Termination proof

1: switching to dependency pairs

app#( app( app( compose , f ) , g ) , x )	→	app#( g , app( f , x ) )
app#( app( app( compose , f ) , g ) , x )	→	app#( f , x )
app#( reverse , l )	→	app#( app( reverse2 , l ) , nil )
app#( reverse , l )	→	app#( reverse2 , l )
app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )	→	app#( app( reverse2 , xs ) , app( app( cons , x ) , l ) )
app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )	→	app#( reverse2 , xs )
app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )	→	app#( app( cons , x ) , l )
app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )	→	app#( cons , x )
last#	→	app#( app( compose , hd ) , reverse )
last#	→	app#( compose , hd )
init#	→	app#( app( compose , reverse ) , app( app( compose , tl ) , reverse ) )
init#	→	app#( compose , reverse )
init#	→	app#( app( compose , tl ) , reverse )
init#	→	app#( compose , tl )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  app#( app( app( compose , f ) , g ) , x )	→	app#( f , x )
  app#( app( app( compose , f ) , g ) , x )	→	app#( g , app( f , x ) )
  app#( reverse , l )	→	app#( app( reverse2 , l ) , nil )
  app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )	→	app#( app( reverse2 , xs ) , app( app( cons , x ) , l ) )
  app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )	→	app#( app( cons , x ) , l )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [last]	=	1
  [hd]	=	0
  [app (x1, x2) ]	=	x1 + x2
  [compose]	=	0
  [tl]	=	0
  [init]	=	3
  [reverse2]	=	0
  [app# (x1, x2) ]	=	2 x1 + 2 x2
  [nil]	=	0
  [reverse]	=	1
  [cons]	=	0
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  app#( app( app( compose , f ) , g ) , x )	→	app#( f , x )
  app#( app( app( compose , f ) , g ) , x )	→	app#( g , app( f , x ) )
  app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )	→	app#( app( reverse2 , xs ) , app( app( cons , x ) , l ) )
  app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )	→	app#( app( cons , x ) , l )

  1.1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [last]	=	1
  [hd]	=	0
  [app (x1, x2) ]	=	x1 + x2
  [compose]	=	1
  [tl]	=	0
  [init]	=	2
  [reverse2]	=	0
  [app# (x1, x2) ]	=	x1 + x2
  [nil]	=	0
  [cons]	=	1
  [reverse]	=	0
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )	→	app#( app( reverse2 , xs ) , app( app( cons , x ) , l ) )
  app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )	→	app#( app( cons , x ) , l )

  1.1.1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [last]	=	3
  [hd]	=	0
  [app (x1, x2) ]	=	x1 + x2
  [compose]	=	0
  [tl]	=	1
  [init]	=	2
  [reverse2]	=	0
  [app# (x1, x2) ]	=	2 x1 + x2
  [nil]	=	0
  [cons]	=	1
  [reverse]	=	0
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  app#( app( reverse2 , app( app( cons , x ) , xs ) ) , l )

  →

  app#( app( cons , x ) , l )

  1.1.1.1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [last]	=	2
  [hd]	=	0
  [app (x1, x2) ]	=	x1 + x2
  [compose]	=	0
  [tl]	=	1
  [init]	=	3
  [reverse2]	=	1
  [app# (x1, x2) ]	=	2 x1
  [nil]	=	0
  [cons]	=	0
  [reverse]	=	1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.1.1.1.1.1: P is empty

  All dependency pairs have been removed.