at1with0 wrote:A lot can be said of unstructured free-form flow. There have been studies done on it and everything.
I personally love the direction my thread has taken.
But having said that, if I may be so bold, let me say something that is on topic. Truth and its relationship to logic.
http://en.wikipedia.org/wiki/Tarski%27s ... ty_theoremTarski's undefinability theorem, stated and proved by Alfred Tarski in 1936, is an important limitative result in mathematical logic, the foundations of mathematics, and in formal semantics. Informally, the theorem states that arithmetical truth cannot be defined in arithmetic.
The theorem applies more generally to any sufficiently strong formal system, showing that truth in the standard model of the system cannot be defined within the system.
Tarski, for those who don't know, was a giant in the field of logic in the first half of the 20th century when lots of interesting facts about logic were being discovered.
I don't understand the proof of this theorem but, intuitively, it makes a lot of sense given that "truth tables" are somewhere in there and using the letters T and F in a truth table does not define what truth is.
He was referring to the Peano axioms, not logic.The constant 0 is assumed to be a natural number:
0 is a natural number.
The next four axioms describe the equality relation.
For every natural number x, x = x. That is, equality is reflexive.
For all natural numbers x and y, if x = y, then y = x. That is, equality is symmetric.
For all natural numbers x, y and z, if x = y and y = z, then x = z. That is, equality is transitive.
For all a and b, if a is a natural number and a = b, then b is also a natural number. That is, the natural numbers are closed under equality.
The remaining axioms define the arithmetical properties of the natural numbers. The naturals are assumed to be closed under a single-valued "successor" function S.
For every natural number n, S(n) is a natural number.
Peano's original formulation of the axioms used 1 instead of 0 as the "first" natural number. This choice is arbitrary, as axiom 5 does not endow the constant 0 with any additional properties. However, because 0 is the additive identity in arithmetic, most modern formulations of the Peano axioms start from 0. Axioms 5 and 6 define a unary representation of the natural numbers: the number 1 is S(0), 2 is S(S(0)) (which is also S(1)), and, in general, any natural number n is Sn(0). The next two axioms define the properties of this representation.
For every natural number n, S(n) = 0 is False. That is, there is no natural number whose successor is 0.
For all natural numbers m and n, if S(m) = S(n), then m = n. That is, S is an injection.
Axioms 1, 6, 7 and 8 imply that the set of natural numbers is infinite, because it contains at least the infinite subset { 0, S(0), S(S(0)), … }, each element of which differs from the rest. To show that every natural number is included in this set requires an additional axiom, which is sometimes called the axiom of induction. This axiom provides a method for reasoning about the set of all natural numbers.
If K is a set such that:
0 is in K, and
for every natural number n, if n is in K, then S(n) is in K,
then K contains every natural number.
The induction axiom is sometimes stated in the following form:
If φ is a unary predicate such that:
φ(0) is true, and
for every natural number n, if φ(n) is true, then φ(S(n)) is true,
then φ(n) is true for every natural number n.
the answer to the earlier question is that the matrix is nondeterminate. Tried it took me a week. So At1 is correct.
i actually am high functioning autistic.
living the dream.