Term Rewriting System R:
[y, n, x]
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
reverse(nil) -> nil
reverse(add(n, x)) -> app(reverse(x), add(n, nil))
shuffle(nil) -> nil
shuffle(add(n, x)) -> add(n, shuffle(reverse(x)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(add(n, x), y) -> APP(x, y)
REVERSE(add(n, x)) -> APP(reverse(x), add(n, nil))
REVERSE(add(n, x)) -> REVERSE(x)
SHUFFLE(add(n, x)) -> SHUFFLE(reverse(x))
SHUFFLE(add(n, x)) -> REVERSE(x)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pair:

APP(add(n, x), y) -> APP(x, y)


Rules:


app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
reverse(nil) -> nil
reverse(add(n, x)) -> app(reverse(x), add(n, nil))
shuffle(nil) -> nil
shuffle(add(n, x)) -> add(n, shuffle(reverse(x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

APP(add(n, x), y) -> APP(x, y)


There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(APP(x1, x2))=  x1 + x2  
  POL(add(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> APP(x1, x2)
add(x1, x2) -> add(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pair:


Rules:


app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
reverse(nil) -> nil
reverse(add(n, x)) -> app(reverse(x), add(n, nil))
shuffle(nil) -> nil
shuffle(add(n, x)) -> add(n, shuffle(reverse(x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS


Dependency Pair:

REVERSE(add(n, x)) -> REVERSE(x)


Rules:


app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
reverse(nil) -> nil
reverse(add(n, x)) -> app(reverse(x), add(n, nil))
shuffle(nil) -> nil
shuffle(add(n, x)) -> add(n, shuffle(reverse(x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

REVERSE(add(n, x)) -> REVERSE(x)


There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(REVERSE(x1))=  x1  
  POL(add(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
REVERSE(x1) -> REVERSE(x1)
add(x1, x2) -> add(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳AFS


Dependency Pair:


Rules:


app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
reverse(nil) -> nil
reverse(add(n, x)) -> app(reverse(x), add(n, nil))
shuffle(nil) -> nil
shuffle(add(n, x)) -> add(n, shuffle(reverse(x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Argument Filtering and Ordering


Dependency Pair:

SHUFFLE(add(n, x)) -> SHUFFLE(reverse(x))


Rules:


app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
reverse(nil) -> nil
reverse(add(n, x)) -> app(reverse(x), add(n, nil))
shuffle(nil) -> nil
shuffle(add(n, x)) -> add(n, shuffle(reverse(x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

SHUFFLE(add(n, x)) -> SHUFFLE(reverse(x))


The following usable rules for innermost can be oriented:

reverse(nil) -> nil
reverse(add(n, x)) -> app(reverse(x), add(n, nil))
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(reverse(x1))=  x1  
  POL(SHUFFLE(x1))=  1 + x1  
  POL(nil)=  0  
  POL(app(x1, x2))=  x1 + x2  
  POL(add(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
SHUFFLE(x1) -> SHUFFLE(x1)
add(x1, x2) -> add(x1, x2)
reverse(x1) -> reverse(x1)
app(x1, x2) -> app(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
           →DP Problem 6
             ↳Dependency Graph


Dependency Pair:


Rules:


app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
reverse(nil) -> nil
reverse(add(n, x)) -> app(reverse(x), add(n, nil))
shuffle(nil) -> nil
shuffle(add(n, x)) -> add(n, shuffle(reverse(x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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