An axiom is defined to be a string of symbols phi a, phi b, phi c, and so on. There is nothing which is an element of a symbol. The union of all the symbols of the set of axioms must be finite. Call the union the alphabet of Q. Now, earlier I stated Q to be the set of all items which satisfy the axioms. That makes no sense if the axioms are just a string of symbols, so we need to introduce a “meaning” function. This function isomorphically maps a set of symbols from the alphabet to an item y. The item y is defined as an item/concept either in this universe or derived from something which is in this universe, or is derived from something that is derived from something that is in this universe, and so on. I term this property being an element of degree n of the universe, where n indicates how many times the “derived from something” is iterated. In other words, y must be something not 100% abstract – somewhere along the way, it had to be derived from something real. For our purposes, a universe is defined as any thing which responds to a “mind” ,i.e., Turing machine. The Turing machine receives its input from the universe, and its ouput alters the universe. Thus, the universe is any permutation of anything whatsoever that is consistent with the given Turing machine. Anyway – Q is thus the set of anything isomorphic in some sense to y. Furthermore, There has to be at least one Turing machine which is an element of Q. Now, what I am attempting to prove is that if we know that a mind M is an element of Q (say, if Q is the set of all minds) that the mere fact that M fully believes that the abstract formulation a.f. (a set of axioms) is the set which Q is based on, and that we know both the equations of M and a.f., then we can derive what the actual axioms are. Here is my idea: M maps his perception of the universe ( an approximation of some sorts based on the actual state of the universe) to an “idea” (some string of symbols). The truth function maps strings to [0,1]. And equivalent maps a string to another string with the same meaning (note that it is reflexive). So, we know what all of these functions map to – thus we can find their inverse. So, because of M’s belief, we have T(M(a.f. is equivalent to Q))=1
M(a.f. is equivalent to Q) = Inverse of T(1)
a.f. is equivalent to Q = Inverse of M(Inverse of T(1))
Now we are stuck – or are we? We have equivalent expressed in a different form the rest. So, change it to Equivalent(a.f.), which is the same as Equivalent(Q). Thus
Equivalent(a.f.) = Inverse of M(Inverse of T(1))
a.f. = Inverse of Equivalent(Inverse of M(Inverse of T(1)))
Now we take the set of all beta which could take place of a.f. in the equation. Call the set Y. We have established that Q must be an element of Y. However, it must also be an element of all beta which satisfy beta = Inverse of Equivalent(Inverse of M^2(Inverse of T(1))), where the two indicates iteration (Call this set Y2.) This is true because M thinks that he thinks that a.f. is equivalent to Q. In fact, Q must be in Yn for all n. Therefore, Q is an element of the intersection of Yn for all n. It also must have M as an element. So, we have greatly limited the possibilities already. It is my belief that for any a.f., there will only be one possibility for Q – but I cannot prove it, try as I might. I would greatly appreciate it if you looked over what I have so far, and then give some tips for what to do now.
Thank you very much for looking over this! I promise that it is all much easier to understand with symbols
. I myself can barely understand it with the words! I hope that you will be able to help! Bye!