Termination proof

1: switching to dependency pairs

f#( cons( f( cons( nil , y ) ) , z ) )	→	copy#( n , y , z )
copy#( 0 , y , z )	→	f#( z )
copy#( s( x ) , y , z )	→	copy#( x , y , cons( f( y ) , z ) )
copy#( s( x ) , y , z )	→	f#( y )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  copy#( s( x ) , y , z )

  →

  copy#( x , y , cons( f( y ) , z ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [copy# (x1, x2, x3) ]	=	3 x1 + x2
  [f (x1) ]	=	x1 + 3
  [s (x1) ]	=	2 x1 + 3
  [copy (x1, x2, x3) ]	=	2 x1 + x2 + 2
  [0]	=	3
  [nil]	=	0
  [cons (x1, x2) ]	=	x1 + 3
  [n]	=	0
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.1.1: P is empty

  All dependency pairs have been removed.