Term Rewriting System R:
[YS, X, XS, Y, L]
app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

APP(cons(X, XS), YS) -> APP(XS, YS)
FROM(X) -> FROM(s(X))
ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, nil))
PREFIX(L) -> PREFIX(L)

Furthermore, R contains four SCCs.

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

Dependency Pair:

APP(cons(X, XS), YS) -> APP(XS, YS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

The following dependency pair can be strictly oriented:

APP(cons(X, XS), YS) -> APP(XS, YS)

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(cons(x1, x2)) =  1 + x2 POL(APP(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Remaining`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

• Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))