Termination proof

1: switching to dependency pairs

f#( j( x , y ) , y )	→	g#( f( x , k( y ) ) )
f#( j( x , y ) , y )	→	f#( x , k( y ) )
f#( j( x , y ) , y )	→	k#( y )
f#( x , h1( y , z ) )	→	h2#( 0 , x , h1( y , z ) )
g#( h2( x , y , h1( z , u ) ) )	→	h2#( s( x ) , y , h1( z , u ) )
h2#( x , j( y , h1( z , u ) ) , h1( z , u ) )	→	h2#( s( x ) , y , h1( s( z ) , u ) )

1.1: dependency graph processor

The dependency pairs are split into 2 component(s).

• The 1st component contains the pair(s)

  f#( j( x , y ) , y )

  →

  f#( x , k( y ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [h (x1) ]	=	x1
  [j (x1, x2) ]	=	2 x1 + x2 + 3
  [i (x1) ]	=	x1
  [k (x1) ]	=	2 x1
  [h2 (x1, x2, x3) ]	=	x1 + x2 + 1
  [f (x1, x2) ]	=	2 x1 + 2 x2 + 2
  [h1 (x1, x2) ]	=	2 x1
  [0]	=	0
  [s (x1) ]	=	0
  [f# (x1, x2) ]	=	x1
  [g (x1) ]	=	x1 + 2
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.1.1: P is empty

  All dependency pairs have been removed.

• The 2nd component contains the pair(s)

  h2#( x , j( y , h1( z , u ) ) , h1( z , u ) )

  →

  h2#( s( x ) , y , h1( s( z ) , u ) )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [h (x1) ]	=	x1 + 1
  [j (x1, x2) ]	=	x1 + 3 x2 + 1
  [i (x1) ]	=	x1
  [k (x1) ]	=	x1
  [h2 (x1, x2, x3) ]	=	x1
  [f (x1, x2) ]	=	x1 + x2 + 1
  [h1 (x1, x2) ]	=	3 x1 + x2 + 1
  [s (x1) ]	=	x1
  [0]	=	0
  [g (x1) ]	=	x1 + 1
  [h2# (x1, x2, x3) ]	=	2 x1 + 3 x2 + 2 x3
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.2.1: P is empty

  All dependency pairs have been removed.