Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/TRCSRSLASHEx4_7_77_Bor03

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a__zeros -> cons₂(0, zeros)
a__tail₁(cons₂(X, XS)) -> mark₁(XS)
mark₁(zeros) -> a__zeros
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
a__zeros -> zeros
a__tail₁(X) -> tail₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__zeros -> cons₂(0, zeros)
a__tail₁(cons₂(X, XS)) -> mark₁(XS)
mark₁(zeros) -> a__zeros
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
a__zeros -> zeros
a__tail₁(X) -> tail₁(X)

Q is empty.

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

MARK₁(tail₁(X)) -> A__TAIL₁(mark₁(X))
A__TAIL₁(cons₂(X, XS)) -> MARK₁(XS)
MARK₁(tail₁(X)) -> MARK₁(X)
MARK₁(zeros) -> A__ZEROS
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__zeros -> cons₂(0, zeros)
a__tail₁(cons₂(X, XS)) -> mark₁(XS)
mark₁(zeros) -> a__zeros
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
a__zeros -> zeros
a__tail₁(X) -> tail₁(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:

MARK₁(tail₁(X)) -> A__TAIL₁(mark₁(X))
A__TAIL₁(cons₂(X, XS)) -> MARK₁(XS)
MARK₁(tail₁(X)) -> MARK₁(X)
MARK₁(zeros) -> A__ZEROS
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__zeros -> cons₂(0, zeros)
a__tail₁(cons₂(X, XS)) -> mark₁(XS)
mark₁(zeros) -> a__zeros
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
a__zeros -> zeros
a__tail₁(X) -> tail₁(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

MARK₁(tail₁(X)) -> A__TAIL₁(mark₁(X))
A__TAIL₁(cons₂(X, XS)) -> MARK₁(XS)
MARK₁(tail₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__zeros -> cons₂(0, zeros)
a__tail₁(cons₂(X, XS)) -> mark₁(XS)
mark₁(zeros) -> a__zeros
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
a__zeros -> zeros
a__tail₁(X) -> tail₁(X)

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