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:

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(cons(x, k), l) → G(k, l, cons(x, k))
G(a, b, c) → F(a, cons(b, c))

The TRS R consists of the following rules:

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(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:

F(cons(x, k), l) → G(k, l, cons(x, k))
G(a, b, c) → F(a, cons(b, c))

The TRS R consists of the following rules:

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(cons(x, k), l) → G(k, l, cons(x, k))
The remaining pairs can at least be oriented weakly.

G(a, b, c) → F(a, cons(b, c))
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 4 + (4)x_2   
POL(G(x1, x2, x3)) = (3)x_1   
POL(F(x1, x2)) = (2)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof

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

G(a, b, c) → F(a, cons(b, c))

The TRS R consists of the following rules:

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.