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)))

Innermost 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)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost 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)))


Strategy:

innermost




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)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost 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)))


Strategy:

innermost




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)))


Strategy:

innermost




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 for innermost 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)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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