Term Rewriting System R:
[x, y, n, u, v, w, z]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
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)
CONCAT(cons(u, v), y) -> CONCAT(v, y)
LESSLEAVES(cons(u, v), cons(w, z)) -> LESSLEAVES(concat(u, v), concat(w, z))
LESSLEAVES(cons(u, v), cons(w, z)) -> CONCAT(u, v)
LESSLEAVES(cons(u, v), cons(w, z)) -> CONCAT(w, z)

Furthermore, R contains seven SCCs.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:

MINUS(s(x), s(y)) -> MINUS(x, y)


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





The following dependency pair can be strictly oriented:

MINUS(s(x), s(y)) -> MINUS(x, y)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MINUS(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 8
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, 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
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





The following dependency pair can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(APP(x1, x2))=  x1  
  POL(add(x1, x2))=  1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 9
Dependency Graph
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, 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
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:

CONCAT(cons(u, v), y) -> CONCAT(v, y)


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





The following dependency pair can be strictly oriented:

CONCAT(cons(u, v), y) -> CONCAT(v, y)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  1 + x2  
  POL(CONCAT(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 10
Dependency Graph
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polynomial Ordering
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





The following dependency pair can be strictly oriented:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))


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

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(QUOT(x1, x2))=  x1  
  POL(0)=  0  
  POL(minus(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
           →DP Problem 11
Dependency Graph
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polynomial Ordering
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





The following dependency pair can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(REVERSE(x1))=  x1  
  POL(add(x1, x2))=  1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
           →DP Problem 12
Dependency Graph
       →DP Problem 6
Polo
       →DP Problem 7
Polo


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polynomial Ordering
       →DP Problem 7
Polo


Dependency Pair:

LESSLEAVES(cons(u, v), cons(w, z)) -> LESSLEAVES(concat(u, v), concat(w, z))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





The following dependency pair can be strictly oriented:

LESSLEAVES(cons(u, v), cons(w, z)) -> LESSLEAVES(concat(u, v), concat(w, z))


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

concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(LESS_LEAVES(x1, x2))=  x2  
  POL(leaf)=  0  
  POL(concat(x1, x2))=  x1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
           →DP Problem 13
Dependency Graph
       →DP Problem 7
Polo


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polynomial Ordering


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules using the Ce-refinement 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 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Polo
           →DP Problem 14
Dependency Graph


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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)))
concat(leaf, y) -> y
concat(cons(u, v), y) -> cons(u, concat(v, y))
lessleaves(x, leaf) -> false
lessleaves(leaf, cons(w, z)) -> true
lessleaves(cons(u, v), cons(w, z)) -> lessleaves(concat(u, v), concat(w, z))





Using the Dependency Graph resulted in no new DP problems.

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