Termination Proof using AProVE

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)))

Termination of R to be shown.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

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)))

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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)))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(reverse(x₁)) = x₁

POL(shuffle(x₁)) = x₁

POL(nil) = 0

POL(app(x₁, x₂)) = x₁ + x₂

POL(APP(x₁, x₂)) = 1 + x₁ + x₂

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

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)))

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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)))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(reverse(x₁)) = x₁

POL(REVERSE(x₁)) = 1 + x₁

POL(shuffle(x₁)) = x₁

POL(nil) = 0

POL(app(x₁, x₂)) = x₁ + x₂

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Narrowing Transformation

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)))

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

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

two new Dependency Pairs are created:

SHUFFLE(add(n, nil)) -> SHUFFLE(nil)
SHUFFLE(add(n, add(n'', x''))) -> SHUFFLE(app(reverse(x''), add(n'', nil)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pair:

SHUFFLE(add(n, add(n'', x''))) -> SHUFFLE(app(reverse(x''), add(n'', nil)))

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)))

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

SHUFFLE(add(n, add(n'', x''))) -> SHUFFLE(app(reverse(x''), add(n'', nil)))

two new Dependency Pairs are created:

SHUFFLE(add(n, add(n'', nil))) -> SHUFFLE(app(nil, add(n'', nil)))
SHUFFLE(add(n, add(n'', add(n''', x')))) -> SHUFFLE(app(app(reverse(x'), add(n''', nil)), add(n'', nil)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

SHUFFLE(add(n, add(n'', add(n''', x')))) -> SHUFFLE(app(app(reverse(x'), add(n''', nil)), add(n'', nil)))
SHUFFLE(add(n, add(n'', nil))) -> SHUFFLE(app(nil, add(n'', nil)))

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)))

The following dependency pairs can be strictly oriented:

SHUFFLE(add(n, add(n'', add(n''', x')))) -> SHUFFLE(app(app(reverse(x'), add(n''', nil)), add(n'', nil)))
SHUFFLE(add(n, add(n'', nil))) -> SHUFFLE(app(nil, add(n'', nil)))

Additionally, the following rules can be oriented:

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)))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(reverse(x₁)) = x₁

POL(shuffle(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.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳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)))

Using the Dependency Graph resulted in no new DP problems.

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

POL(reverse(x₁))	= x₁
POL(shuffle(x₁))	= x₁
POL(nil)	= 0
POL(app(x₁, x₂))	= x₁ + x₂
POL(APP(x₁, x₂))	= 1 + x₁ + x₂
POL(add(x₁, x₂))	= 1 + x₂