Cont: Deeper than primes - Continuation 2

[jsfisher]

"Next" is part of neither your definition for your non-standard successor relation nor for your set-completeness attribute.

"next" is the idea at the basis of my successor relation

It wasn't included in the definition, so it isn't part of the definition.

Perhaps you could show us a set that would be complete under your definitions. Is {A, B} complete?

Yes

Well, that's curious, isn't it?

Didn't you already tell us that {{A, B}} was the successor of {A, B}? And since {A, B} would be the specific case for S in your completeness definition, that would make NOT({{A, B}} IsSuccessorOf {A, B}) the right-hand side of the if-and-only-if expression.

And then since NOT({{A, B}} IsSuccessorOf {A, B}) is logically FALSE, the left-hand side of the if-and-only-if expression must also be FALSE.

Therefore, {A, B} is incomplete.

...according to your definitions. Why do you contract them now?


ETA: Edited to align with the negation actually presented in the original.

 

[Little 10 Toes]

Follow-up question, if {1,2} is complete, why does it need a successor?

 

[doronshadmi]

Is {A} a successor of A?
Is {A, B} a successor of A?
Is A a successor of {}?
Is {B} a successor of {A, B, C}?

In order to avoid confusion (since {S}not-$S holds), the answers are:

Is {A} a successor of A? In case that {A}$A is used.
Is {A, B} a successor of A? No.
Is {B} a successor of {A, B, C}? No.
Is A a successor of {}? In case that A={{}} AND {{}}${} is used.

 

[doronshadmi]

Didn't you already tell us that {{A, B}} was the successor of {A, B}?

Not necessarily, for example: by {{A, B}}not-${A, B}, {{A, B}} is not a successor of {A, B}, and in this case {A, B} is complete.

 

[doronshadmi]

Follow-up question, if {1,2} is complete, why does it need a successor?

Who says it does?

For example: if {{1,2}not-${1,2}, then {1,2} is complete.

 

[jsfisher]

Not necessarily, for example: by {{A, B}}not-${A, B}, {{A, B}} is not a successor of {A, B}.

{{A, B}} either is or it is not, pick one, the successor of {A, B} according to you and your useless definitions.

You chose "is" in this post.

You don't get to change on a whim. {{A, B}} remains the successor of {A, B}.

 

[doronshadmi]

{{A, B}} remains the successor of {A, B}.

Both options are logically defined: ({{A, B}}${A, B}) OR ({{A, B}}not-${A, B})

EDIT:

In case of {{A, B}}${A, B}, incompleteness holds.

In case of {{A, B}}not-${A, B}, incompleteness does not hold.

 

[doronshadmi]

Please look at http://www.internationalskeptics.com/forums/showpost.php?p=11268705&postcount=1323.

 

[doronshadmi]

I wish to be clearer about an expression like {{A, B}}${A, B}.

It is actually:

{ {A, B} }${ A, B },
{ {A, B, {A, B}} }${ A, B, {A, B} },
{ {A, B, {A, B}, {{A, B, {A, B}}} }${ A, B, {A, B}, {{A, B, {A, B}} },

... etc. ad infinitum

 

[jsfisher]

Both options are logically defined: ({{A, B}}${A, B}) OR ({{A, B}}not-${A, B})

Of course both are defined. And by definition, {{A, B}} IsSuccessorOf {A, B} is true, and NOT({{A, B}} IsSuccessorOf {A, B}) is false.

 

[jsfisher]

It is actually...

No. That isn't the way you defined it. It isn't that way now.

 

[doronshadmi]

Of course both are defined.

Both are true and there is no contradiction since there is an OR condition between them.

 

[jsfisher]

Both are true and there is no contradiction since there is an OR condition between them.

If they are both true, than by the very definition of the term, there is a contradiction.

Congratulations. You have concluded that Doronetics is contradictory. It may be dismissed as useless.

 

[doronshadmi]

No. That isn't the way you defined it. It isn't that way now.

It is exactly how it is defined.


{ {A, B} }not-${ A, B } provides completeness.


{ {A, B} }${ A, B },
{ {A, B, {A, B}} }${ A, B, {A, B} },
{ {A, B, {A, B}, {{A, B, {A, B}}} }${ A, B, {A, B}, {{A, B, {A, B}} },

... etc. ad infinitum, provides incompleteness.

 

[doronshadmi]

If they are both true, than by the very definition of the term, there is a contradiction.

Congratulations. You have concluded that Doronetics is contradictory. It may be dismissed as useless.

("is a successor of") OR ("is not a successor of") is not contradictory exactly as
("is a member of") OR ("is not a member of") is not contradictory.

 

[doronshadmi]

jsfisher http://www.internationalskeptics.com/forums/showpost.php?p=11264035&postcount=1249 is very clear about $ mechanism (where not-$ does not provide such mechanism).

 

[jsfisher]

("is a successor of") OR ("is not a successor of") is not contradictory exactly as
("is a member of") OR ("is not a member of") is not contradictory.

Both of those are gibberish.

Isasuccessorof, isamemeberof, and their negations are dyadic relations. You need some set-valued formulae to complete each expression.

You do remember about well-formed formulae, don't you?

 

[doronshadmi]

Both of those are gibberish.

Isasuccessorof, isamemeberof, and their negations are dyadic relations.

Isasuccessorof OR ~Isasuccessorof is not contradictory, exactly as isamemeberof OR ~Isasuccessorof is not contradictory.

More simpler, it is equivalent to T OR ~T.

EDIT:

Please tell me jsfisher, how exactly T OR ~T is contradictory?

 

[jsfisher]

Isasuccessorof OR ~Isasuccessorof is not contradictory, exactly as isamemeberof OR ~Isasuccessorof is not contradictory.

And they are both gibberish for exactly the same reason.

More simpler, it is equivalent to T OR ~T.

T OR ~T is a valid expression were T, presumably, is also a valid expression. Isasuccessorof, on the other hand, isn't a valid expression.

Please tell me jsfisher, how exactly T OR ~T is contradictory?

Why would you think it is?

 

[doronshadmi]

on the other hand, isn't a valid expression.

$ is a valid expression, and so is ~$.

Some example:

In case of {y}~$y, y remains the same (it is completed).

In case of {y}$y, y does not remain the same (it is not completed).

(Some example is seen in http://www.internationalskeptics.com/forums/showpost.php?p=11268767&postcount=1334).

-------------------

You are right about "isamemeberof" since:

"y isamemeberof {y}" is T (holds), where "y ~isamemeberof {y}" is F (it does not hold).

-------------------

So, after all, $ OR ~$ behaves differently than "isamemeberof".