Termination proof

1: switching to dependency pairs

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

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  f#( x , c( y ) )

  →

  f#( y , y )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

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

  one remains with the following pair(s).

  none

  1.1.1.1: P is empty

  All dependency pairs have been removed.

• The 2nd component contains the pair(s)

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

  →

  f#( x , s( c( s( y ) ) ) )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [f (x1, x2) ]	=	0
  [c (x1) ]	=	0
  [s (x1) ]	=	x1 + 1
  [f# (x1, x2) ]	=	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.