Termination proof

1: switching to dependency pairs

rev#( ++( x , y ) )	→	++#( rev( y ) , rev( x ) )
rev#( ++( x , y ) )	→	rev#( y )
rev#( ++( x , y ) )	→	rev#( x )
++#( .( x , y ) , z )	→	++#( y , z )
++#( x , ++( y , z ) )	→	++#( ++( x , y ) , z )
++#( x , ++( y , z ) )	→	++#( x , y )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  rev#( ++( x , y ) )	→	rev#( x )
  rev#( ++( x , y ) )	→	rev#( y )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [++ (x1, x2) ]	=	x1 + x2 + 1
  [rev (x1) ]	=	x1
  [rev# (x1) ]	=	3 x1
  [make (x1) ]	=	3 x1 + 1
  [nil]	=	1
  [. (x1, x2) ]	=	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.

• The 2nd component contains the pair(s)

  ++#( x , ++( y , z ) )	→	++#( ++( x , y ) , z )
  ++#( .( x , y ) , z )	→	++#( y , z )
  ++#( x , ++( y , z ) )	→	++#( x , y )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [++ (x1, x2) ]	=	x1 + x2 + 1
  [rev (x1) ]	=	x1
  [make (x1) ]	=	x1 + 2
  [. (x1, x2) ]	=	0
  [nil]	=	0
  [++# (x1, x2) ]	=	x1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  ++#( .( x , y ) , z )

  →

  ++#( y , z )

  1.1.2.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [++ (x1, x2) ]	=	x1 + x2
  [rev (x1) ]	=	2 x1
  [make (x1) ]	=	x1 + 1
  [. (x1, x2) ]	=	x1 + 1
  [nil]	=	0
  [++# (x1, x2) ]	=	2 x1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.2.1.1: P is empty

  All dependency pairs have been removed.