Term Rewriting System R:
[x, y, z]
f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(x, g(y)) -> F(h(x), i(x, y))
F(x, g(y)) -> I(x, y)
I(x, j(y, z)) -> J(g(y), i(x, z))
I(x, j(y, z)) -> I(x, z)
I(h(x), j(j(y, z), 0)) -> J(i(h(x), j(y, z)), i(x, j(y, z)))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
J(g(x), g(y)) -> J(x, y)

Furthermore, R contains three SCCs.

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

Dependency Pair:

J(g(x), g(y)) -> J(x, y)

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

The following dependency pair can be strictly oriented:

J(g(x), g(y)) -> J(x, y)

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(g(x1)) =  1 + x1 POL(J(x1, x2)) =  x1

resulting in one new DP problem.

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

Dependency Pair:

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pairs:

I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(x, j(y, z)) -> I(x, z)

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

The following dependency pair can be strictly oriented:

I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))

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(I(x1, x2)) =  x1 POL(0) =  0 POL(g(x1)) =  0 POL(h(x1)) =  1 + x1 POL(j(x1, x2)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Inst`

Dependency Pairs:

I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(x, j(y, z)) -> I(x, z)

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

The following dependency pair can be strictly oriented:

I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))

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

j(g(x), g(y)) -> g(j(x, y))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(I(x1, x2)) =  1 + x1 + x2 POL(0) =  1 POL(g(x1)) =  0 POL(h(x1)) =  0 POL(j(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Polo`
`             ...`
`               →DP Problem 6`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Inst`

Dependency Pair:

I(x, j(y, z)) -> I(x, z)

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

The following dependency pair can be strictly oriented:

I(x, j(y, z)) -> I(x, z)

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(I(x1, x2)) =  x2 POL(j(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Polo`
`             ...`
`               →DP Problem 7`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Inst`

Dependency Pair:

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

F(x, g(y)) -> F(h(x), i(x, y))

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

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

F(x, g(y)) -> F(h(x), i(x, y))
one new Dependency Pair is created:

F(h(x''''), g(y')) -> F(h(h(x'''')), i(h(x''''), y'))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Inst`
`           →DP Problem 8`
`             ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pair:

F(h(x''''), g(y')) -> F(h(h(x'''')), i(h(x''''), y'))

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Termination of R could not be shown.
Duration:
0:00 minutes