teoría-del-destructor y análisis-matemático y dualogía-y-lógica-algebraica

Arte:

[Ex][ f(x) es constructor ]

[Ax][ f(x) es destructor ]

Destrocter ponens:

Si ( x [< y || x >] y ) ==>

x [< y <==> x > y

x >] y <==> x < y 

Constrocter ponens:

Si ( x [< y || x >] y ) ==>

x [< y <==> ¬( x > y )

x >] y <==> ¬( x < y )

Demostración:

[<==] Si ¬( x [< y ) ==>

¬( x < y || x = y )

( ¬( x < y ) & x != y ) & ¬( x > y )

( ¬( x < y ) & ¬( x > y ) ) & x != y

¬( x < y || x > y ) & x != y

¬( x != y || x != y ) & x != y

x = y & x != y

[==>] Si x > y ==>

( x [< y || x >] y ) & x > y

( x [< y & x > y || ( x >] y & x > y )

( x [< y & x [< y & x >] y & x != y || ( x [< y & x >] y & x >] y & x != y )

( x = y & x != y ) || ( x = y & x != y ) 

Destrocter ponens:

x = y <==> ( ( f(x) = y & f(x) != x ) || ( f(x) != y & f(x) = x ) )

Constrocter ponens:

x = y <==> ( f(x) = y <==> f(x) = x )

Demostración:

[==>] Si x = y ==>

[==>] Si f(x) = y

f(x) = y & y = x

f(x) = x

[<==] Si f(x) = x

f(x) = x & x = y

f(x) = y

[<==] Si x != y

Si f(x) = y

f(x) = y & y != x

f(x) != x & f(x) = x

Si f(x) = x

f(x) = x & x != y

f(x) != y & f(x) = y

Destrocter ponens: 

Si ( x [< y || x >] y ) ==>

x [< y <==> ...

... [Ek][ x+k = y <==> ( ( f(x)+k = y & f(x) != x ) || ( f(x)+k != y & f(x) = x ) ) ] <==> ...

... ( ( f(x) [< y & f(x) != x ) || ( f(x) > y & f(x) = x ) )

x >] y <==> ...

... [E(-k)][ x+(-k) = y <==> ( ( f(x)+(-k) = y & f(x) != x ) || ( f(x)+(-k) != y & f(x) = x ) ) ] <==> ...

... ( ( f(x) >] y & f(x) != x ) || ( f(x) < y & f(x) = x ) )

Constrocter ponens: 

Si ( x [< y || x >] y ) ==>

x [< y <==> ¬( ( f(x) [< y & f(x) != x ) || ( f(x) > y & f(x) = x ) )

x >] y <==> ¬( ( f(x) >] y & f(x) != x ) || ( f(x) < y & f(x) = x ) )

Demostración:

[==>] Si ( ( f(x) [< y & f(x) != x ) || ( f(x) > y & f(x) = x ) ) 

( ( f(x) [< y || f(x) != x ) & f(x) > y ) || ( ( f(x) [< y || f(x) != x ) & f(x) = x )

( ( ¬( f(x) > y ) & f(x) > y ) || ( f(x) != x  & f(x) > y ) ) || ...

... ( ( f(x) [< y & f(x) = x ) || ( f(x) != x & f(x) = x ) )

( f(x) != x  & f(x) > y ) || ( f(x) [< y & f(x) = x )

( f(x) != x  & f(x) > y ) || ( ¬( f(x) > y ) & f(x) = x )

( f(x) > y & y >] x & f(x) = x ) || ( f(x) != x  & ¬( f(x) > y ) )

( f(x) != x  & ¬( f(x) > y ) )

¬( f(x) = x  & f(x) > y )

¬( f(x) = x  || f(x) > y & y >] x )

¬( f(x) = x || f(x) != x )

[<==] Si ¬( x [< y ) ==>

Si ¬( ( f(x) [< y & f(x) != x ) || ( f(x) > y & f(x) = x ) )

¬( f(x) [< y & f(x) != x ) & ¬( f(x) > y & f(x) = x )

¬( f(x) [< y & f(x) != x ) & ¬( x > y ) & ¬( x [< y )

¬( f(x) [< y & f(x) != x ) & ¬( x > y ) & x > y

Teorema:

[An][Ex][Ey][ x^{4}+y^{2} = n·(n+1) & ¬( x€Z & y€Z ) ]

Demostración:

x = ( (n·(n+1))/2 )^{(1/4)}

y = ( (n·(n+1))/2 )^{(1/2)}

Teorema:

[En][Ax][Ay][ Si x^{4}+y^{2} = n·(n+1) ==> ¬( x€Z & y€Z ) ]

Demostración: [ por destructor ]

Sea n€N ==>

f( x^{4} ) = x

f( y^{2} ) = y

Se define ( x = ( (n·(n+1))/2 ) & y = ( (n·(n+1))/2 ) )

x+y = f( x^{4} )+f( y^{2} ) = x^{4}+y^{2} = n·(n+1)

Teorema:

[En][Ex][Ey][ x^{4}+y^{2} = n·(n+1) & ( x€Z & y€Z ) ]

Demostración:

Se define n = p^{2}

Se define ( x = p & y = p ) ==>

p^{4}+p^{2} = n·(n+1)

Teorema:

[En][Ex][Ey][ Si x^{4}+y^{2} = n·(n+1) ==> ¬( x€Z & y€Z ) ]

Demostración: [ por destructor ]

Sea n€N ==>

Sea ( x€K & y€K ) ==>

f( x^{4} ) = x

f( y^{2} ) = y

( g(x) = ( (n·(n+1))/2 )+(-x)+y & g(y) = ( (n·(n+1))/2 ) )+(-y)+x

h(g(x)) = y & h(g(y)) = x

g(x)+g(y) = x+y = f( x^{4} )+f( y^{2} ) = x^{4}+y^{2} = n·(n+1)

Teorema:

[Ax][ Si f_{n}(x)+(-n) > x ==> f_{n}(x) > x ]

Demostración:

Sea x€K ==>

f_{n}(x) > x+n > x

Teorema:

[Ax][ Si f_{n}(x)+(-n) > x ==> f_{n}(x) > 0 ]

Demostración: [ por destructor ]

Se define x = (-n)+(-p)

f_{n}(x) > x+n

f_{n}(x) > (-p)

f_{n}(x) [< (-p) [< 0

f_{n}(x) [< 0

Teorema:

Si [Ax][Ea(x)][ f(x) = x·a(x) & a(x) es continua ] ==> f(x) es continua

Demostración:

| f(x+h)+(-1)·f(x) | = | ( (x+h)·a(x+h) )+(-1)·( x·a(x) ) | [< |x|·|a(x+h)+(-1)·a(x)|+|h|·|a(x+h)| < u+v < s

Teorema: [ de Weiestrass ]

Si f(x) es continua en [a,b]_{K} ==> f(x) está acotada en [a,b]_{K}

Demostración:

Sea c€[a,b]_{K} ==>

Sea 0 < s < 1 ==>

Sea x€[a,b]_{K} & |x+(-c)| < d ==>

|f(x)| [< | f(x)+(-1)·f(c) |+|f(c)| < s+|f(c)| < 1+|f(c)|

Se define M = max{|f(x)| : a [< x [< b } ==>

[Ec][ c€[a,b]_{K} & M < 1+|f(c)| ]

Sea x€[a,b]_{K} ==>

|f(x)| [< M

Teorema: [ de Weiestrass ]

Si ( f(x) es continua en [a,b]_{K} [ \ ] {w} & lim[x = w][f(x)] = l ) ==> f(x) está acotada en [a,b]_{K}

Demostración:

Sea 0 < s < 1 ==>

Sea x€[a,b]_{K} & |x+(-w)| < d ==>

|f(x)| [< | f(x)+(-l) |+|l| < s+|l| < 1+|l|

Sea k = max{|f(x)| : a [< x [< b } ==>

Se define M = max{k,1+|l|}

Sea x€[a,b]_{K} ==>

|f(x)| [< M

Teorema: [ de Weiestrass ]

Si ( f(x) es continua en K & ...

... lim[x = oo][ f(x) ] = p & lim[x = (-oo)][ f(x) ] = q ) ==> f(x) está acotada en K

Demostración:

Sea 1 > u > 0 ==>

[Ea][Ax][ Si a > x ==> |f(x)+(-q)| < u ]

|f(x)| [< | f(x)+(-q) |+|q| < u+|q| < 1+|q|

Sea 1 > v > 0 ==>

[Eb][Ax][ Si b < x ==> |f(x)+(-p)| < v ]

|f(x)| [< | f(x)+(-p) |+|p| < v+|p| < 1+|p|

Sea k = max{|f(x)| : a [< x [< b } ==>

Se define M = max{1+|q|,k,1+|p|}

Sea x€K ==>

|f(x)| [< M

Teorema:

Si [Ax][ f(x) = |x| ] ==> f(x) es continua

Demostración:

Sea s > 0

Se define 0 < d < s

Sea x€K & |h| < d

| f(x+h)+(-1)·f(x) | = | |x+h|+(-1)·|x| | [< | |x|+|h|+(-1)·|x| |= | |h| | = |h| < d < s

Teorema:

Si [Ax][ 0 [< f(x) [< |x| ] ==> f(x) es continua

Demostración: [ por destructor ]

Se define 0 < s < 1

Sea d > 0 ==>

Se define x > 1 & |h| < d

| f(x+h)+(-1)·f(x) | > | |x+h|+(-1)·|x| | > | |x|+|h|+(-1)·|x| | = | |h| | = |h| = 0 = |g(0)| = |x| = x > 1 > s

Teorema:

Si [Ax][ f(x) = x^{n} ] ==> f(x) es continua

Demostración:

Sea s > 0

Se define 0 < d·|P(|c+d|,|c|)| < s

Sea x€K & |x+(-c)| < d

| f(x)+(-1)·f(c) | = | x^{n}+(-1)·c^{n} | = |x+(-c)|·|P(x,c)| < d·|P(|c+d|,|c|)| < s

Teorema:

Si [Ax][ 0 [< f(x) [< x^{n} ] ==> f(x) es continua

Demostración: [ por destructor ]

Se define 0 < s < 1

Sea d > 0 ==>

Se define x > 1 & |x+(-c)| < d

| f(x)+(-1)·f(c) | > | x^{n}+(-1)·c^{n} | = |x+(-c)|·|P(x,c)| >] d·|P(x,c)| = 0 = g(0) = x > 1 > s

Teorema:

( x·y(x) = 0 & y(x) = f(x)+(-x) ) <==> f(x) = x

( x·y(x) = 0 & y(x) = d_{x...x}^{n}[f(x)]+(-x) ) <==> f(x) = ( 1/(n+1)! )·x^{n+1}

Definición:

[Ey][ ( x@y & y@z ) ] <==> x =[R]= z

Dual-equivalencia[ =[R]= ](A) = { <x,y> : x@y } 

Teorema:

[Ey][ ( f(x)+f(y) = a & f(z)+f(y) = a ) ] <==> f(x) = f(z)

Demostración:

[<==]

Se define (-1)·f(y)+a = f(x) = f(z)

Teorema:

Dual-equivalencia[=](k) es simétrico

Demostración:

Dual-equivalencia[=](k) = {<k,(-k)+a>,<(-k)+a,k>,<(-k),k+a>,<k+a,(-k)>}

Teorema:

[Ey][ ( x+y = 0 & z+y = 0 ) ] <==> x = z

Demostración:

[<==]

Se define (-y) = x = z

Teorema:

Dual-equivalencia[ = : a = 0 ](n) ={<n,(-n)>,<(-n),n>} 

Teorema:

[Ey][ ( x+y = 1 & z+y = 1 ) ] <==> x = z

Demostración:

[<==]

Se define (-y)+1 = x = z

Teorema:

Dual-equivalencia[ = : a = 1 ](1/n) ={<(1/n),( (n+(-1))/n )>,<( (n+(-1))/n ),(1/n)>}

Teorema:

[Ey][ y€[0,m+(-1)]_{N} & ( [Ep][ x = mp+y ] & [Eq][ z = mq+y ] ) ] <==> x =[m]= z

Demostración:

[==>]

( x = mp+y & z = mq+y )

( x+(-1)·mp = y & y = z+(-1)·mq )

x+(-1)·mp = z+(-1)·mq

Se define k = p+(-q) ==>

x+(-z) = mp+(-1)·mq = m·( p+(-q) ) = mk

[<==]

Se define y€[0,m+(-1)]_{N} & y =[m]= x =[m]= z

x+(-y) = mp & z+(-y) = mq

x = mp+y & z = mq+y

Teorema:

Dual-equivalencia[=[2]=]({4k,4k+1,4k+2,4k+3}) = { <4k,0>,<4k+1,1>,<4k+2,0>,<4k+3,1> }