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)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


G2(cons2(x, k), d) -> G2(k, cons2(x, d))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G2(x1, x2)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


F2(a, cons2(x, k)) -> F2(cons2(x, a), k)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  x2
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
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.