Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FLATTEN(unit(x)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> ++'(flatten(x), flatten(y))
FLATTEN(++(x, y)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> FLATTEN(y)
FLATTEN(++(unit(x), y)) -> ++'(flatten(x), flatten(y))
FLATTEN(++(unit(x), y)) -> FLATTEN(x)
FLATTEN(++(unit(x), y)) -> FLATTEN(y)
REV(++(x, y)) -> ++'(rev(y), rev(x))
REV(++(x, y)) -> REV(y)
REV(++(x, y)) -> REV(x)
++'(++(x, y), z) -> ++'(x, ++(y, z))
++'(++(x, y), z) -> ++'(y, z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pair:

++'(++(x, y), z) -> ++'(y, z)

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

++'(++(x, y), z) -> ++'(y, z)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(++'(x₁, x₂)) = x₁

POL(++(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

         ↳Polo

Dependency Pair:

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

Dependency Pairs:

FLATTEN(++(unit(x), y)) -> FLATTEN(y)
FLATTEN(++(unit(x), y)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> FLATTEN(y)
FLATTEN(++(x, y)) -> FLATTEN(x)
FLATTEN(unit(x)) -> FLATTEN(x)

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

FLATTEN(++(unit(x), y)) -> FLATTEN(y)
FLATTEN(++(unit(x), y)) -> FLATTEN(x)
FLATTEN(++(x, y)) -> FLATTEN(y)
FLATTEN(++(x, y)) -> FLATTEN(x)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(unit(x₁)) = x₁

POL(FLATTEN(x₁)) = 1 + x₁

POL(++(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 5

             ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

Dependency Pair:

FLATTEN(unit(x)) -> FLATTEN(x)

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

FLATTEN(unit(x)) -> FLATTEN(x)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(unit(x₁)) = 1 + x₁

POL(FLATTEN(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 5

             ↳Polo

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 3

         ↳Polo

Dependency Pair:

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

Dependency Pair:

REV(++(x, y)) -> REV(y)

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

REV(++(x, y)) -> REV(y)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(REV(x₁)) = x₁

POL(++(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

Dependency Pair:

Rules:

flatten(nil) -> nil
flatten(unit(x)) -> flatten(x)
flatten(++(x, y)) -> ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
flatten(flatten(x)) -> flatten(x)
rev(nil) -> nil
rev(unit(x)) -> unit(x)
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(rev(x)) -> x
++(x, nil) -> x
++(nil, y) -> y
++(++(x, y), z) -> ++(x, ++(y, z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(++'(x₁, x₂))	= x₁
POL(++(x₁, x₂))	= 1 + x₂

POL(unit(x₁))	= x₁
POL(FLATTEN(x₁))	= 1 + x₁
POL(++(x₁, x₂))	= 1 + x₁ + x₂