Termination proof

1: switching to dependency pairs

le#( s( X ) , s( Y ) )	→	le#( X , Y )
minus#( s( X ) , Y )	→	ifMinus#( le( s( X ) , Y ) , s( X ) , Y )
minus#( s( X ) , Y )	→	le#( s( X ) , Y )
ifMinus#( false , s( X ) , Y )	→	minus#( X , Y )
quot#( s( X ) , s( Y ) )	→	quot#( minus( X , Y ) , s( Y ) )
quot#( s( X ) , s( Y ) )	→	minus#( 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#( minus( X , Y ) , s( Y ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [minus (x1, x2) ]	=	x1
  [true]	=	0
  [ifMinus (x1, x2, x3) ]	=	x1
  [quot# (x1, x2) ]	=	3 x1
  [false]	=	0
  [quot (x1, x2) ]	=	x1
  [s (x1) ]	=	2 x1 + 1
  [0]	=	0
  [le (x1, x2) ]	=	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.

• The 2nd component contains the pair(s)

  minus#( s( X ) , Y )	→	ifMinus#( le( s( X ) , Y ) , s( X ) , Y )
  ifMinus#( false , s( X ) , Y )	→	minus#( X , Y )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [true]	=	0
  [minus (x1, x2) ]	=	x1
  [ifMinus (x1, x2, x3) ]	=	x1
  [false]	=	0
  [ifMinus# (x1, x2, x3) ]	=	x1
  [quot (x1, x2) ]	=	2 x1 + 2 x2
  [s (x1) ]	=	x1 + 2
  [0]	=	0
  [le (x1, x2) ]	=	0
  [minus# (x1, x2) ]	=	x1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  minus#( s( X ) , Y )

  →

  ifMinus#( le( s( X ) , Y ) , s( X ) , Y )

  1.1.2.1: dependency graph processor

  The dependency pairs are split into 0 component(s).

• The 3rd component contains the pair(s)

  le#( s( X ) , s( Y ) )

  →

  le#( X , Y )

  1.1.3: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [true]	=	0
  [minus (x1, x2) ]	=	x1
  [ifMinus (x1, x2, x3) ]	=	x1
  [false]	=	0
  [quot (x1, x2) ]	=	x1
  [le# (x1, x2) ]	=	x1
  [s (x1) ]	=	x1 + 1
  [0]	=	0
  [le (x1, x2) ]	=	0
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.3.1: P is empty

  All dependency pairs have been removed.