Termination proof

1: switching to dependency pairs

plus#( s( X ) , Y )	→	plus#( X , Y )
min#( s( X ) , s( Y ) )	→	min#( X , Y )
min#( min( X , Y ) , Z )	→	min#( X , plus( Y , Z ) )
min#( min( X , Y ) , Z )	→	plus#( Y , Z )
quot#( s( X ) , s( Y ) )	→	quot#( min( X , Y ) , s( Y ) )
quot#( s( X ) , s( Y ) )	→	min#( X , Y )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  quot#( s( X ) , s( Y ) )

  →

  quot#( min( X , Y ) , s( Y ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [quot# (x1, x2) ]	=	x1
  [plus (x1, x2) ]	=	2 x1 + x2
  [quot (x1, x2) ]	=	2 x1
  [s (x1) ]	=	x1 + 1
  [0]	=	2
  [Z]	=	0
  [min (x1, x2) ]	=	x1
  [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)

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

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [min# (x1, x2) ]	=	x1
  [plus (x1, x2) ]	=	2 x1
  [quot (x1, x2) ]	=	3
  [s (x1) ]	=	x1
  [0]	=	2
  [Z]	=	0
  [min (x1, x2) ]	=	2 x1 + 1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  min#( s( X ) , s( Y ) )

  →

  min#( X , Y )

  1.1.2.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

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

  one remains with the following pair(s).

  none

  1.1.2.1.1: P is empty

  All dependency pairs have been removed.

• The 3rd component contains the pair(s)

  plus#( s( X ) , Y )

  →

  plus#( X , Y )

  1.1.3: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [plus (x1, x2) ]	=	3 x1 + 2 x2 + 3
  [quot (x1, x2) ]	=	x1
  [s (x1) ]	=	x1 + 2
  [0]	=	0
  [plus# (x1, x2) ]	=	x1
  [Z]	=	0
  [min (x1, x2) ]	=	x1
  [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.