
Re: layer logic: a new dimension to logic?
Posted:
Sep 18, 2011 1:52 PM


Now let´s have a look at the layer set theory, my favourite part of layer theory:
The central idea is to treat ?x is element of set M? (x e M) as a layer statement: It is true in layer t+1 that set x is element of the set M, if the statement A(x) is true in layer t.
Equality of layer sets: W (M1=M2, d+1) = W ( For all t: W(xeM1,t) = W(xeM2,t) , 1 ) Especially: W (M=M, d+1)=w for d>=0.
The empty set 0:
W(x e 0, t+1) := W( W( x e 0, t ) = w , 1 ) = w for t>=0.
The full set All:
W(x e All, t+1) := W( W( x e All, t ) = w v W( x e All, t ) = u v W( x e All, t ) = w , 1 ) = w for t>0 and =u for t=0.
So other than in most set theories in layer theory the full set is a normal set.
Axiom M1 (assignment of statements to sets): W(x e M, t+1) := W ( W ( A(x), t ) =w1 v W ( A(x), t ) =w2 v W ( A(x), t ) =w3 , 1 ) with w1,w2,w3 = w,u,w For every layer set M there exists a layer logic statement A(x) witch fulfils for all t=0,1,2, ?: W(x e M, t+1) = W ( W ( A(x), t ) = w v W ( A(x), t ) = w , 1 ) W(x e M, 0+1) = W ( W ( A(x), 0) = w v W ( A(x), 0 ) = w , 1 ) = W (u=w v u=w, 1 ) = w
Axiom M2 (sets defined by statements): For every layer logic statement A(x) about a layer set x there exits a layer set M so that for all t=0,1,2,3,? holds: W(x e M, t+1) := W ( A(x), t ) (or the expressions of axiom M1).
Definition M3 (definition of meta sets): If F is a logical function (like identity, negation or f.e. FoW(xeM1,t) = W(xeM1,t)=w ) then the following equation defines a meta set M: (M1=M is allowed): W(x e M, t+1) := W ( F o W(x e M1, t), 1 )
Consequences of the axioms and definitions: In layer 0 all sets are u: W( x e M, 0 ) = u (as all statements in layer 0).
In layers > 0: W(x e M, t+1) := w if W ( A(x), t ) = w else W(x e M, t+1) := w
For all x and (normal layer) sets M holds: W(x e M, 1) = u (as W(A(x),0)=u). For all x and meta sets M holds: W(x e M, 1) = w or ?w
Last not least let´s look upon the Russell set:
Classic definition: RC is the set of all sets, that do not have themselves as elements RC:= set of all sets x, with x ?e x
In layer theory: W(x e R, t+1) := W ( W ( x e x, t ) = w v W ( x e x, t ) = u , 1 )
W(x e R, 0+1) = W ( W ( x e x, 0 ) = w v W ( x e x, 0 ) = u , 1 ) = W (u=w v u=u , 1 ) = w
Therefore W(R e R,1) = w W(R e R,2) = W ( W ( R e R, 1 ) = w v W ( R e R, 1 ) = u , 1 ) = W (w=w v w=u , 1 ) = w And so W(R e R,3) = w, W(R e R,4) = w , ?
R is a set with different elements in different layers, but that is no problem in layer set theory.
As All, the set of all sets, is a set in layer theory, it is no surprise, that the diagonalisation of cantor is a problem no more (I just give the main idea):
Be M a set and P(M) its power set and F: M > P(M) a bijection between them (in layer d) Then the set A with W(x e A, t+1) = w := if ( W(x e M,t)=w and W(x e F(x),t)=w ) A is a subset of M and therefore in P(M). So it exists x0 e M with A=F(x0). First case: W(x0 e F(x0),t)=w , then W(x0 e A=F(x0), t+1) = w (no contradiction, as in another layer) Second case: W(x0 e F(x0),t)= w then W(x0 e A=F(x0), t+1) = w (no contradiction, as in another layer) If we have All as M and identity as Bijektion F we get for the set A: W(x e A, t+1) = w := if ( W(x e All,t)=w and W(x e x),t)=w ) = if ( W(x e x),t)=w ) This is the layer Russell set R (I omitted the ´u´value for simplification) and no problem.
So in layer theory we have just one kind of infinity ? and no more cantor´s paradise ?
Yours, Trestone

