Term Rewriting System R:
[x]
a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

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

Dependency Pairs:

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

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

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

A(b(x)) -> B(a(a(x)))
two new Dependency Pairs are created:

A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(u(x''))) -> B(a(x''))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
A(b(u(x''))) -> B(a(x''))
B(c(x)) -> B(b(x))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(x)) -> A(a(x))
C(a(x)) -> A(c(c(x)))
B(c(x)) -> C(b(b(x)))
B(c(x)) -> B(x)

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

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

A(b(x)) -> A(a(x))
two new Dependency Pairs are created:

A(b(b(x''))) -> A(b(a(a(x''))))
A(b(u(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`
`                 ↳Narrowing Transformation`

Dependency Pairs:

A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
A(b(u(x''))) -> B(a(x''))
B(c(x)) -> B(x)
B(c(x)) -> B(b(x))
C(a(x)) -> C(c(x))
B(c(x)) -> C(b(b(x)))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(x)) -> A(c(c(x)))
C(a(x)) -> C(x)

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

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

B(c(x)) -> C(b(b(x)))
two new Dependency Pairs are created:

B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(v(x''))) -> C(b(x''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 4`
`                 ↳Narrowing Transformation`

Dependency Pairs:

A(b(b(x''))) -> A(b(a(a(x''))))
C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(x)) -> A(c(c(x)))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
B(c(x)) -> B(b(x))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(u(x''))) -> A(x'')

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

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

B(c(x)) -> B(b(x))
two new Dependency Pairs are created:

B(c(c(x''))) -> B(c(b(b(x''))))
B(c(v(x''))) -> B(x'')

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

A(b(u(x''))) -> A(x'')
B(c(v(x''))) -> B(x'')
B(c(c(x''))) -> B(c(b(b(x''))))
C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(x)) -> A(c(c(x)))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(b(x''))) -> A(b(a(a(x''))))

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

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

C(a(x)) -> A(c(c(x)))
two new Dependency Pairs are created:

C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(w(x''))) -> A(c(x''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 6`
`                 ↳Narrowing Transformation`

Dependency Pairs:

B(c(v(x''))) -> B(x'')
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(b(x''))) -> A(b(a(a(x''))))
C(a(w(x''))) -> A(c(x''))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(u(x''))) -> A(x'')

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

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

C(a(x)) -> C(c(x))
two new Dependency Pairs are created:

C(a(a(x''))) -> C(a(c(c(x''))))
C(a(w(x''))) -> C(x'')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Polynomial Ordering`

Dependency Pairs:

C(a(w(x''))) -> C(x'')
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(u(x''))) -> B(a(x''))
C(a(w(x''))) -> A(c(x''))
B(c(v(x''))) -> C(b(x''))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
B(c(v(x''))) -> B(x'')

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

The following dependency pairs can be strictly oriented:

A(b(u(x''))) -> A(x'')
A(b(u(x''))) -> B(a(x''))

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

c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
a(b(x)) -> b(a(a(x)))
a(u(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(c(x1)) =  x1 POL(C(x1)) =  x1 POL(v(x1)) =  x1 POL(B(x1)) =  x1 POL(b(x1)) =  x1 POL(a(x1)) =  x1 POL(w(x1)) =  x1 POL(u(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 8`
`                 ↳Polynomial Ordering`

Dependency Pairs:

C(a(w(x''))) -> C(x'')
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(x''))) -> B(c(b(b(x''))))
C(a(w(x''))) -> A(c(x''))
B(c(v(x''))) -> C(b(x''))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
B(c(v(x''))) -> B(x'')

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

The following dependency pairs can be strictly oriented:

C(a(w(x''))) -> C(x'')
C(a(w(x''))) -> A(c(x''))

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

c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
a(b(x)) -> b(a(a(x)))
a(u(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(c(x1)) =  x1 POL(C(x1)) =  x1 POL(v(x1)) =  x1 POL(B(x1)) =  x1 POL(b(x1)) =  x1 POL(a(x1)) =  x1 POL(w(x1)) =  1 + x1 POL(u(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 9`
`                 ↳Polynomial Ordering`

Dependency Pairs:

C(a(a(x''))) -> C(a(c(c(x''))))
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(x''))) -> B(c(b(b(x''))))
B(c(v(x''))) -> C(b(x''))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
B(c(v(x''))) -> B(x'')

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

The following dependency pairs can be strictly oriented:

B(c(v(x''))) -> C(b(x''))
B(c(v(x''))) -> B(x'')

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

c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
a(b(x)) -> b(a(a(x)))
a(u(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(c(x1)) =  x1 POL(C(x1)) =  x1 POL(v(x1)) =  1 + x1 POL(B(x1)) =  x1 POL(b(x1)) =  x1 POL(a(x1)) =  x1 POL(w(x1)) =  1 + x1 POL(u(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 10`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

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

Rules:

a(b(x)) -> b(a(a(x)))
a(u(x)) -> x
b(c(x)) -> c(b(b(x)))
b(v(x)) -> x
c(a(x)) -> a(c(c(x)))
c(w(x)) -> x
u(a(x)) -> x
v(b(x)) -> x
w(c(x)) -> x

Termination of R could not be shown.
Duration:
0:01 minutes