Termination proof

1: switching to dependency pairs

app#( cons( x , l ) , k )	→	app#( l , k )
sum#( cons( x , cons( y , l ) ) )	→	sum#( cons( plus( x , y ) , l ) )
sum#( cons( x , cons( y , l ) ) )	→	plus#( x , y )
sum#( app( l , cons( x , cons( y , k ) ) ) )	→	sum#( app( l , sum( cons( x , cons( y , k ) ) ) ) )
sum#( app( l , cons( x , cons( y , k ) ) ) )	→	app#( l , sum( cons( x , cons( y , k ) ) ) )
sum#( app( l , cons( x , cons( y , k ) ) ) )	→	sum#( cons( x , cons( y , k ) ) )
plus#( s( x ) , y )	→	plus#( x , y )
sum#( plus( cons( 0 , x ) , cons( y , l ) ) )	→	pred#( sum( cons( s( x ) , cons( y , l ) ) ) )
sum#( plus( cons( 0 , x ) , cons( y , l ) ) )	→	sum#( cons( s( x ) , cons( y , l ) ) )

1.1: dependency graph processor

The dependency pairs are split into 4 component(s).

• The 1st component contains the pair(s)

  sum#( app( l , cons( x , cons( y , k ) ) ) )

  →

  sum#( app( l , sum( cons( x , cons( y , k ) ) ) ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [sum# (x1) ]	=	2 x1
  [app (x1, x2) ]	=	2 x1 + 2 x2
  [sum (x1) ]	=	1
  [plus (x1, x2) ]	=	2 x1 + x2 + 2
  [0]	=	0
  [s (x1) ]	=	x1 + 3
  [pred (x1) ]	=	1
  [nil]	=	0
  [cons (x1, x2) ]	=	x1 + 1
  [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)

  sum#( cons( x , cons( y , l ) ) )

  →

  sum#( cons( plus( x , y ) , l ) )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [sum# (x1) ]	=	x1
  [plus (x1, x2) ]	=	2 x1
  [app (x1, x2) ]	=	x1 + 2 x2
  [sum (x1) ]	=	2
  [0]	=	0
  [s (x1) ]	=	0
  [pred (x1) ]	=	2
  [nil]	=	0
  [cons (x1, x2) ]	=	x1 + 2
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.2.1: P is empty

  All dependency pairs have been removed.

• The 3rd component contains the pair(s)

  app#( cons( x , l ) , k )

  →

  app#( l , k )

  1.1.3: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [app (x1, x2) ]	=	x1 + x2
  [sum (x1) ]	=	2
  [plus (x1, x2) ]	=	2 x1 + 3 x2 + 2
  [0]	=	0
  [s (x1) ]	=	3
  [app# (x1, x2) ]	=	2 x1
  [pred (x1) ]	=	2
  [nil]	=	0
  [cons (x1, x2) ]	=	x1 + 1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.3.1: P is empty

  All dependency pairs have been removed.

• The 4th component contains the pair(s)

  plus#( s( x ) , y )

  →

  plus#( x , y )

  1.1.4: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [app (x1, x2) ]	=	2 x1 + x2 + 1
  [sum (x1) ]	=	1
  [plus (x1, x2) ]	=	3 x1 + x2 + 2
  [s (x1) ]	=	x1 + 2
  [0]	=	0
  [plus# (x1, x2) ]	=	x1
  [pred (x1) ]	=	1
  [nil]	=	0
  [cons (x1, x2) ]	=	0
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.4.1: P is empty

  All dependency pairs have been removed.