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:

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

Q is empty.

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

EQ2(s1(X), s1(Y)) -> EQ2(X, Y)
TAKE2(s1(X), cons2(Y, L)) -> TAKE2(X, L)
LENGTH1(cons2(X, L)) -> LENGTH1(L)
INF1(X) -> INF1(s1(X))

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

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:

EQ2(s1(X), s1(Y)) -> EQ2(X, Y)
TAKE2(s1(X), cons2(Y, L)) -> TAKE2(X, L)
LENGTH1(cons2(X, L)) -> LENGTH1(L)
INF1(X) -> INF1(s1(X))

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs.

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

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

LENGTH1(cons2(X, L)) -> LENGTH1(L)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

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.

LENGTH1(cons2(X, L)) -> LENGTH1(L)
Used argument filtering: LENGTH1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Precedence:
trivial



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

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

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

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
          ↳ QDP
          ↳ QDP

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

TAKE2(s1(X), cons2(Y, L)) -> TAKE2(X, L)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

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.

TAKE2(s1(X), cons2(Y, L)) -> TAKE2(X, L)
Used argument filtering: TAKE2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
Used ordering: Precedence:
trivial



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

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

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

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
          ↳ QDP
          ↳ QDP

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

INF1(X) -> INF1(s1(X))

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

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

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

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

EQ2(s1(X), s1(Y)) -> EQ2(X, Y)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

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.

EQ2(s1(X), s1(Y)) -> EQ2(X, Y)
Used argument filtering: EQ2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Precedence:
trivial



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

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

eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))

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.