Termination w.r.t. Q proof of /tmp/tmpxijWhW/Ex3_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

  ↳ 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)

Q is empty.

Using Dependency Pairs [1,15] 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)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.