Termination proof

1: switching to dependency pairs

+#( s( x ) , s( y ) )	→	+#( s( x ) , +( y , 0 ) )
+#( s( x ) , s( y ) )	→	+#( y , 0 )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  +#( s( x ) , s( y ) )

  →

  +#( s( x ) , +( y , 0 ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

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