Termination proof

1: switching to dependency pairs

first#( s( X ) , cons( Y , Z ) )	→	activate#( Z )
activate#( n__first( X1 , X2 ) )	→	first#( activate( X1 ) , activate( X2 ) )
activate#( n__first( X1 , X2 ) )	→	activate#( X1 )
activate#( n__first( X1 , X2 ) )	→	activate#( X2 )
activate#( n__from( X ) )	→	from#( activate( X ) )
activate#( n__from( X ) )	→	activate#( X )
activate#( n__s( X ) )	→	s#( activate( X ) )
activate#( n__s( X ) )	→	activate#( X )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  activate#( n__first( X1 , X2 ) )	→	first#( activate( X1 ) , activate( X2 ) )
  first#( s( X ) , cons( Y , Z ) )	→	activate#( Z )
  activate#( n__first( X1 , X2 ) )	→	activate#( X1 )
  activate#( n__first( X1 , X2 ) )	→	activate#( X2 )
  activate#( n__from( X ) )	→	activate#( X )
  activate#( n__s( X ) )	→	activate#( X )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [from (x1) ]	=	2 x1 + 1
  [activate# (x1) ]	=	x1
  [first (x1, x2) ]	=	3 x1 + x2
  [n__from (x1) ]	=	2 x1 + 1
  [first# (x1, x2) ]	=	x1
  [n__s (x1) ]	=	x1
  [s (x1) ]	=	x1
  [0]	=	1
  [nil]	=	0
  [cons (x1, x2) ]	=	x1
  [n__first (x1, x2) ]	=	3 x1 + x2
  [activate (x1) ]	=	x1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  activate#( n__first( X1 , X2 ) )	→	first#( activate( X1 ) , activate( X2 ) )
  first#( s( X ) , cons( Y , Z ) )	→	activate#( Z )
  activate#( n__first( X1 , X2 ) )	→	activate#( X1 )
  activate#( n__first( X1 , X2 ) )	→	activate#( X2 )
  activate#( n__s( X ) )	→	activate#( X )

  1.1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [from (x1) ]	=	0
  [activate# (x1) ]	=	3 x1 + 1
  [first (x1, x2) ]	=	2 x1 + x2 + 1
  [n__from (x1) ]	=	0
  [first# (x1, x2) ]	=	2 x1 + 2
  [n__s (x1) ]	=	3 x1 + 2
  [s (x1) ]	=	3 x1 + 2
  [0]	=	0
  [nil]	=	0
  [cons (x1, x2) ]	=	2 x1
  [n__first (x1, x2) ]	=	2 x1 + x2 + 1
  [activate (x1) ]	=	x1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.1.1.1: P is empty

  All dependency pairs have been removed.