Termination proof

1: switching to dependency pairs

minus#( +( x , y ) )	→	minus#( minus( minus( x ) ) )
minus#( +( x , y ) )	→	minus#( minus( x ) )
minus#( +( x , y ) )	→	minus#( x )
minus#( +( x , y ) )	→	minus#( minus( minus( y ) ) )
minus#( +( x , y ) )	→	minus#( minus( y ) )
minus#( +( x , y ) )	→	minus#( y )
minus#( *( x , y ) )	→	minus#( minus( minus( x ) ) )
minus#( *( x , y ) )	→	minus#( minus( x ) )
minus#( *( x , y ) )	→	minus#( x )
minus#( *( x , y ) )	→	minus#( minus( minus( y ) ) )
minus#( *( x , y ) )	→	minus#( minus( y ) )
minus#( *( x , y ) )	→	minus#( y )
f#( minus( x ) )	→	minus#( minus( minus( f( x ) ) ) )
f#( minus( x ) )	→	minus#( minus( f( x ) ) )
f#( minus( x ) )	→	minus#( f( x ) )
f#( minus( x ) )	→	f#( x )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  f#( minus( x ) )

  →

  f#( x )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [minus (x1) ]	=	x1 + 2
  [+ (x1, x2) ]	=	2
  [f (x1) ]	=	3 x1
  [f# (x1) ]	=	x1
  [* (x1, x2) ]	=	0
  [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#( +( x , y ) )	→	minus#( minus( x ) )
  minus#( +( x , y ) )	→	minus#( minus( minus( x ) ) )
  minus#( +( x , y ) )	→	minus#( x )
  minus#( +( x , y ) )	→	minus#( minus( minus( y ) ) )
  minus#( +( x , y ) )	→	minus#( minus( y ) )
  minus#( +( x , y ) )	→	minus#( y )
  minus#( *( x , y ) )	→	minus#( minus( minus( x ) ) )
  minus#( *( x , y ) )	→	minus#( minus( x ) )
  minus#( *( x , y ) )	→	minus#( x )
  minus#( *( x , y ) )	→	minus#( minus( minus( y ) ) )
  minus#( *( x , y ) )	→	minus#( minus( y ) )
  minus#( *( x , y ) )	→	minus#( y )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [minus (x1) ]	=	x1
  [+ (x1, x2) ]	=	2 x1 + 2 x2 + 1
  [minus# (x1) ]	=	2 x1
  [f (x1) ]	=	0
  [* (x1, x2) ]	=	2 x1 + 2 x2 + 1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.2.1: P is empty

  All dependency pairs have been removed.