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

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?

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.

-----------------
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.


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

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