Term Rewriting System R:
[YS, X, XS, X1, X2, Y, L]
app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, napp(activate(XS), YS))
app(X1, X2) -> napp(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nnil)), nzWadr(activate(XS), activate(YS)))
prefix(L) -> cons(nil, nzWadr(L, prefix(L)))
nil -> nnil
activate(napp(X1, X2)) -> app(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nnil) -> nil
activate(X) -> X

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

APP(cons(X, XS), YS) -> ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, nnil))
ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
PREFIX(L) -> NIL
PREFIX(L) -> PREFIX(L)
ACTIVATE(napp(X1, X2)) -> APP(X1, X2)
ACTIVATE(nfrom(X)) -> FROM(X)
ACTIVATE(nnil) -> NIL

Furthermore, R contains two SCCs.

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

Dependency Pairs:

ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, nnil))
ACTIVATE(napp(X1, X2)) -> APP(X1, X2)
APP(cons(X, XS), YS) -> ACTIVATE(XS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, napp(activate(XS), YS))
app(X1, X2) -> napp(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nnil)), nzWadr(activate(XS), activate(YS)))
prefix(L) -> cons(nil, nzWadr(L, prefix(L)))
nil -> nnil
activate(napp(X1, X2)) -> app(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nnil) -> nil
activate(X) -> X

The following dependency pair can be strictly oriented:

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)) =  x1 + x2 POL(ZWADR(x1, x2)) =  x1 + x2 POL(n__zWadr(x1, x2)) =  1 + x1 + x2 POL(n__nil) =  0 POL(n__app(x1, x2)) =  x1 POL(ACTIVATE(x1)) =  x1 POL(APP(x1, x2)) =  x1

resulting in one new DP problem.

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

Dependency Pairs:

ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, nnil))
ACTIVATE(napp(X1, X2)) -> APP(X1, X2)
APP(cons(X, XS), YS) -> ACTIVATE(XS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, napp(activate(XS), YS))
app(X1, X2) -> napp(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nnil)), nzWadr(activate(XS), activate(YS)))
prefix(L) -> cons(nil, nzWadr(L, prefix(L)))
nil -> nnil
activate(napp(X1, X2)) -> app(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nnil) -> nil
activate(X) -> X

Using the Dependency Graph the DP problem was split into 1 DP problems.

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

Dependency Pairs:

APP(cons(X, XS), YS) -> ACTIVATE(XS)
ACTIVATE(napp(X1, X2)) -> APP(X1, X2)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, napp(activate(XS), YS))
app(X1, X2) -> napp(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nnil)), nzWadr(activate(XS), activate(YS)))
prefix(L) -> cons(nil, nzWadr(L, prefix(L)))
nil -> nnil
activate(napp(X1, X2)) -> app(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nnil) -> nil
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(napp(X1, X2)) -> APP(X1, X2)

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)) =  x2 POL(n__app(x1, x2)) =  1 + x1 POL(ACTIVATE(x1)) =  x1 POL(APP(x1, x2)) =  x1

resulting in one new DP problem.

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

Dependency Pair:

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

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, napp(activate(XS), YS))
app(X1, X2) -> napp(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nnil)), nzWadr(activate(XS), activate(YS)))
prefix(L) -> cons(nil, nzWadr(L, prefix(L)))
nil -> nnil
activate(napp(X1, X2)) -> app(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nnil) -> nil
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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

The following remains to be proven:
Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, napp(activate(XS), YS))
app(X1, X2) -> napp(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nnil)), nzWadr(activate(XS), activate(YS)))
prefix(L) -> cons(nil, nzWadr(L, prefix(L)))
nil -> nnil
activate(napp(X1, X2)) -> app(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nnil) -> nil
activate(X) -> X

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