I did not say that a Gödel statement does contradict ZFC.

I meant that computing a halting condition for any given statement can take so much time, so that the series of processed statements converges. Citing from the paper:

1. Z … enumerates, one after the other, each of the provable statements in ZFC

2. Z … halts if it ever reaches a contradiction,

3. Because this machine will enumerate every provable statement in ZFC, it will run forever if and only if ZFC is consistent.

Let’s take that step 2. requires t(i) time, so at step i the machine’s performance equals to 1 / t(i) statements in the unit of time. The total amount of processed statements would be equal to a sum of series 1 / t(i).

The statement 3. implicitly claims that this sum is infinite (“every provable statement”), whereas it will converge to a certain finite number, due to increase of contradiction check complexity.

]]>First, there’s never any violation of the Incompleteness Theorem, not if the machine runs forever and not if it halts.

Second, a Gödel statement doesn’t contradict ZFC (why would it?).

Third, the Z machine never makes any non-obvious semantic judgments about what statements do or don’t contradict ZFC. If we ignore the complications caused by the use of Friedman’s statement, such a machine simply halts if it finds a ZFC proof of 0=1 and runs forever if it doesn’t: that’s all.

]]>In that case the machine will run forever and still will not violate the Second Incompleteness Theorem.

]]>Scott, I implicitly made you the arbiter of how the question resolves. 🙂

Comments welcome, on either metaculus.com or here.

]]>there is the existence of an odd perfect number.

In pseudo C:

int i, s, d;

i = 1;

while(1){

s = 0;

for(d = 1; d < i; d++) if(i%d == 0) s = s+d;

if(s == i) return;

i = i+2;

} ]]>