Termination Proof using AProVE

Term Rewriting System R:
[x]
a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A(d(x)) -> B(a(x))
A(d(x)) -> A(x)
B(c(x)) -> A(b(x))
B(c(x)) -> B(x)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

B(c(x)) -> B(x)
A(d(x)) -> A(x)
B(c(x)) -> A(b(x))
A(d(x)) -> B(a(x))

Rules:

a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Strategy:

innermost

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

A(d(x)) -> B(a(x))

two new Dependency Pairs are created:

A(d(d(x''))) -> B(d(c(b(a(x'')))))
A(d(c(x''))) -> B(x'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

A(d(c(x''))) -> B(x'')
A(d(x)) -> A(x)
B(c(x)) -> A(b(x))
B(c(x)) -> B(x)

Rules:

a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Strategy:

innermost

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

B(c(x)) -> A(b(x))

two new Dependency Pairs are created:

B(c(c(x''))) -> A(c(d(a(b(x'')))))
B(c(d(x''))) -> A(x'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pairs:

A(d(x)) -> A(x)
B(c(d(x''))) -> A(x'')
B(c(x)) -> B(x)
A(d(c(x''))) -> B(x'')

Rules:

a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Strategy:

innermost

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

A(d(x)) -> A(x)

two new Dependency Pairs are created:

A(d(d(x''))) -> A(d(x''))
A(d(d(c(x'''')))) -> A(d(c(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

A(d(d(c(x'''')))) -> A(d(c(x'''')))
A(d(d(x''))) -> A(d(x''))
B(c(x)) -> B(x)
A(d(c(x''))) -> B(x'')
B(c(d(x''))) -> A(x'')

Rules:

a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Strategy:

innermost

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

B(c(x)) -> B(x)

two new Dependency Pairs are created:

B(c(c(x''))) -> B(c(x''))
B(c(c(d(x'''')))) -> B(c(d(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

B(c(c(d(x'''')))) -> B(c(d(x'''')))
B(c(c(x''))) -> B(c(x''))
A(d(d(x''))) -> A(d(x''))
B(c(d(x''))) -> A(x'')
A(d(c(x''))) -> B(x'')
A(d(d(c(x'''')))) -> A(d(c(x'''')))

Rules:

a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Strategy:

innermost

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

A(d(c(x''))) -> B(x'')

three new Dependency Pairs are created:

A(d(c(c(d(x''''))))) -> B(c(d(x'''')))
A(d(c(c(c(x''''))))) -> B(c(c(x'''')))
A(d(c(c(c(d(x'''''')))))) -> B(c(c(d(x''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

A(d(c(c(c(d(x'''''')))))) -> B(c(c(d(x''''''))))
B(c(c(x''))) -> B(c(x''))
A(d(c(c(c(x''''))))) -> B(c(c(x'''')))
A(d(c(c(d(x''''))))) -> B(c(d(x'''')))
A(d(d(c(x'''')))) -> A(d(c(x'''')))
A(d(d(x''))) -> A(d(x''))
B(c(d(x''))) -> A(x'')
B(c(c(d(x'''')))) -> B(c(d(x'''')))

Rules:

a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Strategy:

innermost

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

B(c(d(x''))) -> A(x'')

five new Dependency Pairs are created:

B(c(d(d(d(x''''))))) -> A(d(d(x'''')))
B(c(d(d(d(c(x'''''')))))) -> A(d(d(c(x''''''))))
B(c(d(d(c(c(d(x''''''))))))) -> A(d(c(c(d(x'''''')))))
B(c(d(d(c(c(c(x''''''))))))) -> A(d(c(c(c(x'''''')))))
B(c(d(d(c(c(c(d(x'''''''')))))))) -> A(d(c(c(c(d(x''''''''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

B(c(d(d(c(c(c(d(x'''''''')))))))) -> A(d(c(c(c(d(x''''''''))))))
A(d(c(c(c(x''''))))) -> B(c(c(x'''')))
B(c(d(d(c(c(c(x''''''))))))) -> A(d(c(c(c(x'''''')))))
B(c(d(d(c(c(d(x''''''))))))) -> A(d(c(c(d(x'''''')))))
B(c(d(d(d(c(x'''''')))))) -> A(d(d(c(x''''''))))
A(d(c(c(d(x''''))))) -> B(c(d(x'''')))
A(d(d(c(x'''')))) -> A(d(c(x'''')))
A(d(d(x''))) -> A(d(x''))
B(c(d(d(d(x''''))))) -> A(d(d(x'''')))
B(c(c(d(x'''')))) -> B(c(d(x'''')))
B(c(c(x''))) -> B(c(x''))
A(d(c(c(c(d(x'''''')))))) -> B(c(c(d(x''''''))))

Rules:

a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Strategy:

innermost

The following dependency pairs can be strictly oriented:

B(c(d(d(c(c(c(d(x'''''''')))))))) -> A(d(c(c(c(d(x''''''''))))))
A(d(c(c(c(x''''))))) -> B(c(c(x'''')))
B(c(d(d(c(c(c(x''''''))))))) -> A(d(c(c(c(x'''''')))))
B(c(d(d(c(c(d(x''''''))))))) -> A(d(c(c(d(x'''''')))))
B(c(d(d(d(c(x'''''')))))) -> A(d(d(c(x''''''))))
A(d(c(c(d(x''''))))) -> B(c(d(x'''')))
B(c(d(d(d(x''''))))) -> A(d(d(x'''')))
B(c(c(d(x'''')))) -> B(c(d(x'''')))
B(c(c(x''))) -> B(c(x''))
A(d(c(c(c(d(x'''''')))))) -> B(c(c(d(x''''''))))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c(x₁)) = 1 + x₁

POL(B(x₁)) = x₁

POL(d(x₁)) = x₁

POL(A(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pairs:

A(d(d(c(x'''')))) -> A(d(c(x'''')))
A(d(d(x''))) -> A(d(x''))

Rules:

a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 9

                 ↳Polynomial Ordering

Dependency Pair:

A(d(d(x''))) -> A(d(x''))

Rules:

a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Strategy:

innermost

The following dependency pair can be strictly oriented:

A(d(d(x''))) -> A(d(x''))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(d(x₁)) = 1 + x₁

POL(A(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 10

                 ↳Dependency Graph

Dependency Pair:

Rules:

a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

POL(c(x₁))	= 1 + x₁
POL(B(x₁))	= x₁
POL(d(x₁))	= x₁
POL(A(x₁))	= x₁