Termination Proof using AProVE

Term Rewriting System R:
[x]
top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

TOP(free(x)) -> TOP(check(new(x)))
TOP(free(x)) -> CHECK(new(x))
TOP(free(x)) -> NEW(x)
NEW(free(x)) -> NEW(x)
OLD(free(x)) -> OLD(x)
CHECK(free(x)) -> CHECK(x)
CHECK(new(x)) -> NEW(check(x))
CHECK(new(x)) -> CHECK(x)
CHECK(old(x)) -> OLD(check(x))
CHECK(old(x)) -> CHECK(x)

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Remaining

Dependency Pair:

NEW(free(x)) -> NEW(x)

Rules:

top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

The following dependency pair can be strictly oriented:

NEW(free(x)) -> NEW(x)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(NEW(x₁)) = x₁

POL(free(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Remaining

Dependency Pair:

OLD(free(x)) -> OLD(x)

Rules:

top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

The following dependency pair can be strictly oriented:

OLD(free(x)) -> OLD(x)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(free(x₁)) = 1 + x₁

POL(OLD(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

CHECK(old(x)) -> CHECK(x)
CHECK(new(x)) -> CHECK(x)
CHECK(free(x)) -> CHECK(x)

Rules:

top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

The following dependency pair can be strictly oriented:

CHECK(old(x)) -> CHECK(x)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(old(x₁)) = 1 + x₁

POL(free(x₁)) = x₁

POL(CHECK(x₁)) = x₁

POL(new(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 7

             ↳Polynomial Ordering

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

CHECK(new(x)) -> CHECK(x)
CHECK(free(x)) -> CHECK(x)

Rules:

top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

The following dependency pair can be strictly oriented:

CHECK(new(x)) -> CHECK(x)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(free(x₁)) = x₁

POL(CHECK(x₁)) = x₁

POL(new(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 7

             ↳Polo

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Remaining

Dependency Pair:

CHECK(free(x)) -> CHECK(x)

Rules:

top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

The following dependency pair can be strictly oriented:

CHECK(free(x)) -> CHECK(x)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(free(x₁)) = 1 + x₁

POL(CHECK(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 7

             ↳Polo

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

TOP(free(x)) -> TOP(check(new(x)))

Rules:

top(free(x)) -> top(check(new(x)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

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

POL(old(x₁))	= 1 + x₁
POL(free(x₁))	= x₁
POL(CHECK(x₁))	= x₁
POL(new(x₁))	= x₁