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Post by xxxxxxxxx on Jan 25, 2020 3:09:59 GMT
-(-A)--> A ---> (B --> -A) --> (C --> -B) --> ...
(B --> -A) --> (-A --> -A) --> (A --> B --> A --> B...)
(C --> -B) --> (-B --> -B) --> (B --> C --> B --> C...)
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KGrim
Full Member
Coming back to Arktos...for a little while anyways...just to see how things are doing.
Posts: 442
Likes: 238
Country: USA
Region: South East
Location: East Texas
Ancestry: Scotch-Irish
Politics: Conservative
Religion: Eastern Orthodox
Hero: Jesus
Age: 33 soon to be 34
Philosophy: Hesychasm
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Post by KGrim on Jan 25, 2020 4:14:57 GMT
None of this makes any sense to me.
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Post by xxxxxxxxx on Jan 25, 2020 15:18:12 GMT
None of this makes any sense to me. The negation of negative A leads to A. A leads to B but B is not A therefore B leads to not A. B leads to C but C is not B therefore C leads to not B. B leading to Not A leads to -A as self referencing. This leads leads to A, through double negation, which leads to B. B leads to A, through the double negation of -A, but A also leads to B. The same for C and -B occurs above as well. |
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Post by Eugene 2.0 on Jan 27, 2020 20:00:51 GMT
P1) ~~A -> A
P2) A -> B. -> .~(A & B) -> .B -> ~A
P3) B -> C. -> .~(C & B) -> .C -> ~B
P4) B -> ~A. -> -A. -> (~~) -> A. -> B
P5) B -> A. -> (~~-A). & A -> B
P6) A = (C & -B)
a. If B leads to Not A as self referencing to -A, why B --> ~A is self referencing?
b. What --A would mean then?
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Post by xxxxxxxxx on Jan 27, 2020 22:30:10 GMT
P1) ~~A -> A P2) A -> B. -> .~(A & B) -> .B -> ~A P3) B -> C. -> .~(C & B) -> .C -> ~B P4) B -> ~A. -> -A. -> (~~) -> A. -> B P5) B -> A. -> (~~-A). & A -> B P6) A = (C & -B) a. If B leads to Not A as self referencing to -A, why B --> ~A is self referencing? b. What --A would mean then? B leads to the same point it is defined from. Negative A negates to A. A leads to B. B leads to negative A, B leads to the same point it is defined from, this is considering negative A leads to B and B leads to negative A. You can argue that -A --> B --> -A as well. Or you can argue B self references to D as well. So -(-A) --> A --> B --> C --> D (B --> B) --> D (C --> C) --> F |
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Post by Eugene 2.0 on Jan 27, 2020 22:48:35 GMT
P1) ~~A -> A P2) A -> B. -> .~(A & B) -> .B -> ~A P3) B -> C. -> .~(C & B) -> .C -> ~B P4) B -> ~A. -> -A. -> (~~) -> A. -> B P5) B -> A. -> (~~-A). & A -> B P6) A = (C & -B) a. If B leads to Not A as self referencing to -A, why B --> ~A is self referencing? b. What --A would mean then? B leads to the same point it is defined from. Negative A negates to A. A leads to B. B leads to negative A, B leads to the same point it is defined from, this is considering negative A leads to B and B leads to negative A. You can argue that -A --> B --> -A as well. Or you can argue B self references to D as well. So -(-A) --> A --> B --> C --> D (B --> B) --> D (C --> C) --> F Negation of A is everything that is not A. Everything is A and B. Then A implies B, and B implies A. But B is not A, then not A implies B. Both: B being implied from A, and B being implied from not A is everything. Everything implies B. |
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Post by xxxxxxxxx on Jan 28, 2020 0:46:51 GMT
B leads to the same point it is defined from. Negative A negates to A. A leads to B. B leads to negative A, B leads to the same point it is defined from, this is considering negative A leads to B and B leads to negative A. You can argue that -A --> B --> -A as well. Or you can argue B self references to D as well. So -(-A) --> A --> B --> C --> D (B --> B) --> D (C --> C) --> F Negation of A is everything that is not A. Everything is A and B. Then A implies B, and B implies A. But B is not A, then not A implies B. Both: B being implied from A, and B being implied from not A is everything. Everything implies B. Negation of not A results in A. A leads to B therefore leads to not A again, -A leads to B and B leads to not A. B leads to not A and C, but in leading to not A it leads back to the premise and towards a new premise simultaneously. B leads to both A and -A. It leads to not A through C with leading to C as not B as well. B leads to A through the negation of A and A leads to B. B is going in circles. |
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Post by Eugene 2.0 on Jan 28, 2020 21:44:36 GMT
xxxxxxxxxX is a pure negation in/of itself One negation of X and Two negations of X leads to Three negations of XX. One negation through separate Two negations backs to X as XXX. Being negated twice Three XX results in Three )()()(. Unionization and separation (as Two) of X in its Three forms through Three negations allows () to be restored to its former self. By two rows and two strings One negation of X and Two Negations of X leads to Three Three XXX or XXXXXXXXX. (~ > X) & (X > ~) (~X & ~~X) > ~~~XX (~) > (~)v(~) > ~(~&~) > (~) > X > (X X X) ~~(XX XX XX) > (() () ()) > ( )()()( ) ~~~[ )()()( ] > )( ~{ X XX XXX )( x x x )()( x x x )()()( x x x}
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Post by xxxxxxxxx on Jan 29, 2020 1:03:29 GMT
xxxxxxxxx X is a pure negation in/of itself One negation of X and Two negations of X leads to Three negations of XX. One negation through separate Two negations backs to X as XXX. Being negated twice Three XX results in Three )()()(. Unionization and separation (as Two) of X in its Three forms through Three negations allows () to be restored to its former self. By two rows and two strings One negation of X and Two Negations of X leads to Three Three XXX or XXXXXXXXX. (~ > X) & (X > ~) (~X & ~~X) > ~~~XX (~) > (~)v(~) > ~(~&~) > (~) > X > (X X X) ~~(XX XX XX) > (() () ()) > ( )()()( ) ~~~[ )()()( ] > )( ~{ X XX XXX )( x x x )()( x x x )()()( x x x} -A (-A --> -A) --> (A & -A) A results from a double negation, -A occurs through self referencing. Both occur simultaneously. |
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