Termination Proof using AProVE

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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(concat(u, v), concat(w, z))

Termination of R to be shown.

     ↳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)
LESS_LEAVES(cons(u, v), cons(w, z)) -> LESS_LEAVES(concat(u, v), concat(w, z))
LESS_LEAVES(cons(u, v), cons(w, z)) -> CONCAT(u, v)
LESS_LEAVES(cons(u, v), cons(w, z)) -> CONCAT(w, z)

Furthermore, R contains seven SCCs.

     ↳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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(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 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)

     ↳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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(concat(u, v), concat(w, z))

Using the Dependency Graph resulted in no new DP problems.

     ↳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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(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 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)

     ↳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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(concat(u, v), concat(w, z))

Using the Dependency Graph resulted in no new DP problems.

     ↳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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(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 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)

     ↳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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(concat(u, v), concat(w, z))

Using the Dependency Graph resulted in no new DP problems.

     ↳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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(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))

The following usable rules 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

     ↳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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(concat(u, v), concat(w, z))

Using the Dependency Graph resulted in no new DP problems.

     ↳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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(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 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)

     ↳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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(concat(u, v), concat(w, z))

Using the Dependency Graph resulted in no new DP problems.

     ↳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:

Dependency Pair:
LESS_LEAVES(cons(u, v), cons(w, z)) -> LESS_LEAVES(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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(concat(u, v), concat(w, z))
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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(concat(u, v), concat(w, z))

     ↳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:

Dependency Pair:
LESS_LEAVES(cons(u, v), cons(w, z)) -> LESS_LEAVES(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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(concat(u, v), concat(w, z))
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))
less_leaves(x, leaf) -> false
less_leaves(leaf, cons(w, z)) -> true
less_leaves(cons(u, v), cons(w, z)) -> less_leaves(concat(u, v), concat(w, z))

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