Termination proof

1: switching to dependency pairs

fst#( s( X ) , cons( Y , Z ) )	→	activate#( X )
fst#( s( X ) , cons( Y , Z ) )	→	activate#( Z )
add#( s( X ) , Y )	→	activate#( X )
len#( cons( X , Z ) )	→	activate#( Z )
activate#( n__fst( X1 , X2 ) )	→	fst#( X1 , X2 )
activate#( n__from( X ) )	→	from#( X )
activate#( n__add( X1 , X2 ) )	→	add#( X1 , X2 )
activate#( n__len( X ) )	→	len#( X )

1.1: dependency graph processor

The dependency pairs are split into 1 component(s).

• The 1st component contains the pair(s)

  activate#( n__fst( X1 , X2 ) )	→	fst#( X1 , X2 )
  fst#( s( X ) , cons( Y , Z ) )	→	activate#( X )
  activate#( n__add( X1 , X2 ) )	→	add#( X1 , X2 )
  add#( s( X ) , Y )	→	activate#( X )
  activate#( n__len( X ) )	→	len#( X )
  len#( cons( X , Z ) )	→	activate#( Z )
  fst#( s( X ) , cons( Y , Z ) )	→	activate#( Z )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [from (x1) ]	=	2
  [fst (x1, x2) ]	=	x1 + 2 x2 + 2
  [n__from (x1) ]	=	0
  [0]	=	3
  [add# (x1, x2) ]	=	2 x1
  [len# (x1) ]	=	2 x1 + 1
  [nil]	=	2
  [cons (x1, x2) ]	=	x1 + 2
  [activate (x1) ]	=	x1 + 2
  [len (x1) ]	=	x1 + 2
  [n__add (x1, x2) ]	=	x1 + x2
  [activate# (x1) ]	=	2 x1 + 2
  [fst# (x1, x2) ]	=	2 x1 + 2 x2 + 1
  [n__fst (x1, x2) ]	=	x1 + 2 x2
  [add (x1, x2) ]	=	x1 + x2 + 2
  [s (x1) ]	=	x1 + 2
  [n__len (x1) ]	=	x1
  [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.