Termination w.r.t. Q proof of TRS/TRCSREx3_2_Luc97

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__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
a__dbls₁(nil) -> nil
a__dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
a__sel₂(0, cons₂(X, Y)) -> mark₁(X)
a__sel₂(s₁(X), cons₂(Y, Z)) -> a__sel₂(mark₁(X), mark₁(Z))
a__indx₂(nil, X) -> nil
a__indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(dbl₁(X)) -> a__dbl₁(mark₁(X))
mark₁(dbls₁(X)) -> a__dbls₁(mark₁(X))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(indx₂(X1, X2)) -> a__indx₂(mark₁(X1), X2)
mark₁(from₁(X)) -> a__from₁(X)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__dbl₁(X) -> dbl₁(X)
a__dbls₁(X) -> dbls₁(X)
a__sel₂(X1, X2) -> sel₂(X1, X2)
a__indx₂(X1, X2) -> indx₂(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__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
a__dbls₁(nil) -> nil
a__dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
a__sel₂(0, cons₂(X, Y)) -> mark₁(X)
a__sel₂(s₁(X), cons₂(Y, Z)) -> a__sel₂(mark₁(X), mark₁(Z))
a__indx₂(nil, X) -> nil
a__indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(dbl₁(X)) -> a__dbl₁(mark₁(X))
mark₁(dbls₁(X)) -> a__dbls₁(mark₁(X))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(indx₂(X1, X2)) -> a__indx₂(mark₁(X1), X2)
mark₁(from₁(X)) -> a__from₁(X)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__dbl₁(X) -> dbl₁(X)
a__dbls₁(X) -> dbls₁(X)
a__sel₂(X1, X2) -> sel₂(X1, X2)
a__indx₂(X1, X2) -> indx₂(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(dbl₁(X)) -> MARK₁(X)
A__SEL₂(0, cons₂(X, Y)) -> MARK₁(X)
MARK₁(indx₂(X1, X2)) -> MARK₁(X1)
MARK₁(dbl₁(X)) -> A__DBL₁(mark₁(X))
MARK₁(indx₂(X1, X2)) -> A__INDX₂(mark₁(X1), X2)
MARK₁(from₁(X)) -> A__FROM₁(X)
MARK₁(sel₂(X1, X2)) -> A__SEL₂(mark₁(X1), mark₁(X2))
A__SEL₂(s₁(X), cons₂(Y, Z)) -> A__SEL₂(mark₁(X), mark₁(Z))
MARK₁(dbls₁(X)) -> MARK₁(X)
MARK₁(sel₂(X1, X2)) -> MARK₁(X2)
A__SEL₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Z)
MARK₁(dbls₁(X)) -> A__DBLS₁(mark₁(X))
MARK₁(sel₂(X1, X2)) -> MARK₁(X1)
A__SEL₂(s₁(X), cons₂(Y, Z)) -> MARK₁(X)

The TRS R consists of the following rules:

a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
a__dbls₁(nil) -> nil
a__dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
a__sel₂(0, cons₂(X, Y)) -> mark₁(X)
a__sel₂(s₁(X), cons₂(Y, Z)) -> a__sel₂(mark₁(X), mark₁(Z))
a__indx₂(nil, X) -> nil
a__indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(dbl₁(X)) -> a__dbl₁(mark₁(X))
mark₁(dbls₁(X)) -> a__dbls₁(mark₁(X))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(indx₂(X1, X2)) -> a__indx₂(mark₁(X1), X2)
mark₁(from₁(X)) -> a__from₁(X)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__dbl₁(X) -> dbl₁(X)
a__dbls₁(X) -> dbls₁(X)
a__sel₂(X1, X2) -> sel₂(X1, X2)
a__indx₂(X1, X2) -> indx₂(X1, X2)
a__from₁(X) -> from₁(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

MARK₁(dbl₁(X)) -> MARK₁(X)
A__SEL₂(0, cons₂(X, Y)) -> MARK₁(X)
MARK₁(indx₂(X1, X2)) -> MARK₁(X1)
MARK₁(dbl₁(X)) -> A__DBL₁(mark₁(X))
MARK₁(indx₂(X1, X2)) -> A__INDX₂(mark₁(X1), X2)
MARK₁(from₁(X)) -> A__FROM₁(X)
MARK₁(sel₂(X1, X2)) -> A__SEL₂(mark₁(X1), mark₁(X2))
A__SEL₂(s₁(X), cons₂(Y, Z)) -> A__SEL₂(mark₁(X), mark₁(Z))
MARK₁(dbls₁(X)) -> MARK₁(X)
MARK₁(sel₂(X1, X2)) -> MARK₁(X2)
A__SEL₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Z)
MARK₁(dbls₁(X)) -> A__DBLS₁(mark₁(X))
MARK₁(sel₂(X1, X2)) -> MARK₁(X1)
A__SEL₂(s₁(X), cons₂(Y, Z)) -> MARK₁(X)

The TRS R consists of the following rules:

a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
a__dbls₁(nil) -> nil
a__dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
a__sel₂(0, cons₂(X, Y)) -> mark₁(X)
a__sel₂(s₁(X), cons₂(Y, Z)) -> a__sel₂(mark₁(X), mark₁(Z))
a__indx₂(nil, X) -> nil
a__indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(dbl₁(X)) -> a__dbl₁(mark₁(X))
mark₁(dbls₁(X)) -> a__dbls₁(mark₁(X))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(indx₂(X1, X2)) -> a__indx₂(mark₁(X1), X2)
mark₁(from₁(X)) -> a__from₁(X)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__dbl₁(X) -> dbl₁(X)
a__dbls₁(X) -> dbls₁(X)
a__sel₂(X1, X2) -> sel₂(X1, X2)
a__indx₂(X1, X2) -> indx₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

MARK₁(dbl₁(X)) -> MARK₁(X)
A__SEL₂(0, cons₂(X, Y)) -> MARK₁(X)
MARK₁(indx₂(X1, X2)) -> MARK₁(X1)
MARK₁(sel₂(X1, X2)) -> A__SEL₂(mark₁(X1), mark₁(X2))
A__SEL₂(s₁(X), cons₂(Y, Z)) -> A__SEL₂(mark₁(X), mark₁(Z))
MARK₁(dbls₁(X)) -> MARK₁(X)
MARK₁(sel₂(X1, X2)) -> MARK₁(X2)
A__SEL₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Z)
A__SEL₂(s₁(X), cons₂(Y, Z)) -> MARK₁(X)
MARK₁(sel₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__dbl₁(0) -> 0
a__dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
a__dbls₁(nil) -> nil
a__dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
a__sel₂(0, cons₂(X, Y)) -> mark₁(X)
a__sel₂(s₁(X), cons₂(Y, Z)) -> a__sel₂(mark₁(X), mark₁(Z))
a__indx₂(nil, X) -> nil
a__indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(dbl₁(X)) -> a__dbl₁(mark₁(X))
mark₁(dbls₁(X)) -> a__dbls₁(mark₁(X))
mark₁(sel₂(X1, X2)) -> a__sel₂(mark₁(X1), mark₁(X2))
mark₁(indx₂(X1, X2)) -> a__indx₂(mark₁(X1), X2)
mark₁(from₁(X)) -> a__from₁(X)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__dbl₁(X) -> dbl₁(X)
a__dbls₁(X) -> dbls₁(X)
a__sel₂(X1, X2) -> sel₂(X1, X2)
a__indx₂(X1, X2) -> indx₂(X1, X2)
a__from₁(X) -> from₁(X)

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