A Reformulation of Z-Reg.2
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From: MoeBlee on 16 Jun 2010 18:05 Re: Jun 16, 4:50 pm, zuhair <zaljo...(a)gmail.com>: It's very simple. Do I understand correctly that you claim?: (1) Your system proves Z-R. (2) Unique separation schema (as in Z-R set theory but with uniqueness existential quantifier) proves your uniqueness comprehension schema. But if those were true, then unique separation schema, union, and infinity together would prove the rest of Z-R. But that contradicts the proof (in Z-R, if I recall) that if Z-R is consistent then each axiom (of particular interest here, pairing and power set) of Z-R is independent of the other axioms of Z-R. So, if (1) and (2) are proven, then we have a proof of a contradiction in Z-R set theory. MoeBlee > On Jun 16, 4:15 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > On Jun 16, 2:32 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > Moe Blee in a previous post asked me to show the following: > > > > For ARBITRARY formula Phi (except we know x not free in Phi): > > > > all closures of > > > > ExAy(yex <-> (yez & Phi)). > > > > / > > > > I say the above statement is actually equivalent to the unique > > > separation schema mentioned above. > > > I haven't looked over your new formulations, but last night I did > > figure out a way to formulate your compression schema so that it and > > your union axiom entail the separation(with built in extensionality) > > schema, while the separation schema (with built in extensionality) > > entails your compression schema. > > > However, then it can't be the case that your compression schema, union > > schema, and infinity entail pairing and power set (did I understand > > correctly that you claim they do entail pairing and power set?), for > > then they would not be independent in Z set theory, though it has been > > proven that they are independent in Z set theory. > > > MoeBlee > > Well I don't know what is your formulation yet. But I think we need to > look at what I wrote here. Since there might be a mistake then. > > Anyhow let see the proof of pairing and power. > > (1)The proof of power: > > The formula in comprehension schema reads: > > Aw1...wn Ar As E!x Ay (y e x <-> (( y c r \/ y c s ) /\ > phi(y,w1,...,wn))) > > Now let r=s and let n=0, then we'll have: > > Ar E!x Ay (y e x <-> ( y c r /\ phi(y))) > > Let phi(y)<->y=y > > thus we'll end up with > > Ar E!x Ay (y e x <-> y c r) > > It is clear that x is the power-set of r. Thus power is proved. > > (2)The proof of pairing: > > from comprehension with n=2 ,we have: > > Aw1 Aw2 Ar As E!x Ay (y e x <-> ((y c r \/ y c s) /\ phi(y,w1,w2))) > > let phi(y,w1,w2)<->(y=w1 \/ y=w2) > > then: > > Aw1 Aw2 Ar As E!x Ay (y e x <-> ((y c r \/ y c s) /\ phi(y,w1,w2))) > > let phi(y,w1,w2)<-> (y=w1 \/ y=w2) > > then: > > Aw1 Aw2 Ar As E!x Ay (y e x <-> ((y c r \/ y c s) /\ (y=w1 \/ > y=w2))) > > take r=w1, s=w2 > > then: > > Aw1 Aw2 E!x Ay (y e x <-> ((y c w1 \/ y c w2) /\ (y=w1 \/ y=w2))) > > which is reduced to: > > Aw1 Aw2 E!x Ay (y e x <-> (y=w1 \/ y=w2)) > > which is pairing. > > I don't know what you meant about the issue with the independence of > these axioms in Z, and how that can be affected if they spring from > the same comprehension and union axioms. To me I don't see that this > would affect their > independence at all. > > Zuhair
From: zuhair on 16 Jun 2010 19:33 On Jun 16, 5:05 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > Re: Jun 16, 4:50 pm, zuhair <zaljo...(a)gmail.com>: > > It's very simple. > > Do I understand correctly that you claim?: > > (1) Your system proves Z-R. > > (2) Unique separation schema (as in Z-R set theory but with uniqueness > existential quantifier) proves your uniqueness comprehension schema. No of course not! Unique separation schema as you specified needs Pairing (or Boolean union) and Power with it in order to prove my comprehension schema. > > But if those were true, They are not! then unique separation schema, union, and > infinity together would prove the rest of Z-R. But that contradicts > the proof (in Z-R, if I recall) that if Z-R is consistent then each > axiom (of particular interest here, pairing and power set) of Z-R is > independent of the other axioms of Z-R. > > So, if (1) and (2) are proven, then we have a proof of a contradiction > in Z-R set theory. > > MoeBlee > > > On Jun 16, 4:15 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > On Jun 16, 2:32 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > > Moe Blee in a previous post asked me to show the following: > > > > > For ARBITRARY formula Phi (except we know x not free in Phi): > > > > > all closures of > > > > > ExAy(yex <-> (yez & Phi)). > > > > > / > > > > > I say the above statement is actually equivalent to the unique > > > > separation schema mentioned above. > > > > I haven't looked over your new formulations, but last night I did > > > figure out a way to formulate your compression schema so that it and > > > your union axiom entail the separation(with built in extensionality) > > > schema, while the separation schema (with built in extensionality) > > > entails your compression schema. > > > > However, then it can't be the case that your compression schema, union > > > schema, and infinity entail pairing and power set (did I understand > > > correctly that you claim they do entail pairing and power set?), for > > > then they would not be independent in Z set theory, though it has been > > > proven that they are independent in Z set theory. > > > > MoeBlee > > > Well I don't know what is your formulation yet. But I think we need to > > look at what I wrote here. Since there might be a mistake then. > > > Anyhow let see the proof of pairing and power. > > > (1)The proof of power: > > > The formula in comprehension schema reads: > > > Aw1...wn Ar As E!x Ay (y e x <-> (( y c r \/ y c s ) /\ > > phi(y,w1,...,wn))) > > > Now let r=s and let n=0, then we'll have: > > > Ar E!x Ay (y e x <-> ( y c r /\ phi(y))) > > > Let phi(y)<->y=y > > > thus we'll end up with > > > Ar E!x Ay (y e x <-> y c r) > > > It is clear that x is the power-set of r. Thus power is proved. > > > (2)The proof of pairing: > > > from comprehension with n=2 ,we have: > > > Aw1 Aw2 Ar As E!x Ay (y e x <-> ((y c r \/ y c s) /\ phi(y,w1,w2))) > > > let phi(y,w1,w2)<->(y=w1 \/ y=w2) > > > then: > > > Aw1 Aw2 Ar As E!x Ay (y e x <-> ((y c r \/ y c s) /\ phi(y,w1,w2))) > > > let phi(y,w1,w2)<-> (y=w1 \/ y=w2) > > > then: > > > Aw1 Aw2 Ar As E!x Ay (y e x <-> ((y c r \/ y c s) /\ (y=w1 \/ > > y=w2))) > > > take r=w1, s=w2 > > > then: > > > Aw1 Aw2 E!x Ay (y e x <-> ((y c w1 \/ y c w2) /\ (y=w1 \/ y=w2))) > > > which is reduced to: > > > Aw1 Aw2 E!x Ay (y e x <-> (y=w1 \/ y=w2)) > > > which is pairing. > > > I don't know what you meant about the issue with the independence of > > these axioms in Z, and how that can be affected if they spring from > > the same comprehension and union axioms. To me I don't see that this > > would affect their > > independence at all. > > > Zuhair
From: zuhair on 16 Jun 2010 19:39 On Jun 16, 5:05 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > Re: Jun 16, 4:50 pm, zuhair <zaljo...(a)gmail.com>: > > It's very simple. > > Do I understand correctly that you claim?: > > (1) Your system proves Z-R. > > (2) Unique separation schema (as in Z-R set theory but with uniqueness > existential quantifier) proves your uniqueness comprehension schema. No of course not! Unique separation schema as you specified needs Pairing (or Boolean union) and Power with it in order to prove my comprehension schema. > > But if those were true, They are NOT! >then unique separation schema, union, and > infinity together would prove the rest of Z-R. But that contradicts > the proof (in Z-R, if I recall) that if Z-R is consistent then each > axiom (of particular interest here, pairing and power set) of Z-R is > independent of the other axioms of Z-R. > > So, if (1) and (2) are proven, then we have a proof of a contradiction > in Z-R set theory. > > MoeBlee > > > On Jun 16, 4:15 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > On Jun 16, 2:32 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > > Moe Blee in a previous post asked me to show the following: > > > > > For ARBITRARY formula Phi (except we know x not free in Phi): > > > > > all closures of > > > > > ExAy(yex <-> (yez & Phi)). > > > > > / > > > > > I say the above statement is actually equivalent to the unique > > > > separation schema mentioned above. > > > > I haven't looked over your new formulations, but last night I did > > > figure out a way to formulate your compression schema so that it and > > > your union axiom entail the separation(with built in extensionality) > > > schema, while the separation schema (with built in extensionality) > > > entails your compression schema. > > > > However, then it can't be the case that your compression schema, union > > > schema, and infinity entail pairing and power set (did I understand > > > correctly that you claim they do entail pairing and power set?), for > > > then they would not be independent in Z set theory, though it has been > > > proven that they are independent in Z set theory. > > > > MoeBlee > > > Well I don't know what is your formulation yet. But I think we need to > > look at what I wrote here. Since there might be a mistake then. > > > Anyhow let see the proof of pairing and power. > > > (1)The proof of power: > > > The formula in comprehension schema reads: > > > Aw1...wn Ar As E!x Ay (y e x <-> (( y c r \/ y c s ) /\ > > phi(y,w1,...,wn))) > > > Now let r=s and let n=0, then we'll have: > > > Ar E!x Ay (y e x <-> ( y c r /\ phi(y))) > > > Let phi(y)<->y=y > > > thus we'll end up with > > > Ar E!x Ay (y e x <-> y c r) > > > It is clear that x is the power-set of r. Thus power is proved. > > > (2)The proof of pairing: > > > from comprehension with n=2 ,we have: > > > Aw1 Aw2 Ar As E!x Ay (y e x <-> ((y c r \/ y c s) /\ phi(y,w1,w2))) > > > let phi(y,w1,w2)<->(y=w1 \/ y=w2) > > > then: > > > Aw1 Aw2 Ar As E!x Ay (y e x <-> ((y c r \/ y c s) /\ phi(y,w1,w2))) > > > let phi(y,w1,w2)<-> (y=w1 \/ y=w2) > > > then: > > > Aw1 Aw2 Ar As E!x Ay (y e x <-> ((y c r \/ y c s) /\ (y=w1 \/ > > y=w2))) > > > take r=w1, s=w2 > > > then: > > > Aw1 Aw2 E!x Ay (y e x <-> ((y c w1 \/ y c w2) /\ (y=w1 \/ y=w2))) > > > which is reduced to: > > > Aw1 Aw2 E!x Ay (y e x <-> (y=w1 \/ y=w2)) > > > which is pairing. > > > I don't know what you meant about the issue with the independence of > > these axioms in Z, and how that can be affected if they spring from > > the same comprehension and union axioms. To me I don't see that this > > would affect their > > independence at all. > > > Zuhair
From: zuhair on 16 Jun 2010 20:03 On Jun 16, 4:15 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 16, 2:32 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > Moe Blee in a previous post asked me to show the following: > > > For ARBITRARY formula Phi (except we know x not free in Phi): > > > all closures of > > > ExAy(yex <-> (yez & Phi)). > > > / > > > I say the above statement is actually equivalent to the unique > > separation schema mentioned above. > > I haven't looked over your new formulations, but last night I did > figure out a way to formulate your compression schema so that it and > your union axiom entail the separation(with built in extensionality) > schema, while the separation schema (with built in extensionality) > entails your compression schema. How can you do that without pairing and power??? I don't see a way to do that. Actually it would be rather interesting to see your proof that unique separation with union only can prove my comprehension schema, frankly speaking I don't think you can do that without pairing and power. > > However, then it can't be the case that your compression schema, union > schema, and infinity entail pairing and power set (did I understand > correctly that you claim they do entail pairing and power set?), for > then they would not be independent in Z set theory, though it has been > proven that they are independent in Z set theory. > > MoeBlee
From: zuhair on 16 Jun 2010 21:18
On Jun 15, 8:39 pm, zuhair <zaljo...(a)gmail.com> wrote: > Theory A is the set of all sentences(entailed from FOL with identity > and membership) by the following non logical axioms: > > (1) Comprehension: for n=0,1,2,3,...; if phi(y,w1,...,wn) > is a formula in which y,w1,...,wn *are* its sole free variables > and in which x is not free,then > > Aw1...wn Ar As E!x Ay > ( y e x <-> (( y c r \/ y c s ) /\ phi(y,w1,...,wn))) > > were c is the subset relation. > > (2) Union: Ar Ex Ayer(ycx) > > were c is the subset relation. > > (3) Infinity: as in Z. > > / > > Theorem schema of unique separation: > > for n=1,2,3,...., if pi(y,w1,...,wn) is a formula in which > y,w1,...,wn *are* its sole free variables, and in which x is not free, > then > > Aw1...wn E!x Ay ( y e x <-> ( y e w1 & pi(y,w1,...,wn) ) ). Actually we can release some of the restrictions made on pi(y,w1,...,wn) We can define it as: for n=1,2,3,....., if pi(y,w1,...,wn) is a formula in which y *may* be free or w1 *may* be free; and in which w2,...,wn *are* free, and which contain no other free variable, and in which x is not free, then Aw1...wn E!x Ay ( y e x <-> ( y e w1 & pi(y,w1,...,wn) ) ). this would make pi(y,w1,...,wn) more an arbitrary formula than the one originally specified in this post. The line of proof is the same one as the original proof, only change the specifications of pi to the ones mentioned here. > > Proof: > > Let s=r in the comprehension schema, then for the same specifications > in comprehension we get: > > Aw1...wn Ar E!x Ay > ( y e x <-> (y c r /\ phi(y,w1,...,wn))) > > Let phi(y,w1,...,wn)<-> [y e w1 /\ pi(y,w1,...,wn)] > > were pi(y,w1,...,wn) is a formula in which y,w1,...,wn *are* its sole > free variables and in which x is not free. > > substitute in the above formula and we get: > > Aw1...wn Ar E!x Ay > ( y e x <-> (y c r /\ y e w1 /\ pi(y,w1,...,wn))) > > From union we have > > Aw1 Ew* Ayew1(ycw*) > > now w* is not unique, but we know that for every w1, at least one > w* must exist such that all members of w1 are subsets of w*. > > Now instantiate any one of the w* sets for r in the above formula, we > get: > > Aw1...wn E!x Ay > ( y e x <-> (y c w* /\ y e w1 /\ pi(y,w1,...,wn))) > > Now since every member of w1 is subset of w* (by definition of w* > sets). then we have > > (y c w* /\ y e w1) -> y e w1 > > thus the above formula would be reduced to: > > Aw1...wn E!x Ay > ( y e x <-> (y e w1 /\ pi(y,w1,...,wn))) > > Theorem schema of unique separation proved. > > Zuhair |