Termination proof

1: switching to dependency pairs

sort#( cons( x , y ) )	→	insert#( x , sort( y ) )
sort#( cons( x , y ) )	→	sort#( y )
insert#( x , cons( v , w ) )	→	choose#( x , cons( v , w ) , x , v )
choose#( x , cons( v , w ) , 0 , s( z ) )	→	insert#( x , w )
choose#( x , cons( v , w ) , s( y ) , s( z ) )	→	choose#( x , cons( v , w ) , y , z )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  sort#( cons( x , y ) )

  →

  sort#( y )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [sort# (x1) ]	=	x1
  [insert (x1, x2) ]	=	x1 + 1
  [0]	=	0
  [s (x1) ]	=	0
  [sort (x1) ]	=	2 x1
  [nil]	=	0
  [cons (x1, x2) ]	=	x1 + 1
  [choose (x1, ..., x4) ]	=	x1 + 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: P is empty

  All dependency pairs have been removed.

• The 2nd component contains the pair(s)

  choose#( x , cons( v , w ) , 0 , s( z ) )	→	insert#( x , w )
  insert#( x , cons( v , w ) )	→	choose#( x , cons( v , w ) , x , v )
  choose#( x , cons( v , w ) , s( y ) , s( z ) )	→	choose#( x , cons( v , w ) , y , z )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [insert (x1, x2) ]	=	x1 + 2
  [insert# (x1, x2) ]	=	2 x1 + 2
  [choose# (x1, ..., x4) ]	=	2 x1
  [0]	=	0
  [s (x1) ]	=	0
  [sort (x1) ]	=	x1
  [nil]	=	0
  [cons (x1, x2) ]	=	x1 + 2
  [choose (x1, ..., x4) ]	=	x1 + 2
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  choose#( x , cons( v , w ) , s( y ) , s( z ) )

  →

  choose#( x , cons( v , w ) , y , z )

  1.1.2.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [insert (x1, x2) ]	=	3
  [choose# (x1, ..., x4) ]	=	x1
  [s (x1) ]	=	x1 + 1
  [0]	=	0
  [sort (x1) ]	=	3
  [nil]	=	0
  [cons (x1, x2) ]	=	0
  [choose (x1, ..., x4) ]	=	3
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.2.1.1: P is empty

  All dependency pairs have been removed.