Termination w.r.t. Q proof of TRS/TRCSREx6_Luc98

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__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(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:

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> 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)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

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(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> 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)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

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