Termination proof

1: switching to dependency pairs

le#( s( x ) , s( y ) )	→	le#( x , y )
minus#( s( x ) , y )	→	if_minus#( le( s( x ) , y ) , s( x ) , y )
minus#( s( x ) , y )	→	le#( s( x ) , y )
if_minus#( false , s( x ) , y )	→	minus#( x , y )
gcd#( s( x ) , s( y ) )	→	if_gcd#( le( y , x ) , s( x ) , s( y ) )
gcd#( s( x ) , s( y ) )	→	le#( y , x )
if_gcd#( true , s( x ) , s( y ) )	→	gcd#( minus( x , y ) , s( y ) )
if_gcd#( true , s( x ) , s( y ) )	→	minus#( x , y )
if_gcd#( false , s( x ) , s( y ) )	→	gcd#( minus( y , x ) , s( x ) )
if_gcd#( false , s( x ) , s( y ) )	→	minus#( y , x )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  if_gcd#( true , s( x ) , s( y ) )	→	gcd#( minus( x , y ) , s( y ) )
  gcd#( s( x ) , s( y ) )	→	if_gcd#( le( y , x ) , s( x ) , s( y ) )
  if_gcd#( false , s( x ) , s( y ) )	→	gcd#( minus( y , x ) , s( x ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [true]	=	0
  [minus (x1, x2) ]	=	2 x1
  [if_gcd# (x1, x2, x3) ]	=	3 x1 + 3 x2
  [gcd# (x1, x2) ]	=	3 x1 + 3 x2 + 1
  [false]	=	0
  [if_minus (x1, x2, x3) ]	=	2 x1
  [s (x1) ]	=	3 x1 + 3
  [0]	=	0
  [le (x1, x2) ]	=	0
  [gcd (x1, x2) ]	=	x1 + x2 + 3
  [if_gcd (x1, x2, x3) ]	=	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)

  minus#( s( x ) , y )	→	if_minus#( le( s( x ) , y ) , s( x ) , y )
  if_minus#( 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
  [false]	=	0
  [if_minus (x1, x2, x3) ]	=	x1
  [s (x1) ]	=	2 x1 + 2
  [0]	=	0
  [if_minus# (x1, x2, x3) ]	=	x1
  [le (x1, x2) ]	=	0
  [gcd (x1, x2) ]	=	x1 + x2 + 2
  [if_gcd (x1, x2, x3) ]	=	x1 + x2
  [minus# (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).

  none

  1.1.2.1: P is empty

  All dependency pairs have been removed.

• 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
  [false]	=	0
  [if_minus (x1, x2, x3) ]	=	x1
  [le# (x1, x2) ]	=	x1
  [s (x1) ]	=	2 x1 + 1
  [0]	=	0
  [le (x1, x2) ]	=	2 x1
  [gcd (x1, x2) ]	=	2 x1 + x2 + 1
  [if_gcd (x1, x2, x3) ]	=	x1 + x2 + x3
  [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.