Termination proof

1: switching to dependency pairs

rev#( ls )	→	r1#( ls , empty )
r1#( cons( x , k ) , a )	→	r1#( k , cons( x , a ) )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  r1#( cons( x , k ) , a )

  →

  r1#( k , cons( x , a ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [rev (x1) ]	=	2 x1 + 3
  [empty]	=	0
  [cons (x1, x2) ]	=	x1 + 3
  [r1 (x1, x2) ]	=	2 x1 + x2
  [r1# (x1, x2) ]	=	3 x1 + x2
  [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.