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
         ↳FwdInst


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 w.r.t. to the implicit AFS 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
         ↳FwdInst


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
         ↳FwdInst


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 w.r.t. to the implicit AFS 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
         ↳FwdInst


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
         ↳Forward Instantiation Transformation


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)))





On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

QSORT(.(x, y)) -> QSORT(lowers(x, y))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳FwdInst
           →DP Problem 6
             ↳Narrowing Transformation


Dependency Pair:

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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QSORT(.(x, y)) -> QSORT(greaters(x, y))
two new Dependency Pairs are created:

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

The transformation is resulting in no new DP problems.


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