Termination w.r.t. Q proof of TRS/TRCSREx6_9_Luc02c

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__2nd₁(cons1₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__2nd₁(cons₂(X, X1)) -> a__2nd₁(cons1₂(mark₁(X), mark₁(X1)))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons1₂(X1, X2)) -> cons1₂(mark₁(X1), mark₁(X2))
a__2nd₁(X) -> 2nd₁(X)
a__from₁(X) -> from₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__2nd₁(cons1₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__2nd₁(cons₂(X, X1)) -> a__2nd₁(cons1₂(mark₁(X), mark₁(X1)))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons1₂(X1, X2)) -> cons1₂(mark₁(X1), mark₁(X2))
a__2nd₁(X) -> 2nd₁(X)
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(2nd₁(X)) -> MARK₁(X)
A__2ND₁(cons₂(X, X1)) -> MARK₁(X1)
A__2ND₁(cons₂(X, X1)) -> A__2ND₁(cons1₂(mark₁(X), mark₁(X1)))
A__2ND₁(cons1₂(X, cons₂(Y, Z))) -> MARK₁(Y)
A__2ND₁(cons₂(X, X1)) -> MARK₁(X)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(cons1₂(X1, X2)) -> MARK₁(X1)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons1₂(X1, X2)) -> MARK₁(X2)
A__FROM₁(X) -> MARK₁(X)
MARK₁(2nd₁(X)) -> A__2ND₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__2nd₁(cons1₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__2nd₁(cons₂(X, X1)) -> a__2nd₁(cons1₂(mark₁(X), mark₁(X1)))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons1₂(X1, X2)) -> cons1₂(mark₁(X1), mark₁(X2))
a__2nd₁(X) -> 2nd₁(X)
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(2nd₁(X)) -> MARK₁(X)
A__2ND₁(cons₂(X, X1)) -> MARK₁(X1)
A__2ND₁(cons₂(X, X1)) -> A__2ND₁(cons1₂(mark₁(X), mark₁(X1)))
A__2ND₁(cons1₂(X, cons₂(Y, Z))) -> MARK₁(Y)
A__2ND₁(cons₂(X, X1)) -> MARK₁(X)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(cons1₂(X1, X2)) -> MARK₁(X1)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons1₂(X1, X2)) -> MARK₁(X2)
A__FROM₁(X) -> MARK₁(X)
MARK₁(2nd₁(X)) -> A__2ND₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__2nd₁(cons1₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__2nd₁(cons₂(X, X1)) -> a__2nd₁(cons1₂(mark₁(X), mark₁(X1)))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons1₂(X1, X2)) -> cons1₂(mark₁(X1), mark₁(X2))
a__2nd₁(X) -> 2nd₁(X)
a__from₁(X) -> from₁(X)

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