Termination proof

1: switching to dependency pairs

and#( tt , X )	→	activate#( X )
plus#( N , s( M ) )	→	plus#( N , M )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  plus#( N , s( M ) )

  →

  plus#( N , M )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [and (x1, x2) ]	=	x1 + 1
  [plus (x1, x2) ]	=	x1 + 2 x2
  [s (x1) ]	=	x1 + 2
  [0]	=	0
  [plus# (x1, x2) ]	=	x1
  [tt]	=	0
  [activate (x1) ]	=	x1 + 1
  [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.