NST2 and theorems.
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From: zuhair on 10 Nov 2009 20:18 Theorems: theorem: Exist x for all y ( y e x <-> y is a set ) proof: let phi<-> y is a set substitute in classes.QED theorem: Exist! x for all y ( y e x <-> y is a set ) proof: Extensionality Define(V):- x=V <-> for all y ( y e x <-> y is a set ) theorem: Exist x for all y ~ y e x proof:let phi <-> ~y=y substitute in classes.QED theorem: Exist! x for all y ~ y e x proof: Extensionality Define(0):- x=0 <-> for all y ( ~ y e x ) theorem: for every set a Exist x For all y ( y e x <-> y=a ) proof: let phi <-> y=a since a is a set then from classes we'll have the set theorem above.QED theorem: for every set a Exist! x For all y ( y e x <-> y=a ) proof: Extensionality Define: x={a} <-> for all y ( y e x <-> y=a ) theorem: for all x: ~ x e x proof: suppose Exist x: x e x then {x} would not be disjoint of x violating foundation. from that we have ~ V e V. Now it is clear that from classes we can have a class that is an unordered pair of any two sets, from that we can define ordered pair in Kuratowski manner, also we can separation and replacement of sets, unions both finite and infinite also can be defined, but all these are classes the set-hood of which is not yet determined. theorem: 0 is a set. proof: let phi<-> ~y=y substitute in Sets and we get c={y|~y=y} Now ~y=y has no parameter and no constant in it so we need to check the following first: for all c,z ((c={y|~y=y} & c subclass z) -> z=z)) which is trivially true. now we check the next condition: ~for all x ( Mx ->Exist y (~y=y & x e y)) since from axiom 3 we do have at least on set existing, then the above condition would be fulfilled since Exist y (~y=y & x e y) is trivially false, then for all x ( Mx ->Exist y (~y=y & x e y)) is trivially false , so its negation is ture. Now we check the proper subformulas condition, now the only proper subformula of ~y=y is the formula y=y so we only need to prove the following: ~for all y (My~y=y<->Myy=y) since there exist at least one set (axiom 3) then the above sentence is trivially true. thus all conditions in the antecedent of SETS schema are fulfilled, thus the class {y|~y=y} is a set, thus 0 is a set. pairing,union,power,separation and replacement of sets all can be proved in this theory, so is infinity . Proofs shall be posted later. So it appears as if MK is a equivalent to this theory? Zuhair
From: zuhair on 10 Nov 2009 20:26 Theorems: theorem: Exist x for all y ( y e x <-> y is a set ) proof: let phi<-> y is a set substitute in classes.QED theorem: Exist! x for all y ( y e x <-> y is a set ) proof: Extensionality Define(V):- x=V <-> for all y ( y e x <-> y is a set ) theorem: Exist x for all y ~ y e x proof:let phi <-> ~y=y substitute in classes.QED theorem: Exist! x for all y ~ y e x proof: Extensionality Define(0):- x=0 <-> for all y ( ~ y e x ) theorem: for every set a Exist x For all y ( y e x <-> y=a ) proof: let phi <-> y=a since a is a set then from classes we'll have the set theorem above.QED theorem: for every set a Exist! x For all y ( y e x <-> y=a ) proof: Extensionality Define: x={a} <-> for all y ( y e x <-> y=a ) theorem: for all x: ~ x e x proof: suppose Exist x: x e x then {x} would not be disjoint of x violating foundation. from that we have ~ V e V. Now it is clear that from classes we can have a class that is an unordered pair of any two sets, from that we can define ordered pair in Kuratowski manner, also we can separation and replacement of sets,unions both finite and infinite also can be defined, but all these are classes the set-hood of which is not yet determined. theorem: 0 is a set. proof: let phi<-> ~y=y substitute in Sets and we get c={y|~y=y} Now ~y=y has no parameter and no constant in it so we need to check the following first: for all c,z ((c={y|~y=y} & c subclass z) -> z=z)) which is trivially true. now we check the next condition: ~for all x ( Mx ->Exist y (~y=y & x e y)) since from axiom 3 we do have at least on set existing, then the above condition would be fulfilled since Exist y (~y=y & x e y) is trivially false, then for all x ( Mx ->Exist y (~y=y & x e y)) is trivially false , so its negation is ture. Now we check the proper subformulas condition, now the only proper subformula of ~y=y is the formula y=y so we only need to prove the following: ~for all y (My~y=y<->Myy=y) since there exist at least one set (axiom 3) then the above sentence is trivially true. thus all conditions in the antecedent of SETS schema are fulfilled, thus the class {y|~y=y} is a set, thus 0 is a set. pairing,union,power,separation and replacement of sets all can be proved in this theory, so is infinity . Proofs shall be posted later. So it appears as if MK is a equivalent to this theory? Zuhair
From: zuhair on 10 Nov 2009 20:27 Theorems: theorem: Exist x for all y ( y e x <-> y is a set ) proof: let phi<-> y is a set substitute in classes.QED theorem: Exist! x for all y ( y e x <-> y is a set ) proof: Extensionality Define(V):- x=V <-> for all y ( y e x <-> y is a set ) theorem: Exist x for all y ~ y e x proof:let phi <-> ~y=y substitute in classes.QED theorem: Exist! x for all y ~ y e x proof: Extensionality Define(0):- x=0 <-> for all y ( ~ y e x ) theorem: for every set a Exist x For all y ( y e x <-> y=a ) proof: let phi <-> y=a since a is a set then from classes we'll have the set theorem above.QED theorem: for every set a Exist! x For all y ( y e x <-> y=a ) proof: Extensionality Define: x={a} <-> for all y ( y e x <-> y=a ) theorem: for all x: ~ x e x proof: suppose Exist x: x e x then {x} would not be disjoint of x violating foundation. from that we have ~ V e V. Now it is clear that from classes we can have a class that is an unordered pair of any two sets, from that we can define ordered pair in Kuratowski manner, also we can separation and replacement of sets,unions both finite and infinite also can be defined, but all these are classes the set-hood of which is not yet determined. theorem: 0 is a set. proof: let phi<-> ~y=y substitute in Sets and we get c={y|~y=y} Now ~y=y has no parameter and no constant in it so we need to check the following first: for all c,z ((c={y|~y=y} & c subclass z) -> z=z)) which is trivially true. now we check the next condition: ~for all x ( Mx ->Exist y (~y=y & x e y)) since from axiom 3 we do have at least on set existing, then the above condition would be fulfilled since Exist y (~y=y & x e y) is trivially false, then for all x ( Mx ->Exist y (~y=y & x e y)) is trivially false , so its negation is true. Now we check the proper subformulas condition, now the only proper subformula of ~y=y is the formula y=y so we only need to prove the following: ~for all y (My~y=y<->Myy=y) since there exist at least one set (axiom 3) then the above sentence is trivially true. thus all conditions in the antecedent of SETS schema are fulfilled, thus the class {y|~y=y} is a set, thus 0 is a set. pairing,union,power,separation and replacement of sets all can be proved in this theory, so is infinity . Proofs shall be posted later. So it appears as if MK is a equivalent to this theory? Zuhair
From: zuhair on 11 Nov 2009 18:44
On Nov 10, 8:16 pm, zuhair <zaljo...(a)yahoo.com> wrote: > NST2: is the set of all sentences entailed (from FOL with identity > and e) by the following non logical axioms below the following > definition of the notion "set" denoted by M: > > Define(set): x is a set <-> Exist y ( x e y ) > > Define(M): Mx <-> x is a set. > > 1. Extensionality: for all z ( z e y <-> z e x) -> x=y > > 2. Foundation: Exist y y e x -> Exist y (y e x & y disjoint x) > > 3. Classes: If phi is a formula in which at least y is free, and in > which x is not free, then all closures of > > Exist x for all y ( y e x <-> My phi ) > > are axioms. > > Define: x={y|phi} <-> for all y ( y e x <-> My phi ) > > 4. Set existence: Exist x: x is a set. > > Q is a proper substring of phi, means that Q is a substring of phi > and Q is not the formula phi. > > Q is said to be a proper subformula of phi if > Q is a proper substring of phi that is a formula, that has the same > principal free variable as phi, and the set of all parameters of Q is > a subset of the set of all parameters of phi. > > 5. Sets:If phi is a formula in which at least y or z are free, and in > which c is not free, with parameters x1,...,xn, and constants > k1,...,kp , and Q1,...,Qm are all proper subformulas of phi, then all > closures of: > > Mx1,...,Mxn & Mk1,...,Mkp > for all c,z ((c={y|phi} & c subclass z) -> ~ phi(z)) & > ~for all x ( Mx ->Exist y (phi(y) & x e y)) & > ~for all y (Myphi<->MyQ1)&...&~for all y(Myphi<->MyQm) > > -> Exist c (Mc & for all y ( y e c <-> phi ) > > are axioms. > > Theory definition finished/ > > Theorems with proofs shall be posted later. > > Zuhair This theory is inconsistent, simply take Phi to be " y is singleton and for all z ( z e y -> z is singleton )" Then we'll have the set {y|phi} were phi is the above formula, rendering the theory inconsistent since the union of this set is the class of all sets which will be a set, now this theory prove that the subset of any set is a set, thus the Russell class would be a set, which is an obvious contradiction! Zuhair |