Termination proof

1: switching to dependency pairs

exp#( x , s( y ) )	→	*#( x , exp( x , y ) )
exp#( x , s( y ) )	→	exp#( x , y )
*#( s( x ) , y )	→	*#( x , y )
-#( s( x ) , s( y ) )	→	-#( x , y )

1.1: dependency graph processor

The dependency pairs are split into 3 component(s).

• The 1st component contains the pair(s)

  exp#( x , s( y ) )

  →

  exp#( x , y )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [exp (x1, x2) ]	=	2 x1 + 1
  [+ (x1, x2) ]	=	0
  [exp# (x1, x2) ]	=	3 x1
  [- (x1, x2) ]	=	3 x1 + 3 x2 + 3
  [s (x1) ]	=	2 x1 + 1
  [0]	=	3
  [* (x1, x2) ]	=	3
  [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)

  *#( s( x ) , y )

  →

  *#( x , y )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [exp (x1, x2) ]	=	2 x1 + 1
  [*# (x1, x2) ]	=	3 x1
  [+ (x1, x2) ]	=	0
  [- (x1, x2) ]	=	3 x1 + 3 x2 + 3
  [s (x1) ]	=	2 x1 + 1
  [0]	=	3
  [* (x1, x2) ]	=	3
  [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.

• The 3rd component contains the pair(s)

  -#( s( x ) , s( y ) )

  →

  -#( x , y )

  1.1.3: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [exp (x1, x2) ]	=	2
  [+ (x1, x2) ]	=	0
  [- (x1, x2) ]	=	3 x1 + 2 x2
  [s (x1) ]	=	2 x1 + 1
  [0]	=	0
  [-# (x1, x2) ]	=	x1
  [* (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

  1.1.3.1: P is empty

  All dependency pairs have been removed.