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:

f2(a, empty) -> g2(a, empty)
f2(a, cons2(x, k)) -> f2(cons2(x, a), k)
g2(empty, d) -> d
g2(cons2(x, k), d) -> g2(k, cons2(x, d))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

f2(a, empty) -> g2(a, empty)
f2(a, cons2(x, k)) -> f2(cons2(x, a), k)
g2(empty, d) -> d
g2(cons2(x, k), d) -> g2(k, cons2(x, d))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

f2(a, empty) -> g2(a, empty)
f2(a, cons2(x, k)) -> f2(cons2(x, a), k)
g2(empty, d) -> d
g2(cons2(x, k), d) -> g2(k, cons2(x, d))

The set Q consists of the following terms:

f2(x0, empty)
f2(x0, cons2(x1, x2))
g2(empty, x0)
g2(cons2(x0, x1), x2)


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

G2(cons2(x, k), d) -> G2(k, cons2(x, d))
F2(a, empty) -> G2(a, empty)
F2(a, cons2(x, k)) -> F2(cons2(x, a), k)

The TRS R consists of the following rules:

f2(a, empty) -> g2(a, empty)
f2(a, cons2(x, k)) -> f2(cons2(x, a), k)
g2(empty, d) -> d
g2(cons2(x, k), d) -> g2(k, cons2(x, d))

The set Q consists of the following terms:

f2(x0, empty)
f2(x0, cons2(x1, x2))
g2(empty, x0)
g2(cons2(x0, x1), x2)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

G2(cons2(x, k), d) -> G2(k, cons2(x, d))
F2(a, empty) -> G2(a, empty)
F2(a, cons2(x, k)) -> F2(cons2(x, a), k)

The TRS R consists of the following rules:

f2(a, empty) -> g2(a, empty)
f2(a, cons2(x, k)) -> f2(cons2(x, a), k)
g2(empty, d) -> d
g2(cons2(x, k), d) -> g2(k, cons2(x, d))

The set Q consists of the following terms:

f2(x0, empty)
f2(x0, cons2(x1, x2))
g2(empty, x0)
g2(cons2(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP

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

G2(cons2(x, k), d) -> G2(k, cons2(x, d))

The TRS R consists of the following rules:

f2(a, empty) -> g2(a, empty)
f2(a, cons2(x, k)) -> f2(cons2(x, a), k)
g2(empty, d) -> d
g2(cons2(x, k), d) -> g2(k, cons2(x, d))

The set Q consists of the following terms:

f2(x0, empty)
f2(x0, cons2(x1, x2))
g2(empty, x0)
g2(cons2(x0, x1), x2)

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.

G2(cons2(x, k), d) -> G2(k, cons2(x, d))
Used argument filtering: G2(x1, x2)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

f2(a, empty) -> g2(a, empty)
f2(a, cons2(x, k)) -> f2(cons2(x, a), k)
g2(empty, d) -> d
g2(cons2(x, k), d) -> g2(k, cons2(x, d))

The set Q consists of the following terms:

f2(x0, empty)
f2(x0, cons2(x1, x2))
g2(empty, x0)
g2(cons2(x0, x1), x2)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof

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

F2(a, cons2(x, k)) -> F2(cons2(x, a), k)

The TRS R consists of the following rules:

f2(a, empty) -> g2(a, empty)
f2(a, cons2(x, k)) -> f2(cons2(x, a), k)
g2(empty, d) -> d
g2(cons2(x, k), d) -> g2(k, cons2(x, d))

The set Q consists of the following terms:

f2(x0, empty)
f2(x0, cons2(x1, x2))
g2(empty, x0)
g2(cons2(x0, x1), x2)

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.

F2(a, cons2(x, k)) -> F2(cons2(x, a), k)
Used argument filtering: F2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof

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

f2(a, empty) -> g2(a, empty)
f2(a, cons2(x, k)) -> f2(cons2(x, a), k)
g2(empty, d) -> d
g2(cons2(x, k), d) -> g2(k, cons2(x, d))

The set Q consists of the following terms:

f2(x0, empty)
f2(x0, cons2(x1, x2))
g2(empty, x0)
g2(cons2(x0, x1), x2)

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