Term Rewriting System R:
[x, y]
app(f, app(s, x)) -> app(f, x)
app(g, app(app(cons, 0), y)) -> app(g, y)
app(g, app(app(cons, app(s, x)), y)) -> app(s, x)
app(h, app(app(cons, x), y)) -> app(h, app(g, app(app(cons, x), y)))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

APP(f, app(s, x)) -> APP(f, x)
APP(g, app(app(cons, 0), y)) -> APP(g, y)
APP(h, app(app(cons, x), y)) -> APP(h, app(g, app(app(cons, x), y)))
APP(h, app(app(cons, x), y)) -> APP(g, app(app(cons, x), y))

Furthermore, R contains three SCCs.

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

Dependency Pair:

APP(f, app(s, x)) -> APP(f, x)

Rules:

app(f, app(s, x)) -> app(f, x)
app(g, app(app(cons, 0), y)) -> app(g, y)
app(g, app(app(cons, app(s, x)), y)) -> app(s, x)
app(h, app(app(cons, x), y)) -> app(h, app(g, app(app(cons, x), y)))

The following dependency pair can be strictly oriented:

APP(f, app(s, x)) -> APP(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(APP(x1, x2)) =  x2 POL(f) =  0 POL(app(x1, x2)) =  1 + x2

resulting in one new DP problem.

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

Dependency Pair:

Rules:

app(f, app(s, x)) -> app(f, x)
app(g, app(app(cons, 0), y)) -> app(g, y)
app(g, app(app(cons, app(s, x)), y)) -> app(s, x)
app(h, app(app(cons, x), y)) -> app(h, app(g, app(app(cons, 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`
`         ↳Polo`

Dependency Pair:

APP(g, app(app(cons, 0), y)) -> APP(g, y)

Rules:

app(f, app(s, x)) -> app(f, x)
app(g, app(app(cons, 0), y)) -> app(g, y)
app(g, app(app(cons, app(s, x)), y)) -> app(s, x)
app(h, app(app(cons, x), y)) -> app(h, app(g, app(app(cons, x), y)))

The following dependency pair can be strictly oriented:

APP(g, app(app(cons, 0), y)) -> APP(g, 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(0) =  0 POL(g) =  0 POL(cons) =  0 POL(APP(x1, x2)) =  x2 POL(app(x1, x2)) =  1 + x2

resulting in one new DP problem.

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

Dependency Pair:

Rules:

app(f, app(s, x)) -> app(f, x)
app(g, app(app(cons, 0), y)) -> app(g, y)
app(g, app(app(cons, app(s, x)), y)) -> app(s, x)
app(h, app(app(cons, x), y)) -> app(h, app(g, app(app(cons, 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`
`         ↳Polynomial Ordering`

Dependency Pair:

APP(h, app(app(cons, x), y)) -> APP(h, app(g, app(app(cons, x), y)))

Rules:

app(f, app(s, x)) -> app(f, x)
app(g, app(app(cons, 0), y)) -> app(g, y)
app(g, app(app(cons, app(s, x)), y)) -> app(s, x)
app(h, app(app(cons, x), y)) -> app(h, app(g, app(app(cons, x), y)))

The following dependency pair can be strictly oriented:

APP(h, app(app(cons, x), y)) -> APP(h, app(g, app(app(cons, x), y)))

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

app(f, app(s, x)) -> app(f, x)
app(g, app(app(cons, 0), y)) -> app(g, y)
app(g, app(app(cons, app(s, x)), y)) -> app(s, x)
app(h, app(app(cons, x), y)) -> app(h, app(g, app(app(cons, x), y)))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(g) =  0 POL(cons) =  0 POL(s) =  0 POL(h) =  0 POL(app(x1, x2)) =  1 + x1 POL(f) =  0 POL(APP(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 6`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

app(f, app(s, x)) -> app(f, x)
app(g, app(app(cons, 0), y)) -> app(g, y)
app(g, app(app(cons, app(s, x)), y)) -> app(s, x)
app(h, app(app(cons, x), y)) -> app(h, app(g, app(app(cons, x), y)))

Using the Dependency Graph resulted in no new DP problems.

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