Termination proof

1: switching to dependency pairs

flatten#( unit( x ) )	→	flatten#( x )
flatten#( ++( x , y ) )	→	++#( flatten( x ) , flatten( y ) )
flatten#( ++( x , y ) )	→	flatten#( x )
flatten#( ++( x , y ) )	→	flatten#( y )
flatten#( ++( unit( x ) , y ) )	→	++#( flatten( x ) , flatten( y ) )
flatten#( ++( unit( x ) , y ) )	→	flatten#( x )
flatten#( ++( unit( x ) , y ) )	→	flatten#( y )
flatten#( flatten( x ) )	→	flatten#( x )
rev#( ++( x , y ) )	→	++#( rev( y ) , rev( x ) )
rev#( ++( x , y ) )	→	rev#( y )
rev#( ++( x , y ) )	→	rev#( x )
++#( ++( x , y ) , z )	→	++#( x , ++( y , z ) )
++#( ++( x , y ) , z )	→	++#( y , z )

1.1: dependency graph processor

The dependency pairs are split into 3 component(s).

• The 1st component contains the pair(s)

  flatten#( ++( x , y ) )	→	flatten#( x )
  flatten#( unit( x ) )	→	flatten#( x )
  flatten#( ++( x , y ) )	→	flatten#( y )
  flatten#( ++( unit( x ) , y ) )	→	flatten#( x )
  flatten#( ++( unit( x ) , y ) )	→	flatten#( y )
  flatten#( flatten( x ) )	→	flatten#( x )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [++ (x1, x2) ]	=	x1 + x2 + 3
  [unit (x1) ]	=	x1 + 3
  [rev (x1) ]	=	3 x1 + 3
  [flatten# (x1) ]	=	3 x1
  [nil]	=	0
  [flatten (x1) ]	=	2 x1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  flatten#( flatten( x ) )

  →

  flatten#( x )

  1.1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [++ (x1, x2) ]	=	x1 + x2 + 2
  [unit (x1) ]	=	2 x1
  [rev (x1) ]	=	2 x1 + 1
  [flatten# (x1) ]	=	2 x1
  [nil]	=	2
  [flatten (x1) ]	=	2 x1 + 2
  [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.

• The 2nd component contains the pair(s)

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

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [++ (x1, x2) ]	=	x1 + x2 + 1
  [unit (x1) ]	=	x1
  [rev (x1) ]	=	x1
  [rev# (x1) ]	=	x1
  [nil]	=	0
  [flatten (x1) ]	=	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: P is empty

  All dependency pairs have been removed.

• The 3rd component contains the pair(s)

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

  1.1.3: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

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

  one remains with the following pair(s).

  none

  1.1.3.1: P is empty

  All dependency pairs have been removed.