Termination proof

1: switching to dependency pairs

perfectp#( s( x ) )	→	f#( x , s( 0 ) , s( x ) , s( x ) )
f#( s( x ) , 0 , z , u )	→	f#( x , u , minus( z , s( x ) ) , u )
f#( s( x ) , s( y ) , z , u )	→	f#( s( x ) , minus( y , x ) , z , u )
f#( s( x ) , s( y ) , z , u )	→	f#( x , u , z , u )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  f#( s( x ) , s( y ) , z , u )	→	f#( x , u , z , u )
  f#( s( x ) , 0 , z , u )	→	f#( x , u , minus( z , s( x ) ) , u )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [minus (x1, x2) ]	=	x1 + 2
  [true]	=	0
  [if (x1, x2, x3) ]	=	2 x1
  [f (x1, ..., x4) ]	=	3 x1
  [false]	=	0
  [f# (x1, ..., x4) ]	=	3 x1 + 3 x2
  [s (x1) ]	=	3 x1 + 3
  [0]	=	0
  [perfectp (x1) ]	=	3 x1 + 3
  [le (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

  1.1.1.1: P is empty

  All dependency pairs have been removed.