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)

Innermost 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`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`

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)

Strategy:

innermost

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

V(a(a(x))) -> V(x)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

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

Strategy:

innermost

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

V(a(a(a(a(x''))))) -> V(a(a(x'')))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 4`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(c(x1)) =  0 POL(V(x1)) =  1 + x1 POL(d(x1)) =  0 POL(a(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳FwdInst`

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Forward Instantiation Transformation`

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)

Strategy:

innermost

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

W(a(a(x))) -> W(x)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Forward Instantiation Transformation`

Dependency Pair:

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

Strategy:

innermost

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

W(a(a(a(a(x''))))) -> W(a(a(x'')))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Polynomial Ordering`

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(c(x1)) =  0 POL(d(x1)) =  0 POL(a(x1)) =  1 + x1 POL(W(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 8`
`                 ↳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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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