Termination Proof using AProVE

Term Rewriting System R:
[x]
w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

W(r(x)) -> W(x)
B(r(x)) -> B(x)
B(w(x)) -> W(b(x))
B(w(x)) -> B(x)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

W(r(x)) -> W(x)

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

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

W(r(x)) -> W(x)

one new Dependency Pair is created:

W(r(r(x''))) -> W(r(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

W(r(r(x''))) -> W(r(x''))

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

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

W(r(r(x''))) -> W(r(x''))

one new Dependency Pair is created:

W(r(r(r(x'''')))) -> W(r(r(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

W(r(r(r(x'''')))) -> W(r(r(x'''')))

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

W(r(r(r(x'''')))) -> W(r(r(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(W(x₁)) = 1 + x₁

POL(r(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

B(r(x)) -> B(x)

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

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

B(r(x)) -> B(x)

one new Dependency Pair is created:

B(r(r(x''))) -> B(r(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

B(r(r(x''))) -> B(r(x''))

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

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

B(r(r(x''))) -> B(r(x''))

one new Dependency Pair is created:

B(r(r(r(x'''')))) -> B(r(r(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

B(r(r(r(x'''')))) -> B(r(r(x'''')))

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

B(r(r(r(x'''')))) -> B(r(r(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(B(x₁)) = 1 + x₁

POL(r(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Forward Instantiation Transformation

Dependency Pair:

B(w(x)) -> B(x)

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

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

B(w(x)) -> B(x)

one new Dependency Pair is created:

B(w(w(x''))) -> B(w(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 10

             ↳Forward Instantiation Transformation

Dependency Pair:

B(w(w(x''))) -> B(w(x''))

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

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

B(w(w(x''))) -> B(w(x''))

one new Dependency Pair is created:

B(w(w(w(x'''')))) -> B(w(w(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 10

             ↳FwdInst

...

               →DP Problem 11

                 ↳Polynomial Ordering

Dependency Pair:

B(w(w(w(x'''')))) -> B(w(w(x'''')))

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

B(w(w(w(x'''')))) -> B(w(w(x'''')))

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

w(r(x)) -> r(w(x))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(B(x₁)) = 1 + x₁

POL(w(x₁)) = 1 + x₁

POL(r(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 10

             ↳FwdInst

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pair:

Rules:

w(r(x)) -> r(w(x))
b(r(x)) -> r(b(x))
b(w(x)) -> w(b(x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(B(x₁))	= 1 + x₁
POL(w(x₁))	= 1 + x₁
POL(r(x₁))	= 0