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
  ↳ 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))

Q is empty.

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))

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

↳ 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F2(cons2(x, k), l) -> G3(k, l, cons2(x, k))
G3(a, b, c) -> F2(a, cons2(b, c))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = 1 + 2·x1   
POL(G3(x1, x2, x3)) = 2 + 3·x1   
POL(cons2(x1, x2)) = 2 + 3·x1·x2 + 2·x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ PisEmptyProof

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