Deeper than primes

[doronshadmi]

Nice try to change the subject! We're talking about (P) ≠ (~P). We're not talking about 1 ≠ 0 being the same as 1 = 1. The True/False values of "1 ≠ 0" and "1 = 1" are the same.

EDIT: Yeah I know that this post is late, but with doronshadmi's history of re-re-re-edits, I can make sure he won't have time to change a responce.

A great success of staying ignorant.

 

[doronshadmi]

Ooh, I used to like these when I was at school. I know it's not a Mexican riding a bike, or frying an egg. Is it the side view of a badminton court?

No you did not use "like these" when you was at school, because if this was a part of your education, you had the chance to get _|_.

The fact is that you don't get it.

 

[doronshadmi]

It is called negation, in case you keep missing it. It is a unary operation not a ‘comparison‘, but then I do not expect you to understand that.

Unary operation, or unary connective is useful only if there is comparosion of input with output.

Without this proprty your framework is useless.

 

[doronshadmi]

In a two value system where 0 = ~1 thus 1 = ~0, yes it is. In case the process escapes you here it is. The following are just different ways of writing the same logically equal statement in the two value system described.

1 ≠ 0
~ (1 = 0)
~1 = 0
1 ~= 0
1 = ~0
1 = 1

Wrong.

(P ≠ Q) = ((1 ≠ 0) = (1 ~= 0) = (~ (1 = 0)) = (~1 ~= ~0))
≠
(P = Q) = ((~1 = 0) = (0 = 0) = (1 = ~0) = (1 = 1))

Let us focused on ~ (1 = 0)

If you claim that ((~ (1 = 0)) = (~1 = 0) = (1 = ~0)), then (~ (1 = 0)) ≠ ((1 ≠ 0) = (1 ~= 0))

In other words, 1≠0 is not reducible to 1=1 (or 0=0), and your ignorance is exposed again.

 

[doronshadmi]

The minimal base value of Logic is 2, for example:

2^0 = 1
2^1 = 2
...

By OM 0 is Atom (the source of total isolation and total connectivity).

By OM 1 of 2^0 means that P | Q (total isolation) or P Q (total connectivity) are not comparable, and therefore not researchable.

By OM 2 of 2^1 means that P | Q (total isolation) or P Q (total connectivity) are comparable, and therefore researchable.

2^1 is at least P_|_Q , where P=P or Q=Q are local comparison and P≠Q is a non-local comparison.

In both cases = or ≠ relations are non-local with respect to P or Q elements.

By 2^0 we get (P|Q | P Q) or (P|Q P Q), which is not reseachable.

By 2^1 we get P_|_Q, which is reseachable.

P≠Q is difference's comparison (non-local comparison).

P=Q is sameness's comparison (local comparison).

 

[doronshadmi]

By Classical Logic:

2^0 means that P is isolated from NOT-P and vice versa.

2^1 means that we are using unary operation, where a single input has a single output, and 2 is the two possible cases of NOT truth table:

P NOT-P
F T 
T F

In 2^1 is researchable because it is compareable.

 

[The Man]

Unary operation, or unary connective is useful only if there is comparosion of input with output.

Without this proprty your framework is useless.

The output (~P) is entirely dependent on the input (P), the requirement of your “comparosion” for such a dependence is entirely yours. Your “framework” is useless simply because you can demonstrate no, well, use. The uses of logical negation and that input to output dependence however are demonstrated in the very computer you are now using to read this.

 

[The Man]

Wrong.

(P ≠ Q) = ((1 ≠ 0) = (1 ~= 0) = (~ (1 = 0)) = (~1 ~= ~0))
≠
(P = Q) = ((~1 = 0) = (0 = 0) = (1 = ~0) = (1 = 1))



Let us focused on ~ (1 = 0)

If you claim that ((~ (1 = 0)) = (~1 = 0) = (1 = ~0)), then (~ (1 = 0)) ≠ ((1 ≠ 0) = (1 ~= 0))

In other words, 1≠0 is not reducible to 1=1 (or 0=0), and your ignorance is exposed again.

As stated, in the two value system described it is, your simple and repeated ignorance of that fact does not change that fact.

 

[jsfisher]

By Classical Logic:

2^0 means that P is isolated from NOT-P and vice versa.

2^1 means that we are using unary operation, where a single input has a single output, and 2 is the two possible cases of NOT truth table:

P NOT-P
F T 
T F

In 2^1 is researchable because it is compareable.

No, no, no, but the table is ok, and no.

 

[The Man]

The minimal base value of Logic is 2, for example:

2^0 = 1
2^1 = 2
...

By OM 0 is Atom (the source of total isolation and total connectivity).

By OM 1 of 2^0 means that P | Q (total isolation) or P Q (total connectivity) are not comparable, and therefore not researchable.

By OM 2 of 2^1 means that P | Q (total isolation) or P Q (total connectivity) are comparable, and therefore researchable.

2^1 is at least P_|_Q , where P=P or Q=Q are local comparison and P≠Q is a non-local comparison.

In both cases = or ≠ relations are non-local with respect to P or Q elements.

By 2^0 we get (P|Q | P Q) or (P|Q P Q), which is not reseachable.

By 2^1 we get P_|_Q, which is reseachable.

P≠Q is difference's comparison (non-local comparison).

P=Q is sameness's comparison (local comparison).

If you mean the minimal application of logic is a in a two value system then again that is simply trivial as the rest of your post is again simply nonsensical gibberish.

 

[Little 10 Toes]

A great success of staying ignorant.

Classy. Shall we start riffing on each other's mother?

Wrong.

(P ≠ Q) = ((1 ≠ 0) = (1 ~= 0) = (~ (1 = 0)) = (~1 ~= ~0))
≠
(P = Q) = ((~1 = 0) = (0 = 0) = (1 = ~0) = (1 = 1))

Let's break down the values. Since we are using a two-value system, it will be quite easy to show why they all equal each other. You have introduced P, Q, 1, and 0. I will replace P with 1 and Q with 0. Results will either be True or False. I will seperate the two "equations" using square brackets. Also replaced ~= with ≠

Original equation
[ (P ≠ Q) = ((1 ≠ 0) = (1 ~= 0) = (~ (1 = 0)) = (~1 ~= ~0)) ] ≠ [ (P = Q) = ((~1 = 0) = (0 = 0) = (1 = ~0) = (1 = 1))]

Subsituted Equation
[ (1 ≠ 0) = ((1 ≠ 0) = (1 ≠ 0) = (~ (1 = 0)) = (~1 ≠ ~0)) ] ≠ [ (1 = 0) = ((~1 = 0) = (0 = 0) = (1 = ~0) = (1 = 1)) ]

Working left to right to "reduce" the equation
[ (T) = ((T) = (T) = (~(F)) = (T)) ] ≠ [ (F) = ((T) = (T) = (T) = (T)) ]
[ (T) = ((T) = (T) = (T) = (T)) ] ≠ [ (F) = ((T) = (T) = (T) = (T)) ]
[ (T) = (T) ] ≠ [ (F) = (T) ]
[ T ] ≠ [ F ]

Guess what, you're right! The two are not the same!!!!

Let us focused on ~ (1 = 0)

If you claim that ((~ (1 = 0)) = (~1 = 0) = (1 = ~0)), then (~ (1 = 0)) ≠ ((1 ≠ 0) = (1 ~= 0))

Still sticking with results are True or False here. Still using ≠ to replace ~=. Breaking down the two equations side by side:

[ ((~ (1 = 0)) = (~1 = 0) = (1 = ~0)) ] = [ (~ (1 = 0)) ≠ ((1 ≠ 0) = (1 ≠ 0)) ]
[ (( ~ (F)) = (T) = (T)) ] = [ ( ~(F)) ≠ ((T) = (T)) ]
[ ((T) = (T) = (T)) ] = [ (T) ≠ (T) ]
[ (T) ] = [ (F) ]
F
Nope, your claim does not work.

I'd like to thank everyone ~doronshadmi for the help. Soon, I'll be on my way to be a Junior Rank semi-amateur Logician, 3rd Class, Wombat Division!

 

[doronshadmi]

The output (~P) is entirely dependent on the input (P), the requirement of your “comparosion” for such a dependence is entirely yours.

The output (~P) is entirely dependent on the input P and its connection to ~.

Furthermore, the researchability of this framework entirely dependents on the comparison of P(input) with ~P(output).

Also classical Logic can't avoid it, but instead of explicitly address it, it is used (you have no choice) but ignored.

 

[doronshadmi]

As stated, in the two value system described it is, your simple and repeated ignorance of that fact does not change that fact.

The Man "(1≠0) is not reducible to (0=0) or (1=1)" is the fact that you can't comprehend.

 

[doronshadmi]

No, no, no, but the table is ok, and no.

Yes, yes, yes , and the table is ok, and yes.

 

[dafydd]

Na ya de a da,my daddy's bigger than your daddy.

 

[doronshadmi]

If you claim that ((~ (1 = 0)) = (~1 = 0) = (1 = ~0)), then (~ (1 = 0)) ≠ ((1 ≠ 0) = (1 ~= 0))

Nope, your claim does not work.

X = (~ (1 = 0))

A = ((~1 = 0) = (1 = ~0))

B = ((1 ≠ 0) = (1 ~= 0))

Now, [if X=A , then X≠B] = [If you claim that ((~ (1 = 0)) = (~1 = 0) = (1 = ~0)), then (~ (1 = 0)) ≠ ((1 ≠ 0) = (1 ~= 0))]

In other words:

Your [ ((~ (1 = 0)) = (~1 = 0) = (1 = ~0)) ] = [ (~ (1 = 0)) ≠ ((1 ≠ 0) = (1 ≠ 0)) ]

≠

My [if X=A , then X≠B]

 

[doronshadmi]

Na ya de a da,my daddy's bigger than your daddy.

What do you wish to say?

 

[doronshadmi]

If you mean the minimal application of logic is a in a two value system then again that is simply trivial

Simple is not Trivial.

Triviality is entirely the result of your ignorance of the must have foundations that enable Logic in the first place, and the ignorance of Comparison is the core of it.

You are using it right now, but can't comprehend it.

 

[jsfisher]

X = (~ (1 = 0))

Ok, if you wish, but we can reduce that to just X = 1.

A = ((~1 = 0) = (1 = ~0))

Ok, A = 1.

B = ((1 ≠ 0) = (1 ~= 0))

Ok, and B = 1.

Now, [if X=A , then X≠B]

Nope. That is clearly not the case.

... = [If you claim that ((~ (1 = 0)) = (~1 = 0) = (1 = ~0)), then (~ (1 = 0)) ≠ ((1 ≠ 0) = (1 ~= 0))]

No, wrong again.

(1 = 0) is 0, so ~(1=0) would be 1.
(1 ≠ 0) is 1.
(1 ~= 0) is the same statement with slightly different notation.
So, (1 ≠ 0) = (1 ~= 0) is 1.
So, the entire THEN part reduces to 0 since ~(1 = 0) and (1 ≠ 0) = (1 ~= 0) reduce to the same value, and the not-equals fails.

In other words...

...you are wrong again.

 

[doronshadmi]

Ok, if you wish, but we can reduce that to just X = 1.



Ok, A = 1.



Ok, and B = 1.



Nope. That is clearly not the case.



No, wrong again.

(1 = 0) is 0, so ~(1=0) would be 1.
(1 ≠ 0) is 1.
(1 ~= 0) is the same statement with slightly different notation.
So, (1 ≠ 0) = (1 ~= 0) is 1.
So, the entire THEN part reduces to 0 since ~(1 = 0) and (1 ≠ 0) = (1 ~= 0) reduce to the same value, and the not-equals fails.



...you are wrong again.

I am talking about the inability to reduce (1≠0) to (1=1) or (0=0) ( The Man claims that (1≠0) = (1=1) ).

Again, I am not talking on the fact that (1≠0),(1=1) or (0=0) are true statements.

I am talking about the difference of ≠ form = , where (1≠0) is not self referential, where (1=1) or (0=0) are self referential.

EDIT: In other words: (1≠0) ≠ ((1=1) or (0=0)) and you are not listening to my argument (as already shown in http://www.internationalskeptics.com/forums/showpost.php?p=5339246&postcount=7069).