Termination w.r.t. Q proof of TRS/TRCSREx3_3_25_Bor03

Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X), YS) → cons(X)
from(X) → cons(X)
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X), cons(Y)) → cons(app(Y, cons(X)))
prefix(L) → cons(nil)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X), YS) → cons(X)
from(X) → cons(X)
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X), cons(Y)) → cons(app(Y, cons(X)))
prefix(L) → cons(nil)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X), YS) → cons(X)
from(X) → cons(X)
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X), cons(Y)) → cons(app(Y, cons(X)))
prefix(L) → cons(nil)

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0), x1)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0), cons(x1))
prefix(x0)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ZWADR(cons(X), cons(Y)) → APP(Y, cons(X))

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X), YS) → cons(X)
from(X) → cons(X)
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X), cons(Y)) → cons(app(Y, cons(X)))
prefix(L) → cons(nil)

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0), x1)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0), cons(x1))
prefix(x0)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

ZWADR(cons(X), cons(Y)) → APP(Y, cons(X))

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X), YS) → cons(X)
from(X) → cons(X)
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X), cons(Y)) → cons(app(Y, cons(X)))
prefix(L) → cons(nil)

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0), x1)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0), cons(x1))
prefix(x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

ZWADR(cons(X), cons(Y)) → APP(Y, cons(X))

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X), YS) → cons(X)
from(X) → cons(X)
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X), cons(Y)) → cons(app(Y, cons(X)))
prefix(L) → cons(nil)

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0), x1)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0), cons(x1))
prefix(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.