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 Après la philosophie vient la logique

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Astaghfirullah
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Date d'inscription : 17/08/2008

MessageSujet: Après la philosophie vient la logique   2/2/2010, 22:46

Déduction, Deduction, Deduktion …

French Thoughts:
Un arriviste
Un arriviste, ne sais pas c'est quoi un arriviste?
Un arriviste, ne sais pas, ne comprends pas
Et ne crois pas qu'il est un arriviste,
C'est pour cela qu'il est.

English Translation:
An arriviste, is someone who doesn't know, what actually an arriviste is?
An arriviste, doesn't know, doesn't understand
And doesn't think he's an arriviste,
This is why he's.

German Translation:
Ein Arrivist, ist jemand, der nicht weißt, was ein Arrivist ist?
ist jemand, der nicht weiß, nicht versteht
und nicht glaubt, dass er Arrivist ist,
Deswegen ist er ja auch.

Portuguese Translation:
Um Arrivista, esta um homem,
Que não sabia o que um arrivista esta?
Esta um homem, que não sabia, não entendia
E não acredita que ele fosse um arrivista.
Assim ele esta.

Italian Translation:
Un Arrivista, è un uomo che no sapeva ho che è un arrivista?
È un uomo, che no sapeva, no capisce
E no credeva che lei era un arrivista.
Così lei era.

Mathematic Translation:
We refer to:
"An arriviste" as "X"
"True" as "1"
"False" as "0"
Let's be a: M ≅{ True,False }≈{ 1,0 }
Let's be "not" a function where: "not(X) a person who's not X"
We assume also that: not ( True ) ≅False ≈0 and
not( False )≅True ≈1
So in our Universe: T ≅{ X,not(X) }
x ∈{X,not(X)}
Now let's see the value that x: i.e. X and not(X) get by all the functions.
Function Mapping:
Function Definition x ≅X x ≅not(X)
f(x): ”Does x knows the meaning of X” f(X)=0 f(not(X)) ∈ {0,1}
g(x): ”Does x knows that he's X” g(X)=0 g(not(X)) ∈ {0,1}
m(x): ”Does x understand that he's X” m(X)=0 m(not(X)) ∈ {0,1}
n(x): ”Does x believe that he's X” n(X)=0 n(not(X)) ∈ {0,1}
h(x):”Does x is X” h(X)=1 h(not(X)) = 0

We define our function G, which represent our thoughts as:
G(x)≅f(x)^g(x)^m(x)^n(x) 
x ≅X x ≅not(X)
G(X) =f(X)^g(X)^m(X)^n(X) = 0
(See the table above) G(not(X)) ∈ {0,1}
(Could be anything)

Our goal is to find the most appropriate function that fills our thoughts and express our reasoning.
Let's first follow our intuition:
First Try:
The final function H represents our deduction or statement as:
If the function G is false and the function h is right then the deduction is right, and if instead the function G is right and the function h is false then the function is also right, all other possibility are wrong.
Let's try to prove our hypothesis using its correct mapping in the logic:
H(x) ≅G(x) not↔h(x)
≅not( ( G(x)→h(x) ) ^ ( h(x)→G(x) ) )
≅not( ( not(G(x) ) v h(x) ) ^ ( not(h(x) ) v G(x) ) )
≅not(not(G(x) ) v h(x) ) v not( not(h(x) ) v G(x) )
≅(not(not(G(x) )) ^ not(h(x)) ) v ( not(not(h(x) )) ^ not(G(x)) )
Substitute "not(not(G(x)))" with "G(x)" respectively also h(x) because it's the identity, the function is invariant.
≅( G(x) ^ not(h(x)) ) v ( h(x) v not(G(x)) )
≅( ( f(x)^g(x)^m(x)^n(x) ) ^ not(h(x)) ) v ( h(x) ^ not( f(x)^g(x)^m(x)^n(x) ) )
≅( f(x)^g(x)^m(x)^n(x)^not(h(x)) ) v (h(x) ^
(( not(f(x) ) v not(g(x) ) v not(m(x) ) v not(n(x) )) )
Let's now see the result of our function H:
H(X)≅( f(X)^g(X)^m(X)^n(X)^not(h(X)) ) v (h(X) ^
( not(f(X) ) v not(g(X) ) v not(m(X) ) v not(n(X) ) ))
≅( ( 0^0^0^0^not(1) ) v (1 ^ ( not(0) v not(0) v not(0) v not(0) ) )
≅( 0^0^0^0^0 ) v (1 ^ ( 1 v 1 v 1 v 1 ) )
≅0 v (1 ^ 1)
≅0 v 1
≅1 ≈True
Let's now try to minimize the function H to see its behavior with the value not(X):
≅( G(x) ^ not(h(x)) ) v ( h(x) ^ not(G(x)) )
≅( ( G(x)^not(h(x) ) ) v h(x) ) ^ ( ( G(x) ^ not(h(x)) ) v not(G(x)) )
≅(G(x) v h(x)) ^ ( not(h(x)) v h(x) ) ^ ( ( G(x) v not(G(x) ) ) ^
( not(h(x)) v not(G(x)) )
Replace not(not(G(x))) with the Identity, respectively h(x)
≅(( G(x) v h(x) ) ^ 1) ^ (1 ^ ( not(h(x)) v not(G(x)) ) )
≅( G(x) v h(x) ) ^ ( not(h(x)) v not(G(x)) )
≅( G(x) v h(x) ) ^ not( h(x) ^ G(x) )

Let's now replace x with X
≅( G(X) v h(X) ) ^ not( h(X) ^ G(X) )
≅( G(X) v 1 ) ^ not( 1 ^ G(X) )
≅1 ^ not( G(X) )
≅1 ^ not(0) ; See table above
≅1 ^ 1
≅1 ; correct
Let's now replace x with not(X)
≅( G(not(X)) v h(not(X)) ) ^ not( h(not(X)) ^ G(not(X)) )
≅( G(not(X)) v 0 ) ^ not( 0 ^ G(not(X)) )
≅G(not(X)) ^ not(0)
≅G(not(X)) ^ 1
≅G(not(X) ) ∈{ 1,0 } ; correct
=> This implies the correctness of our thoughts.
Then 0 not ↔1 is correct; for x equal to X
Then 1 not ↔0 is also correct; for x equal to not(X)
Then 0 not ↔0 is not correct; for x equal to not(X)
Then 1 not ↔1 is not correct; are not required because it never happens
Denote that this function with the last steps above, demonstrate and in globe more than the original thoughts permitted.
Because the case of 1 not ↔1 never happens in our thoughts.
In this case the function H is more general then the original thoughts. And a more restrictive function should be found, but I think this function doesn't exist.

After the Philosophy came the logic.
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