Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/TRCSRSLASHEx3_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__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(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__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> MARK₁(X)
MARK₁(app₂(X1, X2)) -> A__APP₂(mark₁(X1), mark₁(X2))
MARK₁(prefix₁(X)) -> MARK₁(X)
MARK₁(zWadr₂(X1, X2)) -> MARK₁(X1)
A__APP₂(nil, YS) -> MARK₁(YS)
MARK₁(zWadr₂(X1, X2)) -> MARK₁(X2)
A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> MARK₁(Y)
A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> A__APP₂(mark₁(Y), cons₂(mark₁(X), nil))
MARK₁(app₂(X1, X2)) -> MARK₁(X1)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(prefix₁(X)) -> A__PREFIX₁(mark₁(X))
MARK₁(zWadr₂(X1, X2)) -> A__ZWADR₂(mark₁(X1), mark₁(X2))
MARK₁(from₁(X)) -> MARK₁(X)
A__APP₂(cons₂(X, XS), YS) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)
MARK₁(app₂(X1, X2)) -> MARK₁(X2)

The TRS R consists of the following rules:

a__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(X)

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:

A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> MARK₁(X)
MARK₁(app₂(X1, X2)) -> A__APP₂(mark₁(X1), mark₁(X2))
MARK₁(prefix₁(X)) -> MARK₁(X)
MARK₁(zWadr₂(X1, X2)) -> MARK₁(X1)
A__APP₂(nil, YS) -> MARK₁(YS)
MARK₁(zWadr₂(X1, X2)) -> MARK₁(X2)
A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> MARK₁(Y)
A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> A__APP₂(mark₁(Y), cons₂(mark₁(X), nil))
MARK₁(app₂(X1, X2)) -> MARK₁(X1)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(prefix₁(X)) -> A__PREFIX₁(mark₁(X))
MARK₁(zWadr₂(X1, X2)) -> A__ZWADR₂(mark₁(X1), mark₁(X2))
MARK₁(from₁(X)) -> MARK₁(X)
A__APP₂(cons₂(X, XS), YS) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)
MARK₁(app₂(X1, X2)) -> MARK₁(X2)

The TRS R consists of the following rules:

a__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> MARK₁(X)
MARK₁(app₂(X1, X2)) -> A__APP₂(mark₁(X1), mark₁(X2))
MARK₁(zWadr₂(X1, X2)) -> MARK₁(X1)
MARK₁(prefix₁(X)) -> MARK₁(X)
A__APP₂(nil, YS) -> MARK₁(YS)
MARK₁(zWadr₂(X1, X2)) -> MARK₁(X2)
A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> MARK₁(Y)
A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> A__APP₂(mark₁(Y), cons₂(mark₁(X), nil))
MARK₁(app₂(X1, X2)) -> MARK₁(X1)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(zWadr₂(X1, X2)) -> A__ZWADR₂(mark₁(X1), mark₁(X2))
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__APP₂(cons₂(X, XS), YS) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)
MARK₁(app₂(X1, X2)) -> MARK₁(X2)

The TRS R consists of the following rules:

a__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(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 strictly oriented and are deleted.

A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> MARK₁(X)
MARK₁(zWadr₂(X1, X2)) -> MARK₁(X1)
A__APP₂(nil, YS) -> MARK₁(YS)
MARK₁(zWadr₂(X1, X2)) -> MARK₁(X2)
A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> MARK₁(Y)
A__ZWADR₂(cons₂(X, XS), cons₂(Y, YS)) -> A__APP₂(mark₁(Y), cons₂(mark₁(X), nil))
MARK₁(app₂(X1, X2)) -> MARK₁(X1)
A__APP₂(cons₂(X, XS), YS) -> MARK₁(X)
MARK₁(app₂(X1, X2)) -> MARK₁(X2)

The remaining pairs can at least by weakly be oriented.

MARK₁(app₂(X1, X2)) -> A__APP₂(mark₁(X1), mark₁(X2))
MARK₁(prefix₁(X)) -> MARK₁(X)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(zWadr₂(X1, X2)) -> A__ZWADR₂(mark₁(X1), mark₁(X2))
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)

Used ordering: Combined order from the following AFS and order.
A__ZWADR₂(x1, x2) = A__ZWADR₂(x1, x2)
cons₂(x1, x2) = x1
MARK₁(x1) = x1
app₂(x1, x2) = app₂(x1, x2)
A__APP₂(x1, x2) = A__APP₂(x1, x2)
mark₁(x1) = x1
zWadr₂(x1, x2) = zWadr₂(x1, x2)
prefix₁(x1) = x1
nil = nil
s₁(x1) = x1
from₁(x1) = x1
A__FROM₁(x1) = x1
a__from₁(x1) = x1
a__prefix₁(x1) = x1
a__app₂(x1, x2) = a__app₂(x1, x2)
a__zWadr₂(x1, x2) = a__zWadr₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

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

The following usable rules [14] were oriented:

mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
a__app₂(nil, YS) -> mark₁(YS)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
a__prefix₁(X) -> prefix₁(X)
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__from₁(X) -> from₁(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(app₂(X1, X2)) -> A__APP₂(mark₁(X1), mark₁(X2))
MARK₁(zWadr₂(X1, X2)) -> A__ZWADR₂(mark₁(X1), mark₁(X2))
MARK₁(prefix₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__FROM₁(X) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(prefix₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(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 strictly oriented and are deleted.

MARK₁(s₁(X)) -> MARK₁(X)

The remaining pairs can at least by weakly be oriented.

MARK₁(prefix₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)

Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
s₁(x1) = s₁(x1)
prefix₁(x1) = x1
from₁(x1) = x1
A__FROM₁(x1) = A__FROM₁(x1)
mark₁(x1) = x1
cons₂(x1, x2) = x1
a__from₁(x1) = x1
a__prefix₁(x1) = x1
app₂(x1, x2) = app₂(x1, x2)
a__app₂(x1, x2) = a__app₂(x1, x2)
nil = nil
a__zWadr₂(x1, x2) = a__zWadr₂(x1, x2)
zWadr₂(x1, x2) = zWadr₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

[MARK₁, s₁, A_{_{FROM_1]}} > nil
[a_{_zWadr₂}, zWadr_2] > [app₂, a_{_{app_2]}} > nil

The following usable rules [14] were oriented:

mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
a__app₂(nil, YS) -> mark₁(YS)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
a__prefix₁(X) -> prefix₁(X)
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__from₁(X) -> from₁(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

MARK₁(prefix₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__FROM₁(X) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(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 strictly oriented and are deleted.

MARK₁(from₁(X)) -> MARK₁(X)
A__FROM₁(X) -> MARK₁(X)

The remaining pairs can at least by weakly be oriented.

MARK₁(prefix₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = x1
prefix₁(x1) = x1
from₁(x1) = from₁(x1)
A__FROM₁(x1) = A__FROM₁(x1)
mark₁(x1) = x1
cons₂(x1, x2) = x1
a__from₁(x1) = a__from₁(x1)
a__prefix₁(x1) = x1
app₂(x1, x2) = app₂(x1, x2)
a__app₂(x1, x2) = a__app₂(x1, x2)
nil = nil
s₁(x1) = s
a__zWadr₂(x1, x2) = a__zWadr₂(x1, x2)
zWadr₂(x1, x2) = zWadr₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

[from₁, A_{_FROM₁}, a_{_{from_1]}}
[a_{_zWadr₂}, zWadr_2]

The following usable rules [14] were oriented:

mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
a__app₂(nil, YS) -> mark₁(YS)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
a__prefix₁(X) -> prefix₁(X)
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__from₁(X) -> from₁(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

MARK₁(prefix₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPOrderProof

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

MARK₁(prefix₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(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 strictly oriented and are deleted.

MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The remaining pairs can at least by weakly be oriented.

MARK₁(prefix₁(X)) -> MARK₁(X)

Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
prefix₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ QDPOrderProof

                                ↳ QDP

                                  ↳ QDPOrderProof

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

MARK₁(prefix₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(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 strictly oriented and are deleted.

MARK₁(prefix₁(X)) -> MARK₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
prefix₁(x1) = prefix₁(x1)

Lexicographic Path Order [19].
Precedence:

prefix₁ > MARK₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ 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__app₂(nil, YS) -> mark₁(YS)
a__app₂(cons₂(X, XS), YS) -> cons₂(mark₁(X), app₂(XS, YS))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__zWadr₂(nil, YS) -> nil
a__zWadr₂(XS, nil) -> nil
a__zWadr₂(cons₂(X, XS), cons₂(Y, YS)) -> cons₂(a__app₂(mark₁(Y), cons₂(mark₁(X), nil)), zWadr₂(XS, YS))
a__prefix₁(L) -> cons₂(nil, zWadr₂(L, prefix₁(L)))
mark₁(app₂(X1, X2)) -> a__app₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(zWadr₂(X1, X2)) -> a__zWadr₂(mark₁(X1), mark₁(X2))
mark₁(prefix₁(X)) -> a__prefix₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__app₂(X1, X2) -> app₂(X1, X2)
a__from₁(X) -> from₁(X)
a__zWadr₂(X1, X2) -> zWadr₂(X1, X2)
a__prefix₁(X) -> prefix₁(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.