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)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


F2(cons2(x, k), l) -> G3(k, l, cons2(x, k))
The remaining pairs can at least by weakly be oriented.

G3(a, b, c) -> F2(a, cons2(b, c))
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  F1(x1)
cons2(x1, x2)  =  cons1(x2)
G3(x1, x2, x3)  =  G1(x1)

Lexicographic Path Order [19].
Precedence:
[F1, G1] > cons1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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 [13,14,18] contains 0 SCCs with 1 less node.