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__sel2(0, cons2(X, Y)) -> mark1(X)
a__sel2(s1(X), cons2(Y, Z)) -> a__sel2(mark1(X), mark1(Z))
mark1(from1(X)) -> a__from1(mark1(X))
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
a__from1(X) -> from1(X)
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__sel2(0, cons2(X, Y)) -> mark1(X)
a__sel2(s1(X), cons2(Y, Z)) -> a__sel2(mark1(X), mark1(Z))
mark1(from1(X)) -> a__from1(mark1(X))
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
a__from1(X) -> from1(X)
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:

A__SEL2(0, cons2(X, Y)) -> MARK1(X)
MARK1(s1(X)) -> MARK1(X)
MARK1(sel2(X1, X2)) -> A__SEL2(mark1(X1), mark1(X2))
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__SEL2(s1(X), cons2(Y, Z)) -> A__SEL2(mark1(X), mark1(Z))
MARK1(sel2(X1, X2)) -> MARK1(X2)
A__SEL2(s1(X), cons2(Y, Z)) -> MARK1(Z)
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(sel2(X1, X2)) -> MARK1(X1)
A__SEL2(s1(X), cons2(Y, Z)) -> MARK1(X)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__sel2(0, cons2(X, Y)) -> mark1(X)
a__sel2(s1(X), cons2(Y, Z)) -> a__sel2(mark1(X), mark1(Z))
mark1(from1(X)) -> a__from1(mark1(X))
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
a__from1(X) -> from1(X)
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:

A__SEL2(0, cons2(X, Y)) -> MARK1(X)
MARK1(s1(X)) -> MARK1(X)
MARK1(sel2(X1, X2)) -> A__SEL2(mark1(X1), mark1(X2))
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__SEL2(s1(X), cons2(Y, Z)) -> A__SEL2(mark1(X), mark1(Z))
MARK1(sel2(X1, X2)) -> MARK1(X2)
A__SEL2(s1(X), cons2(Y, Z)) -> MARK1(Z)
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(sel2(X1, X2)) -> MARK1(X1)
A__SEL2(s1(X), cons2(Y, Z)) -> MARK1(X)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__sel2(0, cons2(X, Y)) -> mark1(X)
a__sel2(s1(X), cons2(Y, Z)) -> a__sel2(mark1(X), mark1(Z))
mark1(from1(X)) -> a__from1(mark1(X))
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
a__from1(X) -> from1(X)
a__sel2(X1, X2) -> sel2(X1, X2)

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