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.

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 contains 2 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

A__EQ2(s1(X), s1(Y)) -> A__EQ2(X, Y)
Used argument filtering: A__EQ2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ 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
            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MARK1(take2(X1, X2)) -> MARK1(X2)
MARK1(take2(X1, X2)) -> MARK1(X1)
Used argument filtering: MARK1(x1)  =  x1
take2(x1, x2)  =  take2(x1, x2)
inf1(x1)  =  x1
length1(x1)  =  x1
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof

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

MARK1(inf1(X)) -> MARK1(X)
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MARK1(length1(X)) -> MARK1(X)
Used argument filtering: MARK1(x1)  =  x1
inf1(x1)  =  x1
length1(x1)  =  length1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof

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

MARK1(inf1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MARK1(inf1(X)) -> MARK1(X)
Used argument filtering: MARK1(x1)  =  x1
inf1(x1)  =  inf1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ 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.