Term Rewriting System R:
[x, y, z]
msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MSORT(.(x, y)) -> MIN(x, y)
MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))
MSORT(.(x, y)) -> DEL(min(x, y), .(x, y))
MIN(x, .(y, z)) -> MIN(x, z)
MIN(x, .(y, z)) -> MIN(y, z)
DEL(x, .(y, z)) -> DEL(x, z)

Furthermore, R contains three SCCs. 


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


Dependency Pairs:

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


Rules:


msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MIN(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
         ↳Rw


Dependency Pair:


Rules:


msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(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
         ↳Rw


Dependency Pair:

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


Rules:


msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(DEL(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
         ↳Rw


Dependency Pair:


Rules:


msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(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
         ↳Rewriting Transformation


Dependency Pair:

MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))


Rules:


msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))


Strategy:

innermost




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

MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))
one new Dependency Pair is created:

MSORT(.(x, y)) -> MSORT(if(=(min(x, y), x), y, .(x, del(min(x, y), y))))

The transformation is resulting in no new DP problems.


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