Termination w.r.t. Q proof of TRS/TRCSREx4_7_56_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__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__after₂(0, XS) -> mark₁(XS)
a__after₂(s₁(N), cons₂(X, XS)) -> a__after₂(mark₁(N), mark₁(XS))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(after₂(X1, X2)) -> a__after₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__after₂(X1, X2) -> after₂(X1, X2)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__after₂(0, XS) -> mark₁(XS)
a__after₂(s₁(N), cons₂(X, XS)) -> a__after₂(mark₁(N), mark₁(XS))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(after₂(X1, X2)) -> a__after₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__after₂(X1, X2) -> after₂(X1, X2)

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:

MARK₁(after₂(X1, X2)) -> MARK₁(X2)
MARK₁(s₁(X)) -> MARK₁(X)
A__AFTER₂(0, XS) -> MARK₁(XS)
A__AFTER₂(s₁(N), cons₂(X, XS)) -> MARK₁(XS)
MARK₁(after₂(X1, X2)) -> A__AFTER₂(mark₁(X1), mark₁(X2))
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__AFTER₂(s₁(N), cons₂(X, XS)) -> MARK₁(N)
A__AFTER₂(s₁(N), cons₂(X, XS)) -> A__AFTER₂(mark₁(N), mark₁(XS))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(after₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__after₂(0, XS) -> mark₁(XS)
a__after₂(s₁(N), cons₂(X, XS)) -> a__after₂(mark₁(N), mark₁(XS))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(after₂(X1, X2)) -> a__after₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__after₂(X1, X2) -> after₂(X1, X2)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

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

MARK₁(after₂(X1, X2)) -> MARK₁(X2)
MARK₁(s₁(X)) -> MARK₁(X)
A__AFTER₂(0, XS) -> MARK₁(XS)
A__AFTER₂(s₁(N), cons₂(X, XS)) -> MARK₁(XS)
MARK₁(after₂(X1, X2)) -> A__AFTER₂(mark₁(X1), mark₁(X2))
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__AFTER₂(s₁(N), cons₂(X, XS)) -> MARK₁(N)
A__AFTER₂(s₁(N), cons₂(X, XS)) -> A__AFTER₂(mark₁(N), mark₁(XS))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(after₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__after₂(0, XS) -> mark₁(XS)
a__after₂(s₁(N), cons₂(X, XS)) -> a__after₂(mark₁(N), mark₁(XS))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(after₂(X1, X2)) -> a__after₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__after₂(X1, X2) -> after₂(X1, X2)

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