Termination proof

1: switching to dependency pairs

rev#( cons( x , l ) )	→	rev1#( x , l )
rev#( cons( x , l ) )	→	rev2#( x , l )
rev1#( x , cons( y , l ) )	→	rev1#( y , l )
rev2#( x , cons( y , l ) )	→	rev#( cons( x , rev2( y , l ) ) )
rev2#( x , cons( y , l ) )	→	rev2#( y , l )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  rev#( cons( x , l ) )	→	rev2#( x , l )
  rev2#( x , cons( y , l ) )	→	rev#( cons( x , rev2( y , l ) ) )
  rev2#( x , cons( y , l ) )	→	rev2#( y , l )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [rev2# (x1, x2) ]	=	3 x1 + 3
  [rev2 (x1, x2) ]	=	x1
  [rev (x1) ]	=	x1
  [rev1 (x1, x2) ]	=	2 x1
  [rev# (x1) ]	=	x1 + 2
  [0]	=	0
  [s (x1) ]	=	0
  [nil]	=	0
  [cons (x1, x2) ]	=	3 x1 + 3
  [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)

  rev1#( x , cons( y , l ) )

  →

  rev1#( y , l )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [rev1# (x1, x2) ]	=	x1 + 3 x2
  [rev2 (x1, x2) ]	=	x1 + x2
  [rev (x1) ]	=	x1
  [rev1 (x1, x2) ]	=	0
  [0]	=	0
  [s (x1) ]	=	0
  [nil]	=	0
  [cons (x1, x2) ]	=	x1 + x2 + 1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.2.1: P is empty

  All dependency pairs have been removed.