let n, x, y, z, t be positive Nat; ( ((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t iff ( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) ) )
A1:
((n |^ x) + (n |^ y)) + (n |^ z) = ((n |^ z) + (n |^ y)) + (n |^ x)
;
A2:
((n |^ x) + (n |^ y)) + (n |^ z) = ((n |^ x) + (n |^ z)) + (n |^ y)
;
A3:
((n |^ x) + (n |^ y)) + (n |^ z) = ((n |^ y) + (n |^ z)) + (n |^ x)
;
A4:
((n |^ x) + (n |^ y)) + (n |^ z) = ((n |^ z) + (n |^ x)) + (n |^ y)
;
thus
( not ((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t or ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )
( ( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) ) implies ((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t )proof
assume A5:
((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t
;
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )
per cases
( ( x <= y & y <= z ) or ( x <= z & z <= y ) or ( y <= x & x <= z ) or ( y <= z & z <= x ) or ( z <= x & x <= y ) or ( z <= y & y <= x ) )
;
suppose
(
x <= y &
y <= z )
;
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )hence
( (
n = 2 &
y = x &
z = x + 1 &
t = x + 2 ) or (
n = 2 &
y = x + 1 &
z = x &
t = x + 2 ) or (
n = 2 &
z = y &
x = y + 1 &
t = y + 2 ) or (
n = 3 &
y = x &
z = x &
t = x + 1 ) )
by A5, Th49;
verum end; suppose
(
x <= z &
z <= y )
;
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )hence
( (
n = 2 &
y = x &
z = x + 1 &
t = x + 2 ) or (
n = 2 &
y = x + 1 &
z = x &
t = x + 2 ) or (
n = 2 &
z = y &
x = y + 1 &
t = y + 2 ) or (
n = 3 &
y = x &
z = x &
t = x + 1 ) )
by A2, A5, Th49;
verum end; suppose
(
y <= x &
x <= z )
;
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )then
( (
n = 2 &
y = x &
z = y + 1 &
t = y + 2 ) or (
n = 3 &
y = x &
z = y &
t = y + 1 ) )
by A5, Th49;
hence
( (
n = 2 &
y = x &
z = x + 1 &
t = x + 2 ) or (
n = 2 &
y = x + 1 &
z = x &
t = x + 2 ) or (
n = 2 &
z = y &
x = y + 1 &
t = y + 2 ) or (
n = 3 &
y = x &
z = x &
t = x + 1 ) )
;
verum end; suppose
(
y <= z &
z <= x )
;
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )then
( (
n = 2 &
z = y &
x = y + 1 &
t = y + 2 ) or (
n = 3 &
z = y &
x = y &
t = y + 1 ) )
by A3, A5, Th49;
hence
( (
n = 2 &
y = x &
z = x + 1 &
t = x + 2 ) or (
n = 2 &
y = x + 1 &
z = x &
t = x + 2 ) or (
n = 2 &
z = y &
x = y + 1 &
t = y + 2 ) or (
n = 3 &
y = x &
z = x &
t = x + 1 ) )
;
verum end; suppose
(
z <= x &
x <= y )
;
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )then
( (
n = 2 &
x = z &
y = z + 1 &
t = z + 2 ) or (
n = 3 &
x = z &
y = z &
t = z + 1 ) )
by A4, A5, Th49;
hence
( (
n = 2 &
y = x &
z = x + 1 &
t = x + 2 ) or (
n = 2 &
y = x + 1 &
z = x &
t = x + 2 ) or (
n = 2 &
z = y &
x = y + 1 &
t = y + 2 ) or (
n = 3 &
y = x &
z = x &
t = x + 1 ) )
;
verum end; suppose
(
z <= y &
y <= x )
;
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )then
( (
n = 2 &
y = z &
x = z + 1 &
t = z + 2 ) or (
n = 3 &
y = z &
z = x &
t = z + 1 ) )
by A1, A5, Th49;
hence
( (
n = 2 &
y = x &
z = x + 1 &
t = x + 2 ) or (
n = 2 &
y = x + 1 &
z = x &
t = x + 2 ) or (
n = 2 &
z = y &
x = y + 1 &
t = y + 2 ) or (
n = 3 &
y = x &
z = x &
t = x + 1 ) )
;
verum end; end;
end;
A6:
x + 0 <= x + 1
by XREAL_1:6;
A7:
y + 0 <= y + 1
by XREAL_1:6;
assume
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )
; ((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t
per cases then
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 2 & y = x + 1 & z = x & t = x + 2 ) or ( n = 2 & z = y & x = y + 1 & t = y + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )
;
end;