Termination proof

1: switching to dependency pairs

minus#( s( x ) , s( y ) )	→	minus#( x , y )
quot#( s( x ) , s( y ) )	→	quot#( minus( x , y ) , s( y ) )
quot#( s( x ) , s( y ) )	→	minus#( x , y )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  quot#( s( x ) , s( y ) )

  →

  quot#( minus( x , y ) , s( y ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [minus (x1, x2) ]	=	x1
  [quot# (x1, x2) ]	=	x1
  [quot (x1, x2) ]	=	3 x1
  [s (x1) ]	=	x1 + 1
  [0]	=	3
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.1.1: P is empty

  All dependency pairs have been removed.

• The 2nd component contains the pair(s)

  minus#( s( x ) , s( y ) )

  →

  minus#( x , y )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [minus (x1, x2) ]	=	x1
  [quot (x1, x2) ]	=	x1
  [s (x1) ]	=	x1 + 2
  [0]	=	1
  [minus# (x1, x2) ]	=	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.