Term Rewriting System R:
[x, y, z]
qsort(nil) -> nil
qsort(.(x, y)) -> ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) -> nil
lowers(x, .(y, z)) -> if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) -> nil
greaters(x, .(y, z)) -> if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

QSORT(.(x, y)) -> QSORT(lowers(x, y))
QSORT(.(x, y)) -> LOWERS(x, y)
QSORT(.(x, y)) -> QSORT(greaters(x, y))
QSORT(.(x, y)) -> GREATERS(x, y)
LOWERS(x, .(y, z)) -> LOWERS(x, z)
GREATERS(x, .(y, z)) -> GREATERS(x, z)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo


Dependency Pair:

LOWERS(x, .(y, z)) -> LOWERS(x, z)


Rules:


qsort(nil) -> nil
qsort(.(x, y)) -> ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) -> nil
lowers(x, .(y, z)) -> if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) -> nil
greaters(x, .(y, z)) -> if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))





The following dependency pair can be strictly oriented:

LOWERS(x, .(y, z)) -> LOWERS(x, z)


There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(LOWERS(x1, x2))=  x2  
  POL(.(x1, x2))=  1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 4
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo


Dependency Pair:


Rules:


qsort(nil) -> nil
qsort(.(x, y)) -> ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) -> nil
lowers(x, .(y, z)) -> if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) -> nil
greaters(x, .(y, z)) -> if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Polo


Dependency Pair:

GREATERS(x, .(y, z)) -> GREATERS(x, z)


Rules:


qsort(nil) -> nil
qsort(.(x, y)) -> ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) -> nil
lowers(x, .(y, z)) -> if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) -> nil
greaters(x, .(y, z)) -> if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))





The following dependency pair can be strictly oriented:

GREATERS(x, .(y, z)) -> GREATERS(x, z)


There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(GREATERS(x1, x2))=  x2  
  POL(.(x1, x2))=  1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 5
Dependency Graph
       →DP Problem 3
Polo


Dependency Pair:


Rules:


qsort(nil) -> nil
qsort(.(x, y)) -> ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) -> nil
lowers(x, .(y, z)) -> if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) -> nil
greaters(x, .(y, z)) -> if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polynomial Ordering


Dependency Pairs:

QSORT(.(x, y)) -> QSORT(greaters(x, y))
QSORT(.(x, y)) -> QSORT(lowers(x, y))


Rules:


qsort(nil) -> nil
qsort(.(x, y)) -> ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) -> nil
lowers(x, .(y, z)) -> if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) -> nil
greaters(x, .(y, z)) -> if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))





The following dependency pairs can be strictly oriented:

QSORT(.(x, y)) -> QSORT(greaters(x, y))
QSORT(.(x, y)) -> QSORT(lowers(x, y))


Additionally, the following usable rules using the Ce-refinement can be oriented:

greaters(x, nil) -> nil
greaters(x, .(y, z)) -> if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))
lowers(x, nil) -> nil
lowers(x, .(y, z)) -> if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  0  
  POL(greaters(x1, x2))=  0  
  POL(nil)=  0  
  POL(.(x1, x2))=  1  
  POL(lowers(x1, x2))=  0  
  POL(<=(x1, x2))=  0  
  POL(QSORT(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 6
Dependency Graph


Dependency Pair:


Rules:


qsort(nil) -> nil
qsort(.(x, y)) -> ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) -> nil
lowers(x, .(y, z)) -> if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) -> nil
greaters(x, .(y, z)) -> if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes