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

Termination of R to be shown.

`   R`
`     ↳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.

`   R`
`     ↳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

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:

`   R`
`     ↳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

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:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 3`
`                 ↳Polynomial Ordering`

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

The following dependency pairs can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(c(x1)) =  x1 POL(B(x1)) =  x1 POL(d(x1)) =  1 + x1 POL(A(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 4`
`                 ↳Polynomial Ordering`

Dependency Pair:

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

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(c(x1)) =  1 + x1 POL(B(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 5`
`                 ↳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

Using the Dependency Graph resulted in no new DP problems.

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