Termination proof

1: switching to dependency pairs

g#( x , s( y ) )	→	g#( f( x , y ) , 0 )
g#( s( x ) , y )	→	g#( f( x , y ) , 0 )
g#( f( x , y ) , 0 )	→	g#( x , 0 )
g#( f( x , y ) , 0 )	→	g#( y , 0 )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  g#( f( x , y ) , 0 )	→	g#( x , 0 )
  g#( s( x ) , y )	→	g#( f( x , y ) , 0 )
  g#( f( x , y ) , 0 )	→	g#( y , 0 )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

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

  one remains with the following pair(s).

  g#( s( x ) , y )

  →

  g#( f( x , y ) , 0 )

  1.1.1.1: dependency graph processor

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