Termination w.r.t. Q proof of TRS/TRCSREx4_Zan97

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__f₁(X) -> cons₂(mark₁(X), f₁(g₁(X)))
a__g₁(0) -> s₁(0)
a__g₁(s₁(X)) -> s₁(s₁(a__g₁(mark₁(X))))
a__sel₂(0, cons₂(X, Y)) -> mark₁(X)
a__sel₂(s₁(X), cons₂(Y, Z)) -> a__sel₂(mark₁(X), mark₁(Z))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> a__g₁(mark₁(X))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__g₁(X) -> g₁(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__f₁(X) -> cons₂(mark₁(X), f₁(g₁(X)))
a__g₁(0) -> s₁(0)
a__g₁(s₁(X)) -> s₁(s₁(a__g₁(mark₁(X))))
a__sel₂(0, cons₂(X, Y)) -> mark₁(X)
a__sel₂(s₁(X), cons₂(Y, Z)) -> a__sel₂(mark₁(X), mark₁(Z))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> a__g₁(mark₁(X))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__g₁(X) -> g₁(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:

MARK₁(g₁(X)) -> A__G₁(mark₁(X))
A__SEL₂(0, cons₂(X, Y)) -> MARK₁(X)
MARK₁(f₁(X)) -> A__F₁(mark₁(X))
MARK₁(sel₂(X1, X2)) -> A__SEL₂(mark₁(X1), mark₁(X2))
A__F₁(X) -> MARK₁(X)
A__G₁(s₁(X)) -> MARK₁(X)
A__SEL₂(s₁(X), cons₂(Y, Z)) -> A__SEL₂(mark₁(X), mark₁(Z))
MARK₁(sel₂(X1, X2)) -> MARK₁(X2)
A__G₁(s₁(X)) -> A__G₁(mark₁(X))
A__SEL₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Z)
MARK₁(sel₂(X1, X2)) -> MARK₁(X1)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(g₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__SEL₂(s₁(X), cons₂(Y, Z)) -> MARK₁(X)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(X) -> cons₂(mark₁(X), f₁(g₁(X)))
a__g₁(0) -> s₁(0)
a__g₁(s₁(X)) -> s₁(s₁(a__g₁(mark₁(X))))
a__sel₂(0, cons₂(X, Y)) -> mark₁(X)
a__sel₂(s₁(X), cons₂(Y, Z)) -> a__sel₂(mark₁(X), mark₁(Z))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> a__g₁(mark₁(X))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__g₁(X) -> g₁(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:

MARK₁(g₁(X)) -> A__G₁(mark₁(X))
A__SEL₂(0, cons₂(X, Y)) -> MARK₁(X)
MARK₁(f₁(X)) -> A__F₁(mark₁(X))
MARK₁(sel₂(X1, X2)) -> A__SEL₂(mark₁(X1), mark₁(X2))
A__F₁(X) -> MARK₁(X)
A__G₁(s₁(X)) -> MARK₁(X)
A__SEL₂(s₁(X), cons₂(Y, Z)) -> A__SEL₂(mark₁(X), mark₁(Z))
MARK₁(sel₂(X1, X2)) -> MARK₁(X2)
A__G₁(s₁(X)) -> A__G₁(mark₁(X))
A__SEL₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Z)
MARK₁(sel₂(X1, X2)) -> MARK₁(X1)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(g₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__SEL₂(s₁(X), cons₂(Y, Z)) -> MARK₁(X)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(X) -> cons₂(mark₁(X), f₁(g₁(X)))
a__g₁(0) -> s₁(0)
a__g₁(s₁(X)) -> s₁(s₁(a__g₁(mark₁(X))))
a__sel₂(0, cons₂(X, Y)) -> mark₁(X)
a__sel₂(s₁(X), cons₂(Y, Z)) -> a__sel₂(mark₁(X), mark₁(Z))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> a__g₁(mark₁(X))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__g₁(X) -> g₁(X)
a__sel₂(X1, X2) -> sel₂(X1, X2)

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