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:

app2(nil, YS) -> YS
app2(cons1(X), YS) -> cons1(X)
from1(X) -> cons1(X)
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons1(X), cons1(Y)) -> cons1(app2(Y, cons1(X)))
prefix1(L) -> cons1(nil)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

app2(nil, YS) -> YS
app2(cons1(X), YS) -> cons1(X)
from1(X) -> cons1(X)
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons1(X), cons1(Y)) -> cons1(app2(Y, cons1(X)))
prefix1(L) -> cons1(nil)

Q is empty.

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

ZWADR2(cons1(X), cons1(Y)) -> APP2(Y, cons1(X))

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons1(X), YS) -> cons1(X)
from1(X) -> cons1(X)
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons1(X), cons1(Y)) -> cons1(app2(Y, cons1(X)))
prefix1(L) -> cons1(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:

ZWADR2(cons1(X), cons1(Y)) -> APP2(Y, cons1(X))

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons1(X), YS) -> cons1(X)
from1(X) -> cons1(X)
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons1(X), cons1(Y)) -> cons1(app2(Y, cons1(X)))
prefix1(L) -> cons1(nil)

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