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:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

Q is empty.

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

A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(app(X1, X2)) → A__APP(mark(X1), mark(X2))
MARK(prefix(X)) → MARK(X)
MARK(zWadr(X1, X2)) → MARK(X1)
A__APP(nil, YS) → MARK(YS)
MARK(zWadr(X1, X2)) → MARK(X2)
A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__ZWADR(cons(X, XS), cons(Y, YS)) → A__APP(mark(Y), cons(mark(X), nil))
MARK(app(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(prefix(X)) → A__PREFIX(mark(X))
MARK(zWadr(X1, X2)) → A__ZWADR(mark(X1), mark(X2))
MARK(from(X)) → MARK(X)
A__APP(cons(X, XS), YS) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
MARK(app(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(app(X1, X2)) → A__APP(mark(X1), mark(X2))
MARK(prefix(X)) → MARK(X)
MARK(zWadr(X1, X2)) → MARK(X1)
A__APP(nil, YS) → MARK(YS)
MARK(zWadr(X1, X2)) → MARK(X2)
A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__ZWADR(cons(X, XS), cons(Y, YS)) → A__APP(mark(Y), cons(mark(X), nil))
MARK(app(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(prefix(X)) → A__PREFIX(mark(X))
MARK(zWadr(X1, X2)) → A__ZWADR(mark(X1), mark(X2))
MARK(from(X)) → MARK(X)
A__APP(cons(X, XS), YS) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
MARK(app(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(app(X1, X2)) → A__APP(mark(X1), mark(X2))
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(prefix(X)) → MARK(X)
A__APP(nil, YS) → MARK(YS)
MARK(zWadr(X1, X2)) → MARK(X2)
A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__ZWADR(cons(X, XS), cons(Y, YS)) → A__APP(mark(Y), cons(mark(X), nil))
MARK(app(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(prefix(X)) → A__PREFIX(mark(X))
MARK(zWadr(X1, X2)) → A__ZWADR(mark(X1), mark(X2))
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__APP(cons(X, XS), YS) → MARK(X)
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(app(X1, X2)) → A__APP(mark(X1), mark(X2))
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(prefix(X)) → MARK(X)
A__APP(nil, YS) → MARK(YS)
MARK(zWadr(X1, X2)) → MARK(X2)
A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__ZWADR(cons(X, XS), cons(Y, YS)) → A__APP(mark(Y), cons(mark(X), nil))
MARK(app(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(zWadr(X1, X2)) → A__ZWADR(mark(X1), mark(X2))
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__APP(cons(X, XS), YS) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
MARK(app(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(app(X1, X2)) → A__APP(mark(X1), mark(X2))
MARK(zWadr(X1, X2)) → MARK(X1)
A__APP(nil, YS) → MARK(YS)
MARK(zWadr(X1, X2)) → MARK(X2)
A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(Y)
MARK(app(X1, X2)) → MARK(X1)
MARK(zWadr(X1, X2)) → A__ZWADR(mark(X1), mark(X2))
A__APP(cons(X, XS), YS) → MARK(X)
MARK(app(X1, X2)) → MARK(X2)

The remaining pairs can at least be oriented weakly.

MARK(prefix(X)) → MARK(X)
A__ZWADR(cons(X, XS), cons(Y, YS)) → A__APP(mark(Y), cons(mark(X), nil))
MARK(s(X)) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)

Used ordering: Combined order from the following AFS and order.
A__ZWADR(x1, x2) = A__ZWADR(x1, x2)
cons(x1, x2) = x1
MARK(x1) = MARK(x1)
app(x1, x2) = app(x1, x2)
A__APP(x1, x2) = A__APP(x1, x2)
mark(x1) = x1
zWadr(x1, x2) = zWadr(x1, x2)
prefix(x1) = x1
nil = nil
s(x1) = x1
from(x1) = x1
A__FROM(x1) = A__FROM(x1)
a__prefix(x1) = x1
a__zWadr(x1, x2) = a__zWadr(x1, x2)
a__app(x1, x2) = a__app(x1, x2)
a__from(x1) = x1

Recursive Path Order [2].
Precedence:

[zWadr₂, a_{_{zWadr_2]}} > [A_{_ZWADR₂}, MARK₁, A_{_APP₂}, A_{_{FROM_1]}} > nil
[zWadr₂, a_{_{zWadr_2]}} > [app₂, a_{_{app_2]}} > nil

The following usable rules [14] were oriented:

a__prefix(X) → prefix(X)
mark(s(X)) → s(mark(X))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
a__zWadr(X1, X2) → zWadr(X1, X2)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__from(X) → cons(mark(X), from(s(X)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
a__app(nil, YS) → mark(YS)
mark(nil) → nil
a__from(X) → from(X)
a__zWadr(nil, YS) → nil
mark(from(X)) → a__from(mark(X))
mark(prefix(X)) → a__prefix(mark(X))
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__app(X1, X2) → app(X1, X2)
a__zWadr(XS, nil) → nil
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

MARK(s(X)) → MARK(X)
MARK(prefix(X)) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__ZWADR(cons(X, XS), cons(Y, YS)) → A__APP(mark(Y), cons(mark(X), nil))

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
MARK(prefix(X)) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(prefix(X)) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)

Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
s(x1) = s(x1)
prefix(x1) = x1
from(x1) = x1
A__FROM(x1) = A__FROM(x1)
mark(x1) = x1
cons(x1, x2) = x1
a__prefix(x1) = x1
nil = nil
zWadr(x1, x2) = zWadr(x1, x2)
a__zWadr(x1, x2) = a__zWadr(x1, x2)
a__app(x1, x2) = a__app(x1, x2)
app(x1, x2) = app(x1, x2)
a__from(x1) = x1

Recursive Path Order [2].
Precedence:

[MARK₁, s₁, A_{_{FROM_1]}} > nil
[zWadr₂, a_{_zWadr₂}, a_{_app₂}, app_2] > nil

The following usable rules [14] were oriented:

a__prefix(X) → prefix(X)
mark(s(X)) → s(mark(X))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
a__zWadr(X1, X2) → zWadr(X1, X2)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__from(X) → cons(mark(X), from(s(X)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
a__app(nil, YS) → mark(YS)
mark(nil) → nil
a__from(X) → from(X)
a__zWadr(nil, YS) → nil
mark(from(X)) → a__from(mark(X))
mark(prefix(X)) → a__prefix(mark(X))
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__app(X1, X2) → app(X1, X2)
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
a__zWadr(XS, nil) → nil

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ QDPOrderProof

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

MARK(prefix(X)) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MARK(prefix(X)) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MARK(x1) = x1
prefix(x1) = prefix(x1)
from(x1) = from(x1)
A__FROM(x1) = A__FROM(x1)
mark(x1) = x1
cons(x1, x2) = cons(x1)
a__prefix(x1) = a__prefix(x1)
s(x1) = x1
nil = nil
zWadr(x1, x2) = zWadr(x1, x2)
a__zWadr(x1, x2) = a__zWadr(x1, x2)
a__app(x1, x2) = a__app(x1, x2)
app(x1, x2) = app(x1, x2)
a__from(x1) = a__from(x1)

Recursive Path Order [2].
Precedence:

[from₁, a_{_{from_1]}} > [prefix₁, cons₁, a_{_{prefix_1]}} > nil
[from₁, a_{_{from_1]}} > A_{_FROM₁} > nil
[zWadr₂, a_{_zWadr₂}, a_{_app₂}, app_2] > [prefix₁, cons₁, a_{_{prefix_1]}} > nil

The following usable rules [14] were oriented:

a__prefix(X) → prefix(X)
mark(s(X)) → s(mark(X))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
a__zWadr(X1, X2) → zWadr(X1, X2)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__from(X) → cons(mark(X), from(s(X)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
a__app(nil, YS) → mark(YS)
mark(nil) → nil
a__from(X) → from(X)
a__zWadr(nil, YS) → nil
mark(from(X)) → a__from(mark(X))
mark(prefix(X)) → a__prefix(mark(X))
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__app(X1, X2) → app(X1, X2)
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
a__zWadr(XS, nil) → nil

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.