Term Rewriting System R:
[x]
a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

A(f, 0) -> A(s, 0)
A(d, a(s, x)) -> A(s, a(s, a(d, a(p, a(s, x)))))
A(d, a(s, x)) -> A(s, a(d, a(p, a(s, x))))
A(d, a(s, x)) -> A(d, a(p, a(s, x)))
A(d, a(s, x)) -> A(p, a(s, x))
A(f, a(s, x)) -> A(d, a(f, a(p, a(s, x))))
A(f, a(s, x)) -> A(f, a(p, a(s, x)))
A(f, a(s, x)) -> A(p, a(s, x))

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Narrowing Transformation`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

A(d, a(s, x)) -> A(d, a(p, a(s, x)))

Rules:

a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x

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

A(d, a(s, x)) -> A(d, a(p, a(s, x)))
one new Dependency Pair is created:

A(d, a(s, x'')) -> A(d, x'')

The transformation is resulting in one new DP problem:

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

Dependency Pair:

A(d, a(s, x'')) -> A(d, x'')

Rules:

a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x

The following dependency pair can be strictly oriented:

A(d, a(s, x'')) -> A(d, 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(d) =  0 POL(s) =  0 POL(a(x1, x2)) =  1 + x2 POL(A(x1, x2)) =  x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 3`
`             ↳Polo`
`             ...`
`               →DP Problem 4`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

Rules:

a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`       →DP Problem 2`
`         ↳Narrowing Transformation`

Dependency Pair:

A(f, a(s, x)) -> A(f, a(p, a(s, x)))

Rules:

a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x

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

A(f, a(s, x)) -> A(f, a(p, a(s, x)))
one new Dependency Pair is created:

A(f, a(s, x'')) -> A(f, x'')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 5`
`             ↳Polynomial Ordering`

Dependency Pair:

A(f, a(s, x'')) -> A(f, x'')

Rules:

a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x

The following dependency pair can be strictly oriented:

A(f, a(s, x'')) -> A(f, 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(s) =  0 POL(a(x1, x2)) =  1 + x2 POL(A(x1, x2)) =  x2 POL(f) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 5`
`             ↳Polo`
`             ...`
`               →DP Problem 6`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x

Using the Dependency Graph resulted in no new DP problems.

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