Termination proof

1: switching to dependency pairs

minus#( s( x ) , s( y ) )	→	minus#( x , y )
f#( s( x ) )	→	minus#( s( x ) , g( f( x ) ) )
f#( s( x ) )	→	g#( f( x ) )
f#( s( x ) )	→	f#( x )
g#( s( x ) )	→	minus#( s( x ) , f( g( x ) ) )
g#( s( x ) )	→	f#( g( x ) )
g#( s( x ) )	→	g#( x )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  f#( s( x ) )	→	g#( f( x ) )
  g#( s( x ) )	→	f#( g( x ) )
  f#( s( x ) )	→	f#( x )
  g#( s( x ) )	→	g#( x )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

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

  one remains with the following pair(s).

  f#( s( x ) )

  →

  g#( f( x ) )

  1.1.1.1: dependency graph processor

  The dependency pairs are split into 0 component(s).

• 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) ]	=	2 x1
  [f (x1) ]	=	3 x1 + 1
  [s (x1) ]	=	x1 + 2
  [0]	=	3
  [minus# (x1, x2) ]	=	x1
  [g (x1) ]	=	3 x1 + 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: P is empty

  All dependency pairs have been removed.