Termination w.r.t. Q proof of TRS/TRCSREx7_BLR02

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__head₁(cons₂(X, XS)) -> mark₁(X)
a__2nd₁(cons₂(X, XS)) -> a__head₁(mark₁(XS))
a__take₂(0, XS) -> nil
a__take₂(s₁(N), cons₂(X, XS)) -> cons₂(mark₁(X), take₂(N, XS))
a__sel₂(0, cons₂(X, XS)) -> mark₁(X)
a__sel₂(s₁(N), cons₂(X, XS)) -> a__sel₂(mark₁(N), mark₁(XS))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
a__from₁(X) -> from₁(X)
a__head₁(X) -> head₁(X)
a__2nd₁(X) -> 2nd₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__sel₂(X1, X2) -> sel₂(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__head₁(cons₂(X, XS)) -> mark₁(X)
a__2nd₁(cons₂(X, XS)) -> a__head₁(mark₁(XS))
a__take₂(0, XS) -> nil
a__take₂(s₁(N), cons₂(X, XS)) -> cons₂(mark₁(X), take₂(N, XS))
a__sel₂(0, cons₂(X, XS)) -> mark₁(X)
a__sel₂(s₁(N), cons₂(X, XS)) -> a__sel₂(mark₁(N), mark₁(XS))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
a__from₁(X) -> from₁(X)
a__head₁(X) -> head₁(X)
a__2nd₁(X) -> 2nd₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__sel₂(X1, X2) -> sel₂(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(take₂(X1, X2)) -> MARK₁(X2)
A__SEL₂(0, cons₂(X, XS)) -> MARK₁(X)
MARK₁(2nd₁(X)) -> MARK₁(X)
MARK₁(sel₂(X1, X2)) -> A__SEL₂(mark₁(X1), mark₁(X2))
A__2ND₁(cons₂(X, XS)) -> MARK₁(XS)
MARK₁(take₂(X1, X2)) -> MARK₁(X1)
A__SEL₂(s₁(N), cons₂(X, XS)) -> A__SEL₂(mark₁(N), mark₁(XS))
MARK₁(head₁(X)) -> A__HEAD₁(mark₁(X))
MARK₁(sel₂(X1, X2)) -> MARK₁(X2)
A__SEL₂(s₁(N), cons₂(X, XS)) -> MARK₁(XS)
MARK₁(head₁(X)) -> MARK₁(X)
MARK₁(sel₂(X1, X2)) -> MARK₁(X1)
MARK₁(take₂(X1, X2)) -> A__TAKE₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
A__2ND₁(cons₂(X, XS)) -> A__HEAD₁(mark₁(XS))
MARK₁(from₁(X)) -> MARK₁(X)
A__TAKE₂(s₁(N), cons₂(X, XS)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__HEAD₁(cons₂(X, XS)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(2nd₁(X)) -> A__2ND₁(mark₁(X))
A__FROM₁(X) -> MARK₁(X)
A__SEL₂(s₁(N), cons₂(X, XS)) -> MARK₁(N)

The TRS R consists of the following rules:

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__head₁(cons₂(X, XS)) -> mark₁(X)
a__2nd₁(cons₂(X, XS)) -> a__head₁(mark₁(XS))
a__take₂(0, XS) -> nil
a__take₂(s₁(N), cons₂(X, XS)) -> cons₂(mark₁(X), take₂(N, XS))
a__sel₂(0, cons₂(X, XS)) -> mark₁(X)
a__sel₂(s₁(N), cons₂(X, XS)) -> a__sel₂(mark₁(N), mark₁(XS))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
a__from₁(X) -> from₁(X)
a__head₁(X) -> head₁(X)
a__2nd₁(X) -> 2nd₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__sel₂(X1, X2) -> sel₂(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(take₂(X1, X2)) -> MARK₁(X2)
A__SEL₂(0, cons₂(X, XS)) -> MARK₁(X)
MARK₁(2nd₁(X)) -> MARK₁(X)
MARK₁(sel₂(X1, X2)) -> A__SEL₂(mark₁(X1), mark₁(X2))
A__2ND₁(cons₂(X, XS)) -> MARK₁(XS)
MARK₁(take₂(X1, X2)) -> MARK₁(X1)
A__SEL₂(s₁(N), cons₂(X, XS)) -> A__SEL₂(mark₁(N), mark₁(XS))
MARK₁(head₁(X)) -> A__HEAD₁(mark₁(X))
MARK₁(sel₂(X1, X2)) -> MARK₁(X2)
A__SEL₂(s₁(N), cons₂(X, XS)) -> MARK₁(XS)
MARK₁(head₁(X)) -> MARK₁(X)
MARK₁(sel₂(X1, X2)) -> MARK₁(X1)
MARK₁(take₂(X1, X2)) -> A__TAKE₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
A__2ND₁(cons₂(X, XS)) -> A__HEAD₁(mark₁(XS))
MARK₁(from₁(X)) -> MARK₁(X)
A__TAKE₂(s₁(N), cons₂(X, XS)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__HEAD₁(cons₂(X, XS)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(2nd₁(X)) -> A__2ND₁(mark₁(X))
A__FROM₁(X) -> MARK₁(X)
A__SEL₂(s₁(N), cons₂(X, XS)) -> MARK₁(N)

The TRS R consists of the following rules:

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__head₁(cons₂(X, XS)) -> mark₁(X)
a__2nd₁(cons₂(X, XS)) -> a__head₁(mark₁(XS))
a__take₂(0, XS) -> nil
a__take₂(s₁(N), cons₂(X, XS)) -> cons₂(mark₁(X), take₂(N, XS))
a__sel₂(0, cons₂(X, XS)) -> mark₁(X)
a__sel₂(s₁(N), cons₂(X, XS)) -> a__sel₂(mark₁(N), mark₁(XS))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
a__from₁(X) -> from₁(X)
a__head₁(X) -> head₁(X)
a__2nd₁(X) -> 2nd₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__sel₂(X1, X2) -> sel₂(X1, X2)

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