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

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?

-----------------
Lemma:

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)) )

Proof:

y is in S[x] iff (E z in x)(y = S[z])

w is in S[z] iff (E n in N)(A m in N)(m > n -> w in S_m(z)).

Note that S[z] c N, since each S_m(z) c N.

Thus,

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)) )

Lemma:

y is in T(x) iff (E n in N)(A m in N)(m > n -> (E z in x)(y = S_m(z)))

Proof:

By definition,

y is in T(x) iff (E n in N)(A m in N)( m > n -> y in S_m[x] ).

y is in S_m[x] iff (E z in x)(y = S_m(z)).
------------------

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.

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?

And hey! Don't be so rude. It doesn't serve any purpose.

Han de Bruijn

.