Termination Proof using AProVE

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)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))

Termination of R to be shown.

     ↳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))
ZWADR(cons(X, XS), cons(Y, YS)) -> ZWADR(XS, YS)
PREFIX(L) -> ZWADR(L, prefix(L))
PREFIX(L) -> PREFIX(L)

Furthermore, R contains four SCCs.

     ↳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)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))

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.

     ↳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)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))

Using the Dependency Graph resulted in no new DP problems.

     ↳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)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))
Dependency Pair:
ZWADR(cons(X, XS), cons(Y, YS)) -> ZWADR(XS, YS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))
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)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))

     ↳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)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))
Dependency Pair:
ZWADR(cons(X, XS), cons(Y, YS)) -> ZWADR(XS, YS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))
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)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))

     ↳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)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))
Dependency Pair:
ZWADR(cons(X, XS), cons(Y, YS)) -> ZWADR(XS, YS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))
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)))
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) -> cons(nil, zWadr(L, prefix(L)))

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

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