Termination proof

1: switching to dependency pairs

prime#( s( s( x ) ) )	→	prime1#( s( s( x ) ) , s( x ) )
prime1#( x , s( s( y ) ) )	→	divp#( s( s( y ) ) , x )
prime1#( x , s( s( y ) ) )	→	prime1#( x , s( y ) )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  prime1#( x , s( s( y ) ) )

  →

  prime1#( x , s( y ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [true]	=	0
  [rem (x1, x2) ]	=	x1
  [prime (x1) ]	=	0
  [and (x1, x2) ]	=	0
  [false]	=	0
  [divp (x1, x2) ]	=	2 x1 + 2
  [prime1# (x1, x2) ]	=	2 x1
  [s (x1) ]	=	3 x1 + 3
  [0]	=	0
  [= (x1, x2) ]	=	x1 + 1
  [prime1 (x1, x2) ]	=	0
  [not (x1) ]	=	0
  [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.