Termination proof

1: switching to dependency pairs

plus#( plus( X , Y ) , Z )	→	plus#( X , plus( Y , Z ) )
plus#( plus( X , Y ) , Z )	→	plus#( Y , Z )
times#( X , s( Y ) )	→	plus#( X , times( Y , X ) )
times#( X , s( Y ) )	→	times#( Y , X )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  times#( X , s( Y ) )

  →

  times#( Y , X )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [plus (x1, x2) ]	=	x1
  [s (x1) ]	=	2 x1 + 2
  [times# (x1, x2) ]	=	3 x1 + 2 x2
  [times (x1, x2) ]	=	x1 + x2 + 2
  [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)

  plus#( plus( X , Y ) , Z )	→	plus#( Y , Z )
  plus#( plus( X , Y ) , Z )	→	plus#( X , plus( Y , Z ) )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [plus (x1, x2) ]	=	x1 + x2 + 3
  [s (x1) ]	=	2 x1 + 3
  [plus# (x1, x2) ]	=	x1
  [times (x1, x2) ]	=	3 x1 + 2 x2 + 2
  [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.