Termination proof

1: switching to dependency pairs

s#( f( x , y ) )	→	f#( s( y ) , s( x ) )
s#( f( x , y ) )	→	s#( y )
s#( f( x , y ) )	→	s#( x )
s#( g( x , y ) )	→	g#( s( x ) , s( y ) )
s#( g( x , y ) )	→	s#( x )
s#( g( x , y ) )	→	s#( y )
f#( g( x , y ) , g( u , v ) )	→	g#( f( x , u ) , f( y , v ) )
f#( g( x , y ) , g( u , v ) )	→	f#( x , u )
f#( g( x , y ) , g( u , v ) )	→	f#( y , v )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

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

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [g (x1, x2) ]	=	2 x1 + 3 x2 + 3
  [a]	=	3
  [f (x1, x2) ]	=	3 x1 + 3 x2 + 3
  [s (x1) ]	=	x1
  [s# (x1) ]	=	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.

• The 2nd component contains the pair(s)

  f#( g( x , y ) , g( u , v ) )	→	f#( y , v )
  f#( g( x , y ) , g( u , v ) )	→	f#( x , u )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [g (x1, x2) ]	=	3 x1 + 2 x2 + 1
  [a]	=	3
  [f (x1, x2) ]	=	2 x1 + 2 x2
  [s (x1) ]	=	x1
  [f# (x1, x2) ]	=	2 x1 + x2
  [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.