here, n e N , m e N and S_m[x] defined as above.

Proof
-----
As an excercise for the interested reader ;-)
Bull***.
I've already said that this is false, so how do you dare omit the
proof?

But, as far as I can see, you HAVEN'T proved that they are NOT the same.

In <87d4r3r3ur.fsf@xxxxxxxxxxxxx>, I explicitly anticipated this
argument. Let T(x) = { y in N | (E n in N)(A m in N)( m > n -> y in
S_m[x] ) }
Then we see that
y is in S[x] iff (E z in x)(A w in N)( w in y <->
(E n in N)(A m in N)(m > n -> w in S_m(z)) )
[the corresponding formula in the previous post was not right, but
see
below] while y is in T(x) iff (E n in N)(A m in N)(m > n -> (E z in
x)(y = S_m(z)))
If you want to show that S[x] = T(x), then you sure as hell can't
leave the proof to *this* reader, because I don't see any reason that
it's true.

But, you don't see any reason that it's NOT true either. Am I right?

*You* proved it. *You* showed that S[N] = {N} and T(N) = {}. This
constitutes a proof that S[N] != T(N) and hence that, in general,
S[x] != T(x).

Proof left to reader my ass, Han. That was either just ***-stupid
or dishonest. I had *already* suggested[1] that the lemma is
false. If you think otherwise, then have the balls to prove it.

Yeah, and let Jesse F. Hughes have the balls to _disprove_ it. Read
on.

You provided the proof yourself.

Footnotes: [1] Though, to be honest, my description of S[x] was
simply off.

I agree with you that 90 percent a proof is not a proof, and that I
HAVE to proof the Lemma that was left as "an excercise to the
reader". Though you could have noticed my smiley .. Shall we
conclude that the only flaw in the purported proof is exactly
_there_? Is there anything else wrong? If not, then I _only_ have
to repair this weakest link in the chain. OK?

Yes, if you prove S[N] = T(n) after having already proved that they
are not equal, then you're done.

No idea why you think that 90% of the task through, however. In any
case, the last 10% of an impossible task is the hardest part, I hear.

--
"Four little piggies went to market.
This little piggy wanted ice cream
And Mama said 'Nooooo'!"
--A new Piggy Song, by Quincy Hughes (age 3)
.