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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

V(a(a(x))) -> U(v(x))
V(a(a(x))) -> V(x)
V(a(c(x))) -> U(b(d(x)))
W(a(a(x))) -> U(w(x))
W(a(a(x))) -> W(x)
W(a(c(x))) -> U(b(d(x)))

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`

Dependency Pair:

V(a(a(x))) -> V(x)

Rules:

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

The following dependency pair can be strictly oriented:

V(a(a(x))) -> V(x)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(V(x1)) =  1 + x1 POL(a(x1)) =  1 + x1

resulting in one new DP problem.

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

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`

Dependency Pair:

W(a(a(x))) -> W(x)

Rules:

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

The following dependency pair can be strictly oriented:

W(a(a(x))) -> W(x)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(W(x1)) =  1 + x1 POL(a(x1)) =  1 + x1

resulting in one new DP problem.

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

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

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