Termination proof

1: switching to dependency pairs

-#( s( x ) , s( y ) )	→	-#( x , y )
lt#( s( x ) , s( y ) )	→	lt#( x , y )
div#( s( x ) , s( y ) )	→	if#( lt( x , y ) , 0 , s( div( -( x , y ) , s( y ) ) ) )
div#( s( x ) , s( y ) )	→	lt#( x , y )
div#( s( x ) , s( y ) )	→	div#( -( x , y ) , s( y ) )
div#( 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)

  div#( s( x ) , s( y ) )

  →

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

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [true]	=	0
  [if (x1, x2, x3) ]	=	x1 + x2
  [false]	=	0
  [lt (x1, x2) ]	=	1
  [- (x1, x2) ]	=	x1
  [s (x1) ]	=	x1 + 2
  [0]	=	0
  [div# (x1, x2) ]	=	x1
  [div (x1, x2) ]	=	x1 + 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)

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

  →

  -#( x , y )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [true]	=	3
  [if (x1, x2, x3) ]	=	2 x1 + x2
  [false]	=	1
  [lt (x1, x2) ]	=	2 x1 + x2 + 2
  [- (x1, x2) ]	=	x1
  [s (x1) ]	=	2 x1 + 2
  [0]	=	0
  [div (x1, x2) ]	=	2 x1
  [-# (x1, x2) ]	=	3 x1 + 3 x2
  [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)

  lt#( s( x ) , s( y ) )

  →

  lt#( x , y )

  1.1.3: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [true]	=	3
  [if (x1, x2, x3) ]	=	2 x1 + x2
  [false]	=	1
  [lt (x1, x2) ]	=	2 x1 + x2 + 2
  [- (x1, x2) ]	=	x1
  [s (x1) ]	=	2 x1 + 2
  [0]	=	0
  [lt# (x1, x2) ]	=	3 x1 + 3 x2
  [div (x1, x2) ]	=	2 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.