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(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

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:

G(cons(x, k), d) → G(k, cons(x, d))
F(a, cons(x, k)) → F(cons(x, a), k)
F(a, empty) → G(a, empty)

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

G(cons(x, k), d) → G(k, cons(x, d))
F(a, cons(x, k)) → F(cons(x, a), k)
F(a, empty) → G(a, empty)

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

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

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

G(cons(x, k), d) → G(k, cons(x, d))

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

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.


G(cons(x, k), d) → G(k, cons(x, d))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 4 + x_2   
POL(G(x1, x2)) = (4)x_1 + (5/4)x_2   
The value of delta used in the strict ordering is 11.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

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.

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

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

F(a, cons(x, k)) → F(cons(x, a), k)

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

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(a, cons(x, k)) → F(cons(x, a), k)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 1 + (11/4)x_1 + (5/2)x_2   
POL(F(x1, x2)) = (5/2)x_2   
The value of delta used in the strict ordering is 5/2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

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.