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__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 oriented strictly and are deleted.

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)
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 remaining pairs can at least be oriented weakly.

A__APP₂(nil, YS) -> MARK₁(YS)

Used ordering: Polynomial interpretation [21]:

POL(A__APP₂(x₁, x₂)) = 1 + 3·x₁ + x₂
POL(A__FROM₁(x₁)) = 2 + x₁
POL(A__ZWADR₂(x₁, x₂)) = 2 + 2·x₁ + 3·x₂
POL(MARK₁(x₁)) = 1 + x₁
POL(a__app₂(x₁, x₂)) = 3 + 3·x₁ + x₂
POL(a__from₁(x₁)) = 3 + 3·x₁
POL(a__prefix₁(x₁)) = 3 + x₁
POL(a__zWadr₂(x₁, x₂)) = 2 + 2·x₁ + 3·x₂
POL(app₂(x₁, x₂)) = 3 + 3·x₁ + x₂
POL(cons₂(x₁, x₂)) = 3 + 2·x₁
POL(from₁(x₁)) = 3 + 3·x₁
POL(mark₁(x₁)) = x₁
POL(nil) = 0
POL(prefix₁(x₁)) = 3 + x₁
POL(s₁(x₁)) = 1 + x₁
POL(zWadr₂(x₁, x₂)) = 2 + 2·x₁ + 3·x₂

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__APP₂(nil, YS) -> MARK₁(YS)

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 0 SCCs with 1 less node.