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))

Innermost 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
Argument Filtering and Ordering
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


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


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))


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
       →DP Problem 4
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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))


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 w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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 2
AFS
           →DP Problem 9
Dependency Graph
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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))


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
       →DP Problem 4
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


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


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))


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
AFS
       →DP Problem 4
Argument Filtering and Ordering
       →DP Problem 5
AFS
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


The following usable rules for innermost w.r.t. to the AFS can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2) -> QUOT(x1, x2)
s(x1) -> s(x1)
minus(x1, x2) -> x1


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


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))


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
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Argument Filtering and Ordering
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
AFS
           →DP Problem 12
Dependency Graph
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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))


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
AFS
       →DP Problem 4
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:00 minutes