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(empty, l) -> l
f2(cons2(x, k), l) -> g3(k, l, cons2(x, k))
g3(a, b, c) -> f2(a, cons2(b, c))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

f2(empty, l) -> l
f2(cons2(x, k), l) -> g3(k, l, cons2(x, k))
g3(a, b, c) -> f2(a, cons2(b, c))

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(empty, l) -> l
f2(cons2(x, k), l) -> g3(k, l, cons2(x, k))
g3(a, b, c) -> f2(a, cons2(b, c))

The set Q consists of the following terms:

f2(empty, x0)
f2(cons2(x0, x1), x2)
g3(x0, x1, x2)


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

F2(cons2(x, k), l) -> G3(k, l, cons2(x, k))
G3(a, b, c) -> F2(a, cons2(b, c))

The TRS R consists of the following rules:

f2(empty, l) -> l
f2(cons2(x, k), l) -> g3(k, l, cons2(x, k))
g3(a, b, c) -> f2(a, cons2(b, c))

The set Q consists of the following terms:

f2(empty, x0)
f2(cons2(x0, x1), x2)
g3(x0, x1, x2)

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

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

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

F2(cons2(x, k), l) -> G3(k, l, cons2(x, k))
G3(a, b, c) -> F2(a, cons2(b, c))

The TRS R consists of the following rules:

f2(empty, l) -> l
f2(cons2(x, k), l) -> g3(k, l, cons2(x, k))
g3(a, b, c) -> f2(a, cons2(b, c))

The set Q consists of the following terms:

f2(empty, x0)
f2(cons2(x0, x1), x2)
g3(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(cons2(x, k), l) -> G3(k, l, cons2(x, k))
Used argument filtering: F2(x1, x2)  =  x1
cons2(x1, x2)  =  cons1(x2)
G3(x1, x2, x3)  =  x1
Used ordering: Quasi Precedence: trivial 


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

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

G3(a, b, c) -> F2(a, cons2(b, c))

The TRS R consists of the following rules:

f2(empty, l) -> l
f2(cons2(x, k), l) -> g3(k, l, cons2(x, k))
g3(a, b, c) -> f2(a, cons2(b, c))

The set Q consists of the following terms:

f2(empty, x0)
f2(cons2(x0, x1), x2)
g3(x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.