Double Negation --> P=P --> Double Positives

[Eodnhoj7]

The negation of negative P results in P, with P existing if and only if P is equal to (directed through) P.

P therefore P results in Q as as a variation or tautology of P.

Q is thus equivalent to: -P existing if and only if P-->P results in both P and -P.

P-->P manifests P but also -P considering double positives result in multiple states of P thus gradation.


-(-P) --> [P <--> ((P-->P)<-->(P<--P))] --> [{(P-->P) --> {Q} <-->(-P<-->((P-->P)-->(P,-P))}]

[commonsense]

Not again!

[Eodnhoj7]

Not again!

Then cut out the double positives:


(-(-P) --> P) <--> (P-->P)

Is that simple enough for you?


Actually [(P-->P) --> [(-P)]<--> (-(-P)-->P)]

So never mind, keep the double positive.

Now...is that simple enough for you?

[Arising_uk]

Please hurry up and start this blog you were talking about as you are cluttering up this forum with your nonsense verbiage.

[Eodnhoj7]

Please hurry up and start this blog you were talking about as you are cluttering up this forum with your nonsense verbiage.

No.

[Arising_uk]

Pretty please with knobs on.

[Arising_uk]

The negation of negative P results in P, with P existing if and only if P is equal to (directed through) P. ...

No, P exists if there is a P.

P therefore P results in Q as as a variation or tautology of P. ...

What are you babbling about?

Q is thus equivalent to: -P existing if and only if P-->P results in both P and -P.

Gibberish.

P-->P manifests P but also -P considering double positives result in multiple states of P thus gradation. ...

P->P doesn't 'manifest' (whatever the hell this is supposed to mean?) P, it's just a tautology based upon there being a P. No idea what the rest of the gibberish is about.

-(-P) --> [P <--> ((P-->P)<-->(P<--P))] --> [{(P-->P) --> {Q} <-->(-P<-->((P-->P)-->(P,-P))}]

What is this (P,-P)?

[Eodnhoj7]

The negation of negative P results in P, with P existing if and only if P is equal to (directed through) P. ...

No, P exists if there is a P.

derr...let me point out what I said above:

"The negation of negative P results in P.

This is double negation of intuitionist logic.

-(-P)-->P

That is it....

P therefore P results in Q as as a variation or tautology of P. ...

What are you babbling about?

If P exists if and only if (P-->P) and P-->Q, then (P-->P)-->Q.

Real simple.

Q is thus equivalent to: -P existing if and only if P-->P results in both P and -P.

Gibberish.

Nope.

If Q is not P, as Q is a variation of P, then (P-->Q)-->(P-->-P).



P-->P manifests P but also -P considering double positives result in multiple states of P thus gradation. ...

P->P doesn't 'manifest' (whatever the hell this is supposed to mean?) P, it's just a tautology based upon there being a P. No idea what the rest of the gibberish is about.

You have no idea about anything, I bet you have no idea of your kids are really your own .

Manifest means display, etc...do you want me to teach you the alphabet too?

I will break it down.

((P-->Q) --> (P-->-P)) <--> (-P=Q)

((P-->P)-->Q) --> ((P-->P)-->-P)

-(-P) --> [P <--> ((P-->P)<-->(P<--P))] --> [{(P-->P) --> {Q} <-->(-P<-->((P-->P)-->(P,-P))}]

What is this (P,-P)?

No different than saying 1 orange plus 1 orange equals 2 oranges, but orange is one set of objects

So 1+1= 1(2)

P-->P therefore P and -P

You can throw an & sign in instead of ",".

[Arising_uk]

derr...let me point out what I said above: ...

My apologies I let your terminology affect me.

In classical Propositional Logic(PL)(which I presume is what you're mangling) there is no 'If P exists'. The standard semantics are True and False so the formula is read as 'if P is true'. What do you mean by 'if P exists'?

"The negation of negative P results in P.

This is double negation of intuitionist logic.

-(-P)-->P

That is it.... ...

Is it, how strange as intuitionistic propositional logic(IPL) is distinguished from classical propositional logic(CPL)in that it does not accept double negation as an axiom, so what is this intuitionist logic you are talking about?

Still, in CPL (--P -> P) reads as "If it is not the case that it is not the case that P is true then P is true" and as this is a tautology, as what it effectively says is (P -> P), then I agree it is always true and valid whether P is true or false.

You do understand what the variables are referring to in PL don't you or have you forgotten or never learnt this, they stand for declarative sentences.

If P exists if and only if (P-->P) and P-->Q, then (P-->P)-->Q.

Real simple.

Not really, real simple is P, P->Q ∴ Q which is called modus ponens in PL.

What you've said is, (P <-> (P v ¬P)^(P->Q)) ->((P v ¬P)->Q) which is false where P is false and Q is true.

But you'll still have to say what "P exists" means?

Nope.

If Q is not P, as Q is a variation of P, then (P-->Q)-->(P-->-P).

But if you are going to use variables for the negation of a proposition then you can't really mix and match them as all you are really saying in the above is (P -> Q) -> (P -> Q) or more accurately (P -> ¬P) -> (P -> ¬P) which again is a tautology so true regardless of whether P is true or false.

You have no idea about anything, I bet you have no idea of your kids are really your own . ...

What father ever really does but unless they are a paranoid schizophrenic or suspect infidelity and request a DNA test they are still the parent of the child. Still, not an issue that'll ever be bothering you.

You really should be concentrating on improving your English if you wish to display your claimed IQ .

Manifest means display, etc...do you want me to teach you the alphabet too?

You mean vocabulary or lexicon.

I will break it down.

((P-->Q) --> (P-->-P)) <--> (-P=Q)

I've already told you, "=" is not logical operator in PL. If you are going to use it then please provide a truth table for "=".

((P-->P)-->Q) --> ((P-->P)-->-P)

As I've already told you above, you are deluding yourself if you think you can use "¬" and a variable for "¬P" in the same formula. You have to drop one of them as it's fooling you into thinking they are different objects.

No different than saying 1 orange plus 1 orange equals 2 oranges, but orange is one set of objects

So 1+1= 1(2)

P-->P therefore P and -P

You can throw an & sign in instead of ",".

-(-P) --> [P <--> ((P-->P)<-->(P<--P))] --> [{(P-->P) --> {Q} <-->(-P<-->((P-->P)-->(P,-P))}]

This is such a mess so I can't tell which operator is the main one, is it this one,

-(-P) --> [P <--> ((P-->P)<-->(P<--P))] --> [{(P-->P) --> {Q} <-->(-P<-->((P-->P)-->(P,-P))}]

or this one?

-(-P) --> [P <--> ((P-->P)<-->(P<--P))] --> [{(P-->P) --> {Q} <-->(-P<-->((P-->P)-->(P,-P))}]

Or one of the others.

Not that it'll make much of a difference as I guess it'll be subject to the same issues as all the rest.

[Skepdick]

In classical Propositional Logic(PL)(which I presume is what you're mangling) there is no 'If P exists'. The standard semantics are True and False so the formula is read as 'if P is true'. What do you mean by 'if P exists'?

Propositions are types ( https://en.wikipedia.org/wiki/Curry%E2% ... espondence ).

If your semantic only allows for propositions to be true or false, then I can simply add another layer of indirection/encapsulation/abstraction.
I can propose things about the truth-value of another proposition.

I propose that the proposition "Apples exist" is true.
I propose that the proposition "Pears exist" is false.
I propose that the proposition "God exists" is true

And the meta-logic that corresponds to the above is as follows:

https://repl.it/repls/RichBurlywoodEmbed

Code: Select all

def exists?(object_name)
  begin
    Kernel.const_get(object_name)
  rescue NameError
    return false
  end
  return true
end

class Apple; end
class God; end

puts exists?('Apple')
puts exists?('Pear')
puts exists?('God')

The above can be translated into English as follows: There is an object named 'Apple' and an object named 'God' in my ontology. There is no object named 'Pear' in my ontology.

[Eodnhoj7]

derr...let me point out what I said above: ...

My apologies I let your terminology affect me.

In classical Propositional Logic(PL)(which I presume is what you're mangling) there is no 'If P exists'. The standard semantics are True and False so the formula is read as 'if P is true'. What do you mean by 'if P exists'?

"The negation of negative P results in P.

This is double negation of intuitionist logic.

-(-P)-->P

That is it.... ...

Is it, how strange as intuitionistic propositional logic(IPL) is distinguished from classical propositional logic(CPL)in that it does not accept double negation as an axiom, so what is this intuitionist logic you are talking about?

Your assumption is disconnected.

I am not talking about any one form of logic, and not limiting it to propositional precisely because that is a fallacy. If I begin with propositions then I have to define them and we are left with an infinite regress. If I just assume the definition of a proposition then I am literally connecting and dividing assumptions.

As to "intuitionist double negation":

It doesn't accept double negation...unless aristotelian logic and intuitionist logic are connected by context in which case it does implicitly.

If intuitionist logic does not accept --P, but requires P-->P, and (--P-->--P)<--> (P-->P)

then intuitionist logic, grounded in (P-->P), accepts this as well. Is it the standard interpretation? No.

Is it inevitable? Yes.

Why?

Because intuitionistic identity properties are rooted in aristotelian and vice versa. You cannot say some things exist and others do not without first acknowledging they exist. It is like saying "x" does not exist...but you have to define it ad-infinitum thus it always exists as there is always some new definition to negate.



Still, in CPL (--P -> P) reads as "If it is not the case that it is not the case that P is true then P is true" and as this is a tautology, as what it effectively says is (P -> P), then I agree it is always true and valid whether P is true or false.

Yes it does say (P-->P). However this identity is rooted in double negation.

Like I said, intuitionist logic is rooted in double negation.

-P is P voided. The voiding of void results in being. The void of a voided P results in P.

The assumption of an assumption results in a truth statement. All propositions as assumed are intrinsically empty in nature and as intrinsically empty they are negated.



You do understand what the variables are referring to in PL don't you or have you forgotten or never learnt this, they stand for declarative sentences.

Declaritive sentences are asserted propositions, they are fundamentally assumptions.

If P exists if and only if (P-->P) and P-->Q, then (P-->P)-->Q.

Real simple.

Not really, real simple is P, P->Q ∴ Q which is called modus ponens in PL.

If P therefore P then P therefore Q as P therefore P results in P therefore Q as P cannot exist on it's own terms.


What you've said is, (P <-> (P v ¬P)^(P->Q)) ->((P v ¬P)->Q) which is false where P is false and Q is true.

No Q is a variation of P.

"If the sky exists then color exists." (P-->Q)






But you'll still have to say what "P exists" means?

Nope.

If Q is not P, as Q is a variation of P, then (P-->Q)-->(P-->-P).

But if you are going to use variables for the negation of a proposition then you can't really mix and match them as all you are really saying in the above is (P -> Q) -> (P -> Q) or more accurately (P -> ¬P) -> (P -> ¬P) which again is a tautology so true regardless of whether P is true or false.

Yes, that is the point it is a tautology, but these truth statement you claim are just tautologies and nothing more.

There is no solution or not solution, just tautologies and tautologies are contexts with even the definition of context being one context into many...which is a tautology.

That is the point, it is a loop no matter whether you want to avoid it or not.

The OP title? It is a loop. All the laws are just variations of one assumption going to many then going to one assumption again.

You have no idea about anything, I bet you have no idea of your kids are really your own . ...

What father ever really does but unless they are a paranoid schizophrenic or suspect infidelity and request a DNA test they are still the parent of the child. Still, not an issue that'll ever be bothering you.

You really should be concentrating on improving your English if you wish to display your claimed IQ .

Manifest means display, etc...do you want me to teach you the alphabet too?

You mean vocabulary or lexicon.

Look up both terms in a dictionary and they end up leading back to eachother.

I will break it down.

((P-->Q) --> (P-->-P)) <--> (-P=Q)

I've already told you, "=" is not logical operator in PL. If you are going to use it then please provide a truth table for "=".

((P-->P)-->Q) --> ((P-->P)-->-P)

As I've already told you above, you are deluding yourself if you think you can use "¬" and a variable for "¬P" in the same formula. You have to drop one of them as it's fooling you into thinking they are different objects.

They are different object.

Blue and Not blue are different objects in certain respects and the same in others.

No different than saying 1 orange plus 1 orange equals 2 oranges, but orange is one set of objects

So 1+1= 1(2)

P-->P therefore P and -P

You can throw an & sign in instead of ",".

-(-P) --> [P <--> ((P-->P)<-->(P<--P))] --> [{(P-->P) --> {Q} <-->(-P<-->((P-->P)-->(P,-P))}]

This is such a mess so I can't tell which operator is the main one, is it this one,

That is the point, there are so many contexts inside of contexts the only way to differentiate is to see it as rings within rings.







-(-P) --> [P <--> ((P-->P)<-->(P<--P))] --> [{(P-->P) --> {Q} <-->(-P<-->((P-->P)-->(P,-P))}]

or this one?

-(-P) --> [P <--> ((P-->P)<-->(P<--P))] --> [{(P-->P) --> {Q} <-->(-P<-->((P-->P)-->(P,-P))}]

Or one of the others.

Not that it'll make much of a difference as I guess it'll be subject to the same issues as all the rest.

[Arising_uk]

In classical Propositional Logic(PL)(which I presume is what you're mangling) there is no 'If P exists'. The standard semantics are True and False so the formula is read as 'if P is true'. What do you mean by 'if P exists'?

Propositions are types ( https://en.wikipedia.org/wiki/Curry%E2% ... espondence ).

If your semantic only allows for propositions to be true or false, then I can simply add another layer of indirection/encapsulation/abstraction.
I can propose things about the truth-value of another proposition.

I propose that the proposition "Apples exist" is true.
I propose that the proposition "Pears exist" is false.
I propose that the proposition "God exists" is true

And the meta-logic that corresponds to the above is as follows:

https://repl.it/repls/RichBurlywoodEmbed

Code: Select all

def exists?(object_name)
  begin
    Kernel.const_get(object_name)
  rescue NameError
    return false
  end
  return true
end

class Apple; end
class God; end

puts exists?('Apple')
puts exists?('Pear')
puts exists?('God')

The above can be translated into English as follows: There is an object named 'Apple' and an object named 'God' in my ontology. There is no object named 'Pear' in my ontology.

I'm not sure of your point here, as for sure if pears don't exist then its false that pears exist and the proposition "Pears exist" will be false?

[Eodnhoj7]

In classical Propositional Logic(PL)(which I presume is what you're mangling) there is no 'If P exists'. The standard semantics are True and False so the formula is read as 'if P is true'. What do you mean by 'if P exists'?

Propositions are types ( https://en.wikipedia.org/wiki/Curry%E2% ... espondence ).

If your semantic only allows for propositions to be true or false, then I can simply add another layer of indirection/encapsulation/abstraction.
I can propose things about the truth-value of another proposition.

I propose that the proposition "Apples exist" is true.
I propose that the proposition "Pears exist" is false.
I propose that the proposition "God exists" is true

And the meta-logic that corresponds to the above is as follows:

https://repl.it/repls/RichBurlywoodEmbed

Code: Select all

def exists?(object_name)
  begin
    Kernel.const_get(object_name)
  rescue NameError
    return false
  end
  return true
end

class Apple; end
class God; end

puts exists?('Apple')
puts exists?('Pear')
puts exists?('God')

The above can be translated into English as follows: There is an object named 'Apple' and an object named 'God' in my ontology. There is no object named 'Pear' in my ontology.

I'm not sure of your point here, as for sure if pears don't exist then its false that pears exist and the proposition "Pears exist" will be false?

If you have a true/false statement, and the value is one or the other, then you expand the context and the value not only changes by can be true/false simultaneous.

The sky is blue. T
The sky is blue when no clouds are present. T
The sky is blue when no clouds are present, but clouds are present. F
The sky is blue when no clouds are present, we have had clouds part of the day. T/F

[Skepdick]

I'm not sure of your point here, as for sure if pears don't exist then its false that pears exist and the proposition "Pears exist" will be false?

This is a proposition about pears:
P1: Pears exist
P1 is false.

This is a proposition about another proposition (P1):
P2: P1 is false
P2 is true.

https://repl.it/repls/CrushingUpsetError

Code: Select all

def exists?(object_name)
  begin
    Kernel.const_get(object_name)
  rescue NameError
    return false
  end
  return true
end

puts p1 = exists?('Pear') # => false
puts p2 = (p1 == false) # => true

P2 is not the same proposition as 'Pears don't exist'.

Keeping the language/meta language distinction in the back of your head ( https://en.wikipedia.org/wiki/Metalanguage )

P1 is stated in the object language.
P2 is stated in the metalanguage.

It's a primitive form of Encapsulation

[nothing]

P =/= P
root-P = (+) or (-) P
P = (+/-)P
P = *P
_____
*denotes relative variability (+) or (-) allowing for (e)motion)

*P x *P... = (?)P

Let P = (+/-)1
and expand:

i. 1 x 1 = 1
ii. 1 x -1 = -1
iii. -1 x 1 = -1
iv. -1 x -1 = 1
_________
(1 x 1) = (-1 x -1)

four quadrants:
i. only positive = positive
ii. negative subject = negative relative result
iii. (inversion of ii) = (same as ii)
iv. only negative = (same as i)

implications:
a. only accumulative subjects implies only accumulative results
b. a negative subject (either) implies only negative results
c. only negative subjects implies only accumulative results
follows
d. 'only accumulative' results come either by way of 'only positive' subjects or 'only negative' subjects.

(1 x 1) = (-1 x -1)

__________
viz. a pair of equivalent antithetical dichotomous dipole(s): letter a/aleph/alpha
viz. a theoretically simultaneous-inside-outside view of a single toroid
viz. first fundamental distinction (universal): in(side)/out(side) and second letter b/beis/beta
https://i.ytimg.com/vi/wQ5QydF29oU/maxresdefault.jpg

GENESIS 2:17
But of the tree of the knowledge of good and evil, thou shalt not eat of it: for in the day that thou eatest thereof thou shalt surely die.

If ignorance is accumulated (ie. as food is eaten/digested), and a and c are both accumulative, whereas b negates, then b is the only 'solution' that allows negation of any/all belief-based ignorance, which demands trial:

ii. 1 x -1 = -1
iii. -1 x 1 = -1

All knowing is by way of indefinitely trying all belief, but
not all belief is by way of indefinitely trying to know all.

0 I am (willing to...)
-1 KNOW
(+) any/all (creation)
(-) not to (destruction)
+1 BELIEVE
____________________________
-0 KNOW, any/all not to BELIEVE = KNOWLEDGE (Conscious)
+0 BELIEVE, not to any/all KNOW = IGNORANCE (of)

resolves back into P requiring more than just P:
P = *P
wherein P can be viewed from/as one of two directions
(inside-out and/or outside-in)
thus two views of the same subject. Therefor, if P is known,
it can be used to infer/know anything that is not P
by removing what is known of P from the concerned equation.