Deeper than primes

[doronshadmi]

Without ~(p and ~p) being a tautology you need to demonstrate that you have a way of doing proofs.

Robin, you have missed http://www.internationalskeptics.com/forums/showpost.php?p=6365450&postcount=11706.

We haven't yet covered off the matter of what happens to modus ponens when you set the second term as the negation of the first.

We have covered it because ~(p and ~p) is a tautology which determines that (p and ~p) is not used in order to get some proof.

It is absurd to say that MP and MT are tautologies only for certain values of their terms and not for others, and then say you can use them in proofs.

It is not an absurd at all if not using (p and ~p) is a tautology.

Furthermore, please demonstrate the need of RAA in http://en.wikipedia.org/wiki/Proof_without_words or in http://books.google.com/books?id=cy...m=2&sqi=2&ved=0CBsQ6AEwAQ#v=onepage&q&f=false.

 

[Little 10 Toes]

So it is about time to move on and deal with x = x, such that x does not have a strict id, which is determined by the strict n>1:1 ratio, as demonstrated in http://www.internationalskeptics.com/forums/showpost.php?p=6368575&postcount=11719.

All you have is to grasp http://www.internationalskeptics.com/forums/showpost.php?p=6367054&postcount=11713.

First of all "p" is not a qubit. Second, ratios don't go over 100%. And if you were doing a 2,1,0 ratio, you would write 1:1:1. Again, math failure. No, let's not move on. How about you answer one of my previous questions.

So after several times of you having to clarify what your previous answer is, we're at the point where your definition of a domain is "A realm where the measured and the measurer are interacted". Clear as mud.

Define "interacted".

 

[Robin]

Robin, you have missed http://www.internationalskeptics.com/forums/showpost.php?p=6365450&postcount=11706

You are right - let me quote it:

Direct proofs are different than indirect proofs, becuse they are not using p AND ~P at all, and this is exactly the meaning of ~(p and ~p), which means: "not using (p and ~p) in direct proofs".

Are you serious? I mean really? Are you serious???

Do you really think that?

Furthermore, please demonstrate the need of RAA in...

You might want to try actually reading your cites. Read the back cover of the book you cited where it says

"...most mathematicians would agree that they are not proofs in the formal sense...proofs without words are pictures or diagrams that help the reader see why a particular mathematical statement is true, and also to see how one could begin to go about proving it true".​

Now you may have missed my request to show the proof you claim to have posted. State what it is proving and how the conclusion is demonstrated.

I have asked several times. Why are you avoiding this?

 

[doronshadmi]

You are right - let me quote it:


Are you serious? I mean really? Are you serious???

Yes. Do you have a problem to understand that the tautology ~(p AND ~p) means, using formal proof techniques, where (p AND ~p) is not used?

You might want to try actually reading your cites. Read the back cover of the book you cited where it says

"...most mathematicians would agree that they are not proofs in the formal sense...proofs without words are pictures or diagrams that help the reader see why a particular mathematical statement is true, and also to see how one could begin to go about proving it true".​

"...most mathematicians would agree that they are not proofs in the formal sense..." or in other words, you are talking about an agreement about this subject and not about rigorous proof about this subject.

 

[doronshadmi]

First of all "p" is not a qubit. Second, ratios don't go over 100%. And if you were doing a 2,1,0 ratio, you would write 1:1:1. Again, math failure. No, let's not move on. How about you answer one of my previous questions.

3:1 is not 1:1:1

 

[The Man]

No it supports my claim as follows:

No it doesn’t .

1) We are talking only about Cardinality.

2) σ is any infinite cardinal and μ is any cardinal.

“and given an infinite cardinal σ and a cardinal μ, there will be a cardinal κ such that μ + κ = σ if and only if μ ≤ σ. It will be unique (and equal to σ) if and only if μ < σ.”

Let’s not forget about κ Doron.

3) The case σ – μ (where μ=0) is trivial, because 0 is not a successor of any cardinal, and we can't use the con-commutative property of subtraction, because there are no negative cardinals (in that case 0 – σ, is an invalid expression).

No Doron what makes “0 – σ” “an invalid expression” (particularly when μ < σ) is that subtraction is non-commutative.

If μ=0 (and thus μ < σ ) then κ = σ.

4) The non-trivial case of σ – μ is considered only if μ>0, and since σ > μ, we can't use the con-commutative property of subtraction also in this case, because there are no negative cardinals (in that case μ – σ, is an invalid expression).

Doron the restriction of μ £ σ is a result of the “con-commutative property of subtraction” otherwise the result could be negative.

If μ < σ then κ = σ

Your “non-trivial case” gives the same results as your “trivial” case. So do you consider your “non-trivial case” to be “indistinguishable” from your “trivial” case, as they both give the same results? Or is it simply that you don’t apply your own criteria for determining what you consider “indistinguishable” consistently?

5) So we are left only with σ – μ, where μ>0, and this is exactly the non-trivial case that we are dealing with.

Did you just forget about the case where μ = σ, or are you simply ignoring it?

6) By following (1) to (5) σ – μ = σ + μ = σ , and at this non-trivial case "-" is indistinguishable from "+".

So multiplication “is indistinguishable from” division because 5*1=5/1=5?

So by understanding (1) to (6) we immediately understand that "-" no longer gives definite results (it does not have definite results that are different form the results that are derive from "+"), exactly because it is indistinguishable from "+".

Again read the cited references


“σ - μ” gives a definite result Doron, “κ = σ ” when μ < σ. However, when μ = σ the definite results simply depend on the actual definitions of the sets involved. Again read your own cited reference, but this time try to actually understand it.

In other words, you have only a ratio of 1:2 (50% 1 and 50% 0),

Nope, try reading the entire post instead of just one line.

which is a local strict value between 0 and 1 local and strict values, where I also have a ratio of 2:1 (100% 1 and 100% 0), where in both cases there is non-locality among local non-strict values or local strict values.

Doron “50% 1 and 50% 0” is a TRUE:FALSE ratio of 1:1. As Little 10 Toes already noted “ratios don't go over 100%” so your “100% 1 and 100% 0” simply evaluates to a 100% contradiction .

For example:

This form represents non-locality among local non-strict values (symmetry):

1   1 
0   0 
|___|

This form represents non-locality among local strict values (asymmetry):

1   0 
|___| 
|

The current logical foundations of the mathematical science are based only on the second form,
where "superposition" is probability of (50% 1 and 50% 0), which is actually another local strict value between 0 and 1 local and strict values.

Again try actually reading the whole post this time.

But in both forms Non-locality and Locality are involved.

Again stop simply trying to posit aspects of your own failed reasoning onto others.

 

[doronshadmi]

“50% 1 and 50% 0” is a TRUE:FALSE ratio of 1:1.

According to your reasoning also “0% 1 and 0% 0” is a TRUE:FALSE ratio of 1:1.

How "lovely".

It will be unique (and equal to σ) if and only if μ < σ.”

So cardinal κ is non-unique (and therefore can't be surly = σ) if μ = σ, according to this quote.

Wellcome to the kingdom of non-strict cardinals, in this case.

No Doron what makes “0 – σ” “an invalid expression” (particularly when μ < σ) is that subtraction is non-commutative.

No, if we deal with Cardinality and μ < σ, then the non-commutative aspect of subtraction gives an invalid expression
like “μ – σ” and a valid expression like “σ – μ”.

Another example of an invalid expression is "1/(n>1)=x" if we claim that x is a natural number.

Since invalid expressions are gibberish, then we actually use only valid expressions.

Under the consideration of using only valid mathematical expressions in the case of Cardinality, "+" and "-" are indistinguishable operations.

Your “non-trivial case” gives the same results as your “trivial” case.

Only if invalid mathematical expression is also a valid mathematical expression.

Again stop simply trying to posit aspects of your own failed reasoning onto others.

Again stop simply trying to force aspects of some limited reasoning onto others.

Again try actually reading the whole post this time.

Again, try actually to read and understand http://www.internationalskeptics.com/forums/showpost.php?p=6367054&postcount=11713, because reading-only is not enough.

 

[Little 10 Toes]

According to your reasoning also “0% 1 and 0% 0” is a TRUE:FALSE ratio of 1:1.

How "lovely".

I'm sure Juila Child used that example all the time. "Now remember, for every cup of flour used in cakes, make sure you put in 0 teaspoons of vanilla extract and 0 tablespoons of baking powder. That way you'll get the perfect ratio."

 

[epix]

First of all "p" is not a qubit. Second, ratios don't go over 100%. And if you were doing a 2,1,0 ratio, you would write 1:1:1. Again, math failure. No, let's not move on. How about you answer one of my previous questions.

When Doron finds another proof of him not being able to handle math for 6-year-old kids, he will invent his new rules how to treat ratios and percentages "w.r.t." OM to make up for the total failure of basic reasoning that makes the cake dough.

If there is a case of a non-standard treatment, the writers always issue a big notice to alert the reader about it. Doron never does that. Only after he finds out that his formulas no worky in the kitchens and college dorms, he starts explaining what he actually meant. When he's done, he continues to serve his strongly counter-intuitive soup.

 

[Robin]

Yes. Do you have a problem to understand that the tautology ~(p AND ~p) means, using formal proof techniques, where (p AND ~p) is not used?

It doesn't even remotely mean that. Not by the wildest stretch of the imagination.

I can't imagine how someone could get it so utterly, disasterously wrong.

Anyway, let me repeat this from my last post:

Now you may have missed my request to show the proof you claim to have posted. State what it is proving and how the conclusion is demonstrated.

I have asked several times. Why are you avoiding this?​

You don't actually have any proofs in your mathematics, do you?

 

[doronshadmi]

Definition A: That has no predecessor has "the minimal magnitude of existence".

Example: Emptiness has "the minimal magnitude of existence".

The Cardinality of Emptiness is 0.

Definition B: That has no successor has "the maximal magnitude of existence".

Example: Fullness (the opposite of Emptiness) has "the maximal magnitude of existence".

The Cardinality of Fullness is ∞.

Definition C: That has predecessor AND successor has "an intermediate magnitude of existence".

The cardinality of that has "an intermediate magnitude of existence" is x, such that 0 < x < ∞. (where < is "greater than").

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

X is a placeholder for 0, x or ∞.


Definition D: In the case of Cardinality, "That has no predecessor, its cardinality is not a successor".

Example: X+0=X

Details:

5+0=5, where 0 is not a successor.

|infinite collection|+0=|infinite collection|, where 0 is not a successor.

∞+0= ∞, where 0 is not a successor.


Definition E: In the case of Cardinality, "That has no successor, its cardinality is not a predecessor".

Example: ∞-X= ∞

Details:

∞-5= ∞, where ∞ is not a predecessor.

∞-|infinite collection|= ∞, where ∞ is not a predecessor.

∞-∞= ∞, where ∞ is not a predecessor.


Definition F: A set exists at least as "That has no successor" AND "That has no predecessor".

Example: {} is a set , such that the outer "{""}" represents "That has no successor" AND the emptiness between the outer "{""}" represents "That has no predecessor".


Definition G: The cardinality of the members of a given set is defined by their "magnitude of existence" w.r.t Definition A or Definition B.

Examples: |{}|=0, |{{}}|=1, |{x,y,"banana"}|=3, |{1,2,3,4,5,…}|=non-strict cardinality, because no diverges collection of members has the magnitude of existence of "That has no successor", |{[0,1]}|=non-strict cardinality, because no converges collection of members has the magnitude of existence of "That has no successor".

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

Theorem: No set is a member of itself.

Proof:

According to Definition F, a set exists even if there are no members.

According to Definition G, any Cardinality of a given set, which is defined by the magnitude of existence of some collection (or its absence),
is < ∞.

Any collection is based on members (or their absence) so no collection of members or any given member has the cardinality of ∞.

In that case the magnitude of existence of a set is not identical to any magnitude of existence of any member, and we can conclude that no set is a member of itself.

Q.E.D

In this proof we have used MP:

MP.

If P, then Q.
P.
Therefore, Q.

If

P="the cardinality of a collection (where any collection is based on members (or their absence)) < ∞",

Then.

Q=" No set is a member of itself."

P.
Therefore, Q.

 

[doronshadmi]

That way you'll get the perfect ratio."

Which provides 0:0 (0 cakes).

 

[doronshadmi]

If there is a case of a non-standard treatment, the writers always issue a big notice to alert the reader about it. Doron never does that.

What you claim is simply worng, as can be clearly seen, for example, in:

http://www.scribd.com/doc/18453171/International-Journal-of-Pure-and-Applied-Mathematics-Volume-49

http://www.scribd.com/doc/16542245/OMPT

http://www.scribd.com/doc/21967511/...considerations-of-Some-Mathematical-Paradigms

Only after he finds out that his formulas no worky in the kitchens and college dorms, he starts explaining what he actually meant.

Only after I find that I am still not understood, I use search for new ways to explain OM.

When he's done, he continues to serve his strongly counter-intuitive soup.

Strongly counter-intuitive soup is the Cantorian claim that non-finite collections have strict Cardinality.

All we need is to get the notion of Fullness ("that has no successor").

 

[doronshadmi]

It doesn't even remotely mean that. Not by the wildest stretch of the imagination.

I can't imagine how someone could get it so utterly, disasterously wrong.

I see that you have missed the part of http://www.internationalskeptics.com/forums/showpost.php?p=6369257&postcount=11724, which shows that your argument is nothing but an agreement about the considered subject and not a rigorous proof of it.

Why are you avoiding this fact? Is it becuse you are not aware of the difference between Visual Special Learning and Auditory Sequential Learning? ( http://www.edfac.unimelb.edu.au/eldi/selage/documents/GLT-Visualspatialgift.pdf )( http://www.scribd.com/doc/21954904/Dear-Reader-Let-Us-Explore-the-Traditional-Point-of-View )( http://www.scribd.com/doc/17039028/OMDP )?

Anyway, let me repeat this from my last post:

Now you may have missed my request to show the proof you claim to have posted. State what it is proving and how the conclusion is demonstrated.

I have asked several times. Why are you avoiding this?​

You don't actually have any proofs in your mathematics, do you?

http://www.internationalskeptics.com/forums/showpost.php?p=6372664&postcount=11731

 

[The Man]

According to your reasoning also “0% 1 and 0% 0” is a TRUE:FALSE ratio of 1:1.

Sure, as would your “100% 1 and 100% 0” or 25% 1 and 25% 0 and it is not my “reasoning” it is just the result of the values and ratio being expressed.

However, none of these variants represent a probability distribution totaling 100% while “50% 1 and 50% 0” does. To try to put it in simpler terms for you Doron it simply means the proposition P has an equal probability of evaluation to TRUE as it does to FALSE.

How "lovely".

How inane.

So cardinal κ is non-unique (and therefore can't be surly = σ) if μ = σ, according to this quote.

Again read your own cited reference, it simply depends on the specific definitions of the actual sets involved and κ can “surly = σ” “if μ = σ”.

Again read your own cited reference.

( http://books.google.com/books?id=JdU...action&f=false )

Example 3 on page 87 to be exact.

Wellcome to the kingdom of non-strict cardinals, in this case.

Sorry Doron your fiat-land “kingdom” is strictly just yours.

No, if we deal with Cardinality and μ < σ, then the non-commutative aspect of subtraction gives an invalid expression
like “μ – σ” and a valid expression like “σ – μ”.

Wait, so your stating “No” to the assertion that the reason your expression ““0 – σ” is invalid “is that subtraction is non-commutative” then asserting “the non-commutative aspect of subtraction gives an invalid expression like “μ – σ”” when “we deal with Cardinality and μ < σ” which is your expression ““0 – σ” . Do you ever agree with yourself Doron?

Another example of an invalid expression is "1/(n>1)=x" if we claim that x is a natural number.

Since invalid expressions are gibberish, then we actually use only valid expressions.

Again it is the non-commutative aspect of subtraction that is the reason for the μ ≤ σ restriction.

Under the consideration of using only valid mathematical expressions in the case of Cardinality, "+" and "-" are indistinguishable operations.

So, your “non-trivial case” is “indistinguishable” from your “trivial” case, or is it simply that you don’t apply your own criteria for determining what you consider “indistinguishable” consistently?

Only if invalid mathematical expression is also a valid mathematical expression.

Which of your expressions in your “non-trivial case” or your “trivial” case do you now consider to be an “invalid mathematical expression”?

Again stop simply trying to force aspects of some limited reasoning onto others.

Doron if you don’t like the aspects of your own reasoning, including the “limited reasoning” you evidently simply like to ascribe to others, then find some better reasoning, but stop trying to simply pin your own limitations onto others.

Again, try actually to read and understand http://www.internationalskeptics.com/forums/showpost.php?p=6367054&postcount=11713, because reading-only is not enough.

Again stop simply trying to posit aspects of your own failed reasoning onto others.

 

[Robin]

Definition A: That has no predecessor has "the minimal magnitude of existence".

Example: Emptiness has "the minimal magnitude of existence".

The Cardinality of Emptiness is 0.

Definition B: That has no successor has "the maximal magnitude of existence".

Example: Fullness (the opposite of Emptiness) has "the maximal magnitude of existence".

The Cardinality of Fullness is ∞.

Definition C: That has predecessor AND successor has "an intermediate magnitude of existence".

The cardinality of that has "an intermediate magnitude of existence" is x, such that 0 < x < ∞. (where < is "greater than").

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

X is a placeholder for 0, x or ∞.


Definition D: In the case of Cardinality, "That has no predecessor, its cardinality is not a successor".

Example: X+0=X

Details:

5+0=5, where 0 is not a successor.

|infinite collection|+0=|infinite collection|, where 0 is not a successor.

∞+0= ∞, where 0 is not a successor.


Definition E: In the case of Cardinality, "That has no successor, its cardinality is not a predecessor".

Example: ∞-X= ∞

Details:

∞-5= ∞, where ∞ is not a predecessor.

∞-|infinite collection|= ∞, where ∞ is not a predecessor.

∞-∞= ∞, where ∞ is not a predecessor.


Definition F: A set exists at least as "That has no successor" AND "That has no predecessor".

Example: {} is a set , such that the outer "{""}" represents "That has no successor" AND the emptiness between the outer "{""}" represents "That has no predecessor".


Definition G: The cardinality of the members of a given set is defined by their "magnitude of existence" w.r.t Definition A or Definition B.

Examples: |{}|=0, |{{}}|=1, |{x,y,"banana"}|=3, |{1,2,3,4,5,…}|=non-strict cardinality, because no diverges collection of members has the magnitude of existence of "That has no successor", |{[0,1]}|=non-strict cardinality, because no converges collection of members has the magnitude of existence of "That has no successor".

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

Theorem: No set is a member of itself.

Proof:

According to Definition F, a set exists even if there are no members.

According to Definition G, any Cardinality of a given set, which is defined by the magnitude of existence of some collection (or its absence),
is < ∞.

Any collection is based on members (or their absence) so no collection of members or any given member has the cardinality of ∞.

In that case the magnitude of existence of a set is not identical to any magnitude of existence of any member, and we can conclude that no set is a member of itself.

Q.E.D

In this proof we have used MP:

MP.

If P, then Q.
P.
Therefore, Q.

If

P="the cardinality of a collection (where any collection is based on members (or their absence)) < ∞",

Then.

Q=" No set is a member of itself."

P.
Therefore, Q.

But if MP is not necessarily a tautology (as you have said it is not), obviously it cannot be used in a proof.

Using MP in a proof says that MP is necessarily a tautology, which implies that ~(p and ~p) is also a tautology.

That, in turn, implies that (p and ~ p) is a contradiction.

(On a side note, you are saying that the cardinality of any set can only be one of two possible values, yes?)

 

[doronshadmi]

No Doron what makes “0 – σ” “an invalid expression” (particularly when μ < σ) is that subtraction is non-commutative.

No, if we deal with Cardinality and μ < σ, then the non-commutative aspect of subtraction gives an invalid expression
like “μ – σ” and a valid expression like “σ – μ”.

Wait, so your stating “No” to the assertion that the reason your expression ““0 – σ” is invalid “is that subtraction is non-commutative” then asserting “the non-commutative aspect of subtraction gives an invalid expression like “μ – σ”” when “we deal with Cardinality and μ < σ” which is your expression ““0 – σ” . Do you ever agree with yourself Doron?

Exactly, and as a result we are left only with valid expressions (μ – σ is not one of them), exactly because the considered subject is exactly Cardinality.

By using only valid expressions that are related exactly to Cardinality, the valid expressions are σ – μ = σ + μ = σ > μ > 0, and according to these valid expressions "+" and "-" are indistinguishable, which is the non-trivial case.

EDIT:

If we return to [url]http://www.internationalskeptics.com/forums/showpost.php?p=6349487&postcount=11598[/url] which deals only with Cardinality, then aleph1 – aleph0 = aleph1 + aleph0 = aleph1 > aleph0 > 0, where the result aleph1 is a meaningless result because "+" and "–" are are indistinguishable.

 

[doronshadmi]

But if MP is not necessarily a tautology (as you have said it is not), obviously it cannot be used in a proof.

Using MP in a proof says that MP is necessarily a tautology, which implies that ~(p and ~p) is also a tautology.

That, in turn, implies that (p and ~ p) is a contradiction.

So what? ~(p and ~p) does not mean that I have to use RAA in my proof, and I definitely do not use RAA in my proof.

It does not mean that RAA can't be used in order to get some proof.

What I say is that a lot of proofs can be done without using RAA, and actually if something can be proven by ~RAA and also by RAA, then ~RAA proof is considered as more elegant than the RAA proof, in most cases.

(On a side note, you are saying that the cardinality of any set can only be one of two possible values, yes?)

No, I do not say that. Please look at Definition G and its examples.

 

[The Man]

Exactly, and as a result we are left only with valid expressions (μ – σ is not one of them), exactly because the considered subject is exactly Cardinality.

So you were intentionally disagreeing with yourself when you said “No”, then cited the same reason I did for “μ – σ” not being valid?

By using only valid expressions that are related exactly to Cardinality, the valid expressions are σ – μ = σ + μ = σ > μ > 0, and according to these valid expressions "+" and "-" are indistinguishable, which is the non-trivial case.

Again if you think “"+" and "-" are indistinguishable”, that is just your problem.

So is "μ + σ" one of your "valid expressions"?

 

[doronshadmi]

So you were intentionally disagreeing with yourself when you said “No”, then cited the same reason I did for “μ – σ” not being valid?




Again if you think “"+" and "-" are indistinguishable”, that is just your problem.

So is "μ + σ" one of your "valid expressions"?

You don't get it do you?

If we return to http://www.internationalskeptics.com/forums/showpost.php?p=6349487&postcount=11598 which deals only with Cardinality, then aleph1 – aleph0 = aleph1 + aleph0 = aleph1 > aleph0 > 0, where the result aleph1 is a meaningless result because "+" and "–" are are indistinguishable.

So is "μ + σ" one of your "valid expressions"?

If one claims that σ is strict, then this is an invalid expression.