Termination w.r.t. Q proof of TRS/TRCSREx1_GL02a

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

a__eq₂(0, 0) -> true
a__eq₂(s₁(X), s₁(Y)) -> a__eq₂(X, Y)
a__eq₂(X, Y) -> false
a__inf₁(X) -> cons₂(X, inf₁(s₁(X)))
a__take₂(0, X) -> nil
a__take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
a__length₁(nil) -> 0
a__length₁(cons₂(X, L)) -> s₁(length₁(L))
mark₁(eq₂(X1, X2)) -> a__eq₂(X1, X2)
mark₁(inf₁(X)) -> a__inf₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(length₁(X)) -> a__length₁(mark₁(X))
mark₁(0) -> 0
mark₁(true) -> true
mark₁(s₁(X)) -> s₁(X)
mark₁(false) -> false
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(nil) -> nil
a__eq₂(X1, X2) -> eq₂(X1, X2)
a__inf₁(X) -> inf₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__length₁(X) -> length₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__eq₂(0, 0) -> true
a__eq₂(s₁(X), s₁(Y)) -> a__eq₂(X, Y)
a__eq₂(X, Y) -> false
a__inf₁(X) -> cons₂(X, inf₁(s₁(X)))
a__take₂(0, X) -> nil
a__take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
a__length₁(nil) -> 0
a__length₁(cons₂(X, L)) -> s₁(length₁(L))
mark₁(eq₂(X1, X2)) -> a__eq₂(X1, X2)
mark₁(inf₁(X)) -> a__inf₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(length₁(X)) -> a__length₁(mark₁(X))
mark₁(0) -> 0
mark₁(true) -> true
mark₁(s₁(X)) -> s₁(X)
mark₁(false) -> false
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(nil) -> nil
a__eq₂(X1, X2) -> eq₂(X1, X2)
a__inf₁(X) -> inf₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__length₁(X) -> length₁(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:

A__EQ₂(s₁(X), s₁(Y)) -> A__EQ₂(X, Y)
MARK₁(take₂(X1, X2)) -> MARK₁(X2)
MARK₁(inf₁(X)) -> MARK₁(X)
MARK₁(eq₂(X1, X2)) -> A__EQ₂(X1, X2)
MARK₁(length₁(X)) -> A__LENGTH₁(mark₁(X))
MARK₁(take₂(X1, X2)) -> MARK₁(X1)
MARK₁(inf₁(X)) -> A__INF₁(mark₁(X))
MARK₁(length₁(X)) -> MARK₁(X)
MARK₁(take₂(X1, X2)) -> A__TAKE₂(mark₁(X1), mark₁(X2))

The TRS R consists of the following rules:

a__eq₂(0, 0) -> true
a__eq₂(s₁(X), s₁(Y)) -> a__eq₂(X, Y)
a__eq₂(X, Y) -> false
a__inf₁(X) -> cons₂(X, inf₁(s₁(X)))
a__take₂(0, X) -> nil
a__take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
a__length₁(nil) -> 0
a__length₁(cons₂(X, L)) -> s₁(length₁(L))
mark₁(eq₂(X1, X2)) -> a__eq₂(X1, X2)
mark₁(inf₁(X)) -> a__inf₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(length₁(X)) -> a__length₁(mark₁(X))
mark₁(0) -> 0
mark₁(true) -> true
mark₁(s₁(X)) -> s₁(X)
mark₁(false) -> false
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(nil) -> nil
a__eq₂(X1, X2) -> eq₂(X1, X2)
a__inf₁(X) -> inf₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__length₁(X) -> length₁(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:

A__EQ₂(s₁(X), s₁(Y)) -> A__EQ₂(X, Y)
MARK₁(take₂(X1, X2)) -> MARK₁(X2)
MARK₁(inf₁(X)) -> MARK₁(X)
MARK₁(eq₂(X1, X2)) -> A__EQ₂(X1, X2)
MARK₁(length₁(X)) -> A__LENGTH₁(mark₁(X))
MARK₁(take₂(X1, X2)) -> MARK₁(X1)
MARK₁(inf₁(X)) -> A__INF₁(mark₁(X))
MARK₁(length₁(X)) -> MARK₁(X)
MARK₁(take₂(X1, X2)) -> A__TAKE₂(mark₁(X1), mark₁(X2))

The TRS R consists of the following rules:

a__eq₂(0, 0) -> true
a__eq₂(s₁(X), s₁(Y)) -> a__eq₂(X, Y)
a__eq₂(X, Y) -> false
a__inf₁(X) -> cons₂(X, inf₁(s₁(X)))
a__take₂(0, X) -> nil
a__take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
a__length₁(nil) -> 0
a__length₁(cons₂(X, L)) -> s₁(length₁(L))
mark₁(eq₂(X1, X2)) -> a__eq₂(X1, X2)
mark₁(inf₁(X)) -> a__inf₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(length₁(X)) -> a__length₁(mark₁(X))
mark₁(0) -> 0
mark₁(true) -> true
mark₁(s₁(X)) -> s₁(X)
mark₁(false) -> false
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(nil) -> nil
a__eq₂(X1, X2) -> eq₂(X1, X2)
a__inf₁(X) -> inf₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__length₁(X) -> length₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

A__EQ₂(s₁(X), s₁(Y)) -> A__EQ₂(X, Y)

The TRS R consists of the following rules:

a__eq₂(0, 0) -> true
a__eq₂(s₁(X), s₁(Y)) -> a__eq₂(X, Y)
a__eq₂(X, Y) -> false
a__inf₁(X) -> cons₂(X, inf₁(s₁(X)))
a__take₂(0, X) -> nil
a__take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
a__length₁(nil) -> 0
a__length₁(cons₂(X, L)) -> s₁(length₁(L))
mark₁(eq₂(X1, X2)) -> a__eq₂(X1, X2)
mark₁(inf₁(X)) -> a__inf₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(length₁(X)) -> a__length₁(mark₁(X))
mark₁(0) -> 0
mark₁(true) -> true
mark₁(s₁(X)) -> s₁(X)
mark₁(false) -> false
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(nil) -> nil
a__eq₂(X1, X2) -> eq₂(X1, X2)
a__inf₁(X) -> inf₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__length₁(X) -> length₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

A__EQ₂(s₁(X), s₁(Y)) -> A__EQ₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(A__EQ₂(x₁, x₂)) = 3·x₁ + 3·x₁·x₂ + 3·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__eq₂(0, 0) -> true
a__eq₂(s₁(X), s₁(Y)) -> a__eq₂(X, Y)
a__eq₂(X, Y) -> false
a__inf₁(X) -> cons₂(X, inf₁(s₁(X)))
a__take₂(0, X) -> nil
a__take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
a__length₁(nil) -> 0
a__length₁(cons₂(X, L)) -> s₁(length₁(L))
mark₁(eq₂(X1, X2)) -> a__eq₂(X1, X2)
mark₁(inf₁(X)) -> a__inf₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(length₁(X)) -> a__length₁(mark₁(X))
mark₁(0) -> 0
mark₁(true) -> true
mark₁(s₁(X)) -> s₁(X)
mark₁(false) -> false
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(nil) -> nil
a__eq₂(X1, X2) -> eq₂(X1, X2)
a__inf₁(X) -> inf₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__length₁(X) -> length₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

MARK₁(take₂(X1, X2)) -> MARK₁(X2)
MARK₁(inf₁(X)) -> MARK₁(X)
MARK₁(take₂(X1, X2)) -> MARK₁(X1)
MARK₁(length₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__eq₂(0, 0) -> true
a__eq₂(s₁(X), s₁(Y)) -> a__eq₂(X, Y)
a__eq₂(X, Y) -> false
a__inf₁(X) -> cons₂(X, inf₁(s₁(X)))
a__take₂(0, X) -> nil
a__take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
a__length₁(nil) -> 0
a__length₁(cons₂(X, L)) -> s₁(length₁(L))
mark₁(eq₂(X1, X2)) -> a__eq₂(X1, X2)
mark₁(inf₁(X)) -> a__inf₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(length₁(X)) -> a__length₁(mark₁(X))
mark₁(0) -> 0
mark₁(true) -> true
mark₁(s₁(X)) -> s₁(X)
mark₁(false) -> false
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(nil) -> nil
a__eq₂(X1, X2) -> eq₂(X1, X2)
a__inf₁(X) -> inf₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__length₁(X) -> length₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MARK₁(take₂(X1, X2)) -> MARK₁(X2)
MARK₁(inf₁(X)) -> MARK₁(X)
MARK₁(take₂(X1, X2)) -> MARK₁(X1)
MARK₁(length₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MARK₁(x₁)) = 3·x₁ + 3·x₁²
POL(inf₁(x₁)) = 3 + x₁
POL(length₁(x₁)) = 3 + 3·x₁ + 3·x₁²
POL(take₂(x₁, x₂)) = 3 + 2·x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__eq₂(0, 0) -> true
a__eq₂(s₁(X), s₁(Y)) -> a__eq₂(X, Y)
a__eq₂(X, Y) -> false
a__inf₁(X) -> cons₂(X, inf₁(s₁(X)))
a__take₂(0, X) -> nil
a__take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
a__length₁(nil) -> 0
a__length₁(cons₂(X, L)) -> s₁(length₁(L))
mark₁(eq₂(X1, X2)) -> a__eq₂(X1, X2)
mark₁(inf₁(X)) -> a__inf₁(mark₁(X))
mark₁(take₂(X1, X2)) -> a__take₂(mark₁(X1), mark₁(X2))
mark₁(length₁(X)) -> a__length₁(mark₁(X))
mark₁(0) -> 0
mark₁(true) -> true
mark₁(s₁(X)) -> s₁(X)
mark₁(false) -> false
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(nil) -> nil
a__eq₂(X1, X2) -> eq₂(X1, X2)
a__inf₁(X) -> inf₁(X)
a__take₂(X1, X2) -> take₂(X1, X2)
a__length₁(X) -> length₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.