Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

g(f(x, y)) → f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y))))

The set Q consists of the following terms:

g(f(x0, x1))



↳ QTRS
  ↳ DependencyPairsProof

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

g(f(x, y)) → f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y))))

The set Q consists of the following terms:

g(f(x0, x1))


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(f(x, y)) → G(g(y))
G(f(x, y)) → G(g(x))
G(f(x, y)) → G(x)
G(f(x, y)) → G(y)

The TRS R consists of the following rules:

g(f(x, y)) → f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y))))

The set Q consists of the following terms:

g(f(x0, x1))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing

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

G(f(x, y)) → G(g(y))
G(f(x, y)) → G(g(x))
G(f(x, y)) → G(x)
G(f(x, y)) → G(y)

The TRS R consists of the following rules:

g(f(x, y)) → f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y))))

The set Q consists of the following terms:

g(f(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule G(f(x, y)) → G(g(x)) at position [0] we obtained the following new rules:

G(f(f(x0, x1), y1)) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing

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

G(f(f(x0, x1), y1)) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))
G(f(x, y)) → G(g(y))
G(f(x, y)) → G(x)
G(f(x, y)) → G(y)

The TRS R consists of the following rules:

g(f(x, y)) → f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y))))

The set Q consists of the following terms:

g(f(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule G(f(x, y)) → G(g(y)) at position [0] we obtained the following new rules:

G(f(y0, f(x0, x1))) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ ForwardInstantiation

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

G(f(f(x0, x1), y1)) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))
G(f(x, y)) → G(x)
G(f(y0, f(x0, x1))) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))
G(f(x, y)) → G(y)

The TRS R consists of the following rules:

g(f(x, y)) → f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y))))

The set Q consists of the following terms:

g(f(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule G(f(x, y)) → G(x) we obtained the following new rules:

G(f(f(y_0, y_1), x1)) → G(f(y_0, y_1))
G(f(f(y_0, f(y_1, y_2)), x1)) → G(f(y_0, f(y_1, y_2)))
G(f(f(f(y_0, y_1), y_2), x1)) → G(f(f(y_0, y_1), y_2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ ForwardInstantiation

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

G(f(f(x0, x1), y1)) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))
G(f(f(y_0, y_1), x1)) → G(f(y_0, y_1))
G(f(f(y_0, f(y_1, y_2)), x1)) → G(f(y_0, f(y_1, y_2)))
G(f(f(f(y_0, y_1), y_2), x1)) → G(f(f(y_0, y_1), y_2))
G(f(x, y)) → G(y)
G(f(y0, f(x0, x1))) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))

The TRS R consists of the following rules:

g(f(x, y)) → f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y))))

The set Q consists of the following terms:

g(f(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule G(f(x, y)) → G(y) we obtained the following new rules:

G(f(x0, f(y_0, y_1))) → G(f(y_0, y_1))
G(f(x0, f(f(f(y_0, y_1), y_2), y_3))) → G(f(f(f(y_0, y_1), y_2), y_3))
G(f(x0, f(f(y_0, y_1), y_2))) → G(f(f(y_0, y_1), y_2))
G(f(x0, f(f(y_0, f(y_1, y_2)), y_3))) → G(f(f(y_0, f(y_1, y_2)), y_3))
G(f(x0, f(y_0, f(y_1, y_2)))) → G(f(y_0, f(y_1, y_2)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ MNOCProof

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

G(f(f(x0, x1), y1)) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))
G(f(x0, f(y_0, y_1))) → G(f(y_0, y_1))
G(f(f(y_0, y_1), x1)) → G(f(y_0, y_1))
G(f(x0, f(f(f(y_0, y_1), y_2), y_3))) → G(f(f(f(y_0, y_1), y_2), y_3))
G(f(x0, f(f(y_0, y_1), y_2))) → G(f(f(y_0, y_1), y_2))
G(f(f(y_0, f(y_1, y_2)), x1)) → G(f(y_0, f(y_1, y_2)))
G(f(x0, f(y_0, f(y_1, y_2)))) → G(f(y_0, f(y_1, y_2)))
G(f(x0, f(f(y_0, f(y_1, y_2)), y_3))) → G(f(f(y_0, f(y_1, y_2)), y_3))
G(f(f(f(y_0, y_1), y_2), x1)) → G(f(f(y_0, y_1), y_2))
G(f(y0, f(x0, x1))) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))

The TRS R consists of the following rules:

g(f(x, y)) → f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y))))

The set Q consists of the following terms:

g(f(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ MNOCProof
                        ↳ QDP
                          ↳ NonTerminationProof

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

G(f(f(x0, x1), y1)) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))
G(f(x0, f(y_0, y_1))) → G(f(y_0, y_1))
G(f(f(y_0, y_1), x1)) → G(f(y_0, y_1))
G(f(x0, f(f(f(y_0, y_1), y_2), y_3))) → G(f(f(f(y_0, y_1), y_2), y_3))
G(f(x0, f(f(y_0, y_1), y_2))) → G(f(f(y_0, y_1), y_2))
G(f(f(y_0, f(y_1, y_2)), x1)) → G(f(y_0, f(y_1, y_2)))
G(f(f(f(y_0, y_1), y_2), x1)) → G(f(f(y_0, y_1), y_2))
G(f(x0, f(f(y_0, f(y_1, y_2)), y_3))) → G(f(f(y_0, f(y_1, y_2)), y_3))
G(f(x0, f(y_0, f(y_1, y_2)))) → G(f(y_0, f(y_1, y_2)))
G(f(y0, f(x0, x1))) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))

The TRS R consists of the following rules:

g(f(x, y)) → f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y))))

Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

G(f(f(x0, x1), y1)) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))
G(f(x0, f(y_0, y_1))) → G(f(y_0, y_1))
G(f(f(y_0, y_1), x1)) → G(f(y_0, y_1))
G(f(x0, f(f(f(y_0, y_1), y_2), y_3))) → G(f(f(f(y_0, y_1), y_2), y_3))
G(f(x0, f(f(y_0, y_1), y_2))) → G(f(f(y_0, y_1), y_2))
G(f(f(y_0, f(y_1, y_2)), x1)) → G(f(y_0, f(y_1, y_2)))
G(f(f(f(y_0, y_1), y_2), x1)) → G(f(f(y_0, y_1), y_2))
G(f(x0, f(f(y_0, f(y_1, y_2)), y_3))) → G(f(f(y_0, f(y_1, y_2)), y_3))
G(f(x0, f(y_0, f(y_1, y_2)))) → G(f(y_0, f(y_1, y_2)))
G(f(y0, f(x0, x1))) → G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))

The TRS R consists of the following rules:

g(f(x, y)) → f(f(g(g(x)), g(g(y))), f(g(g(x)), g(g(y))))


s = G(f(f(x0, x1), y1)) evaluates to t =G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1)))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from G(f(f(x0, x1), y1)) to G(f(f(g(g(x0)), g(g(x1))), f(g(g(x0)), g(g(x1))))).