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Post by xxxxxxxxx on Jan 31, 2019 22:33:48 GMT
((P=P) = (-P = -P)) → ¬ (P ∨ -P)
The More I study Logic and Math, the more I realize how completely absurd it is when using "space" as its own proof system (consider "space" is the only universal axiom that axioms due to the simple "dot").....This statement:
((P=P) = (-P = -P)) → ¬ (P ∨ -P)
is complete bullsh't...but it is true. No wonder our faith in mathematics/aristotelian logic, in helping us develop computing power, has led to the state of economic, moral, and cultural (identity) chaos we are in today. This is more absurd than anything even Wittgenstein worked on...unless someone has a different opinion.
Take some time to reread it again and again...it is true and makes no sense what so ever. The only thing that can be observed "rationally" is that the principle of excluded middle does not exist and all statements are "identities" properties necessitate a law of "necessitated middle".
This argument is taken from the "Fallacy of Euclidian Axioms" thread and will address the nature of "degrees" as a subset of the material.
1. All right angles are not equal to each other.
2. If a right angle exists at 90 degrees of 1 unit by 1 unit and the next right angle is .1x.1 progress to .0000....1 of the original size, the final angle effectively exists as a point smaller than the degree with composed the ninety degree of the Right Angle.
3. The question of what constitutes the "degree" ends itself in a paradoxical state considering the degrees which constitute Right Angle A (1x1) are the same as that which constitute Right Angle B ((1>n) → 0) x ((1>n) → 0) observes:
a. The lines which compose Angle B effectively become smaller than the degrees which constitute Angle A.
b. Angle A and Angle B are both composed of 90 degrees; however the 90 degrees which compose Angle B effectively fit into Angle A exist at a number approaching infinity.
c. Angle A is still equal to Angle B.
4. (Steps)
a. (A=B)=(A>B) ∴
b. ((A>A) ∧ (B>B)) ∴ ((A∋B) ∴ ((A∋A) ∧ (B∋B))) ∴
c. ((A=A)=(A>A)=(A∋A) ∧ (B=B)=(B>B)=(B∋B)) ∴
d. (((B → A) ∴ ((A → A) ∧ (B → B)) ∴ (A → B) ∧ (B → A)) ∵
e. ((A=A)=(A>A)=(A∋A) ∧ (B=B)=(B>B)=(B∋B)) ∴
f. ((A=A)=(A>A)=(A∋A)=(A → A) ∧ (B=B)=(B>B)=(B∋B)=(B → B)) ∴
g. .......
h. ((P=P) = (-P = -P)) → ¬ (P ∨ -P)
**** A=B but A is not B; hence A=P and B=-P
****Will Be Continued
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Post by karl on Feb 2, 2019 11:29:51 GMT
((P=P) = (-P = -P)) → ¬ (P ∨ -P) (P equals P) equals (negation of P equals negation of P), and this means that either P or the negation of P has to be false.
You can just skip to the one to the right as the starting point. But, when we look at how it's formally done, then you are not actually starting at the fundamentals. Who says you can just state "P=P" without proving it first? Here's how is done:
(PVP = P) → P=P
Or, in words: "If P or P is true, then P is true, and that means that if P is true then P is true."
And yes, although it's part of Bertrand Russell's principia mathematica, it's meaningless to create a symbolic procedure for going from self-evidence to self-evidence.
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Post by xxxxxxxxx on Feb 2, 2019 16:12:43 GMT
((P=P) = (-P = -P)) → ¬ (P ∨ -P) (P equals P) equals (negation of P equals negation of P), and this means that either P or the negation of P has to be false.
You can just skip to the one to the right as the starting point. But, when we look at how it's formally done, then you are not actually starting at the fundamentals. Who says you can just state "P=P" without proving it first? Here's how is done:
(PVP = P) → P=P
Or, in words: "If P or P is true, then P is true, and that means that if P is true then P is true."
And yes, although it's part of Bertrand Russell's principia mathematica, it's meaningless to create a symbolic procedure for going from self-evidence to self-evidence.
1. The premise of using 2-90 degree right angles, where each angle is "P", but effectively are not equal gives a rational example to negate the principle of identity in it's current Aristotleian Format, thus necessitating that while "identity" is defined as an axiom, the current definition alone is contradictory.
2. P=P is the law of identity. -P=-P is also the law of identity. The law of identity is equal to the law of identity, thus necessitating a negation of the excluded middle.
So the argument is multidimensional based on an axiomatic example of "space" while the language itself is prone to a self-negation.
3. There are no fundamental other than any "starting axiom" is a fundamental when the law of excluded middle is negated. The negation of the law of excluded middle; necessitates all axioms as fundamentally a starting point axiom.
4. (PVP = P) → P=P also effectively negates "or", however relative to Russel's premise it is quite faulty considering all "linear continuums" necessitate all axioms as center-points to further axioms; therefore all axioms inherently have "meaning" in and of themselves.
5. Russel projects to Godel, Godel cycles to Russel. They are two sides of the same coin and effectively observe the "axiom" as linearism or circularity being a point of origin.
6. All symobolism is inherently geometric in these regards:
The problem of logic/math, and all symbolism lies in a contradiction between form and function, the noun and verb fundamentally, where we are left with an alternation between the two resulting in a form of "atomism" where there are multiple types of atomic facts which necessitates a further from of atomism where one type of atom leads to another type with "type" resulting in a tautology leading to a self referencing "atomism" where the atom takes on a nature of the "whole".
Thus, through the inherent atomic nature of one atomic fact being both composed of and composing atomic facts, a "set theory" type of math/logic is integrated into this symbolism.
This dualism between active/passive is solved by a synthesis between the active/passive resulting in the "symbol" as quite literally a point of observation, both projective and circular in nature, to exist as perpetual "meaning through meaning as meaning" where all symbolism as both active and passive exist through a continual recursion with thus continual recursion as "symbolic" in nature and hence containing a simultaneous active/passive nature which is self maintained through the symbols themselves.
Thus the atomic fact exists as symmetrical to the wholism perspective and "being", which in English vocabulary can take a passive and active nature, is proof itself.
1. A,B,C,D are all axioms and as axioms are points of origin and hence unproven:
Ex: A•B•C•D•
2. All axioms are defined by there direction to other axioms, and as such exist as an axiom and point of origin:
Ex: (A• -> B• -> C• -> D•)•
(Insert various combinations)
3. All axioms that exist as a modal state effectively can cycle with axioms that do not exist in a modal state considering the modality defines the none modal and the modal exists through the non modal. For example green describes the tree, and tree describes the green, hence tree green and green tree are connected as one entity and variate in order dependent upon the language.
Ex: (A•○B•)•
Ex: ((A•○B•)• -> C• -> D•)•
(Insert various combinations)
4. Negation is a definition in an axiom and exists as both a statement of relation and gradation. For example "not blue" observes the variables of "not blue" as the answer within the given context.
Considering "blue" is is an axiom, and "not blue" is a negation of this axiom, this negation exists where the axiom exists as a point of inversion from one axiom of "blue" to the many axioms of "not blue" with even just "red", as an answer showing the divergence of blue into red by observing "not blue".
Ex: B¤
This applies to statements as well in the hegelian sense of antithetical:
Ex: (A• -> B• -> C•)¤
5. Implication, where the answer is probable, necessitates the axiom as the actual state amidst potentially other states. This actualized state is a multiplicitious form, due to is gradient nature, of a potential unity:
Ex. (•/B••)•
6. Equivalence observes each phenomena effectively as directed towards each other. Such as 4=2+2 and 2+2=4. Equivalence also shows a simulateous disconnect of the phenomena as what is equivalent is by necessitating fundamentally separate.
Ex: (A• <->•<-> B•)•
7. Conjunction, where both axioms effectively are directed towards eachother and unify into a new axiom through "and" , "both", etc.:
Example (A and B exist at both ends of angle on left side) :
((A•B•)> •)•
8. Disjustion, where both axioms effectively separate from a prior unity do to "either", "or", etc. which observes a prior state of unity amidst the axioms until some choice comes along and separate them.
Example (inversion of conjuction):
(• <(A•B•))•
9. Infer is same as imply and can be considered an inefficient use of language.
10. Identity can be observed as same as equality and can be considered an inefficient use of language.
11. Membership where one axiom exists as a particulate/part of another axiom which is more generalized, with this generality observing a potential state of unity and the axiom as a part effectively acting as an extension that exists in an actualized state can be observed as a fraction.
Ex:
(A•/B••)•
12: Negation of membership where the axiom is not longer a particulate of that general axiom:
(A•/B•¤)•
13: "All x" necessitates the axiom as both a general and particular state:
(A•/A••)•
14: There is at least one B observe x existing as a part of the general A and can observe a form of implied set. Hence A is both a general and particular. At least 1 observes the particular, as part of an implied set as both a general and particular state.
((•/••)/(B•/B••))•
15. There is no axiom still observes a deficiency of that axiom and can be observed as a negation in accords with point 4.
16. Intersection observes the axioms converge and diverge:
Ex: ((A•B•)>•<(A•B•))•
17. Empty set can be observed as no axiom or negation of that axiom.
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Post by karl on Feb 3, 2019 6:25:15 GMT
(P equals P) equals (negation of P equals negation of P), and this means that either P or the negation of P has to be false.
You can just skip to the one to the right as the starting point. But, when we look at how it's formally done, then you are not actually starting at the fundamentals. Who says you can just state "P=P" without proving it first? Here's how is done:
(PVP = P) → P=P
Or, in words: "If P or P is true, then P is true, and that means that if P is true then P is true."
And yes, although it's part of Bertrand Russell's principia mathematica, it's meaningless to create a symbolic procedure for going from self-evidence to self-evidence.
2. P=P is the law of identity. -P=-P is also the law of identity. The law of identity is equal to the law of identity, thus necessitating a negation of the excluded middle.
I'm not sure I understood the last part. What you wrote may be interpreted as a double negative, as if you stated: "thus necessitating a negation of the law of the excluded middle." Is that what you meant? My guess, however, is that you meant the opposite, that the law of identity necessitates the law of the excluded middle. Which one is it?
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Post by xxxxxxxxx on Feb 3, 2019 17:41:15 GMT
2. P=P is the law of identity. -P=-P is also the law of identity. The law of identity is equal to the law of identity, thus necessitating a negation of the excluded middle.
I'm not sure I understood the last part. What you wrote may be interpreted as a double negative, as if you stated: "thus necessitating a negation of the law of the excluded middle." Is that what you meant? My guess, however, is that you meant the opposite, that the law of identity necessitates the law of the excluded middle. Which one is it?
-P=P because -P exists through the law of identity, thus there is no excluded middle. |
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Post by karl on Feb 3, 2019 17:57:20 GMT
I'm not sure I understood the last part. What you wrote may be interpreted as a double negative, as if you stated: "thus necessitating a negation of the law of the excluded middle." Is that what you meant? My guess, however, is that you meant the opposite, that the law of identity necessitates the law of the excluded middle. Which one is it?
-P=P because -P exists through the law of identity, thus there is no excluded middle.
The negation of P equals P because the negation of P exists through the law of identiy? What is that supposed to mean?
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Post by xxxxxxxxx on Feb 3, 2019 18:26:03 GMT
-P=P because -P exists through the law of identity, thus there is no excluded middle.
The negation of P equals P because the negation of P exists through the law of identiy? What is that supposed to mean?
The law of identity is equal to the law of identity, P=P and -P=-P are both the law of identity. |
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Post by karl on Feb 3, 2019 18:35:25 GMT
The negation of P equals P because the negation of P exists through the law of identiy? What is that supposed to mean?
The law of identity is equal to the law of identity, P=P and -P=-P are both the law of identity.
Ok, so your claim is that the law of identiy, which includes both P=P and -P=-P, means there is no excluded middle, which I presume you mean is expressed with: ¬ (P ∨ -P).
This is not something I agree with. ¬ (P ∨ -P) only states that neither P nor -P are false. That doesn't mean that one of them are right. A statement might be right, wrong, or undecided.
P=I will eat a sandwhich for breakfast tomorrow. P=P: If I will eat a sandwhich for breakfast tomorrow, then that means I will eat a sandwhich for breakfast tomorrow.
-P=-P: If I will not eat a sandwhich for breakfast tomorrow, then that means I will not eat a sandwhich for breakfast tomorrow.
¬ (P ∨ -P): It can't be both false, that I will eat a sandwhich for breakfast tomorrow, and that I won't eat a sandwhich for breakfast tomorrow.
All statements are true, and none rule out the excluded middle. For it is not decided whether I will eat a sandwhich for breakfast tomorrow.
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Post by xxxxxxxxx on Feb 3, 2019 18:37:11 GMT
The law of identity is equal to the law of identity, P=P and -P=-P are both the law of identity.
Ok, so your claim is that the law of identiy, which includes both P=P and -P=-P, means there is no excluded middle, which I presume you mean is expressed with: ¬ (P ∨ -P).
This is not something I agree with. ¬ (P ∨ -P) only states that neither P nor -P are false. That doesn't mean that one of them are right. A statement might be right, wrong, or undecided.
P=I will eat a sandwhich for breakfast tomorrow. P=P: If I will eat a sandwhich for breakfast tomorrow, then that means I will eat a sandwhich for breakfast tomorrow.
-P=-P: If I will not eat a sandwhich for breakfast tomorrow, then that means I will not eat a sandwhich for breakfast tomorrow.
¬ (P ∨ -P): It can't be both false, that I will eat a sandwhich for breakfast tomorrow, and that I won't eat a sandwhich for breakfast tomorrow.
All statements are true, and none rule out the excluded middle. For it is not decided whether I will eat a sandwhich for breakfast tomorrow.
P or not P is still a law, one set of laws negates the other. |
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Post by karl on Feb 3, 2019 18:41:06 GMT
Ok, so your claim is that the law of identiy, which includes both P=P and -P=-P, means there is no excluded middle, which I presume you mean is expressed with: ¬ (P ∨ -P).
This is not something I agree with. ¬ (P ∨ -P) only states that neither P nor -P are false. That doesn't mean that one of them are right. A statement might be right, wrong, or undecided.
P=I will eat a sandwhich for breakfast tomorrow. P=P: If I will eat a sandwhich for breakfast tomorrow, then that means I will eat a sandwhich for breakfast tomorrow.
-P=-P: If I will not eat a sandwhich for breakfast tomorrow, then that means I will not eat a sandwhich for breakfast tomorrow.
¬ (P ∨ -P): It can't be both false, that I will eat a sandwhich for breakfast tomorrow, and that I won't eat a sandwhich for breakfast tomorrow.
All statements are true, and none rule out the excluded middle. For it is not decided whether I will eat a sandwhich for breakfast tomorrow.
P or not P is still a law, one set of laws negates the other. I'm not sure what you're trying to express with that. Proving that P and -P can't both be false, doesn't prove that either P or -P are true. So the law of the excluded middle hasn't been proven. |
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Post by xxxxxxxxx on Feb 3, 2019 18:48:35 GMT
P or not P is still a law, one set of laws negates the other. I'm not sure what you're trying to express with that. Proving that P and -P can't both be false, doesn't prove that either P or -P are true. So the law of the excluded middle hasn't been proven. If P=-P it negates the law of excluded middle. |
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Post by karl on Feb 3, 2019 18:51:01 GMT
I'm not sure what you're trying to express with that. Proving that P and -P can't both be false, doesn't prove that either P or -P are true. So the law of the excluded middle hasn't been proven. If P=-P it negates the law of excluded middle.
How did you arrive at P = -P?
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Post by xxxxxxxxx on Feb 3, 2019 18:52:21 GMT
If P=-P it negates the law of excluded middle.
How did you arrive at P = -P?
(P=P)=(-P=-P) |
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Post by karl on Feb 3, 2019 18:57:36 GMT
How did you arrive at P = -P?
(P=P)=(-P=-P) That doesn't lead to P=-P. |
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Post by xxxxxxxxx on Feb 3, 2019 19:19:26 GMT
That doesn't lead to P=-P. So (P=P)=P is not true? |
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