Termination w.r.t. Q proof of /tmp/tmpWZXvas/beans.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

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(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L¹(f(s(s(y)), f(z, w))) → L¹(f(s(0), f(y, f(s(z), w))))
L¹(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(0), nil)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
L¹(f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))
L¹(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

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:

F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L¹(f(s(s(y)), f(z, w))) → L¹(f(s(0), f(y, f(s(z), w))))
L¹(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(0), nil)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
L¹(f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))
L¹(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

          ↳ QDP

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

F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(F(x₁, x₂)) = x₁ + x₂
POL(f(x₁, x₂)) = 2·x₁ + x₂
POL(nil) = 2
POL(s(x₁)) = 1 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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

            ↳ UsableRulesProof

            ↳ UsableRulesProof

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

L¹(f(s(s(y)), f(z, w))) → L¹(f(s(0), f(y, f(s(z), w))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ Narrowing

            ↳ UsableRulesProof

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

L¹(f(s(s(y)), f(z, w))) → L¹(f(s(0), f(y, f(s(z), w))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule L¹(f(s(s(y)), f(z, w))) → L¹(f(s(0), f(y, f(s(z), w)))) at position [0] we obtained the following new rules:

L¹(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L¹(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L¹(f(s(s(x0)), f(s(x1), nil))) → L¹(f(s(0), f(s(x0), f(x1, f(s(0), nil)))))
L¹(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L¹(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))
L¹(f(s(s(s(s(x1)))), f(y1, x3))) → L¹(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L¹(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L¹(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

L¹(f(s(s(x0)), f(s(x1), nil))) → L¹(f(s(0), f(s(x0), f(x1, f(s(0), nil)))))
L¹(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L¹(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))
L¹(f(s(s(s(s(x1)))), f(y1, x3))) → L¹(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L¹(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L¹(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L¹(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L¹(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

            ↳ UsableRulesProof

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

L¹(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L¹(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))
L¹(f(s(s(s(s(x1)))), f(y1, x3))) → L¹(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L¹(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L¹(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L¹(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L¹(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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.

L¹(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L¹(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))

The remaining pairs can at least be oriented weakly.

L¹(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L¹(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))
L¹(f(s(s(s(s(x1)))), f(y1, x3))) → L¹(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L¹(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L¹(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))

Used ordering: Polynomial interpretation [25,35]:

POL(L¹(x₁)) = (3/4)x_1
POL(f(x₁, x₂)) = (1/4)x_2
POL(s(x₁)) = 11/4
POL(0) = 15/4
POL(nil) = 15/4

The value of delta used in the strict ordering is 675/16384.
The following usable rules [17] were oriented:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

            ↳ UsableRulesProof

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

L¹(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L¹(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))
L¹(f(s(s(s(s(x1)))), f(y1, x3))) → L¹(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L¹(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L¹(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

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

L¹(f(s(s(y)), f(z, w))) → L¹(f(s(0), f(y, f(s(z), w))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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