Termination w.r.t. Q proof of /tmp/tmpuDwxJN/aprove2.srs

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:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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:

A(a(a(x1))) → B(x1)
F(h(x1)) → F(s(h(x1)))
B(a(x1)) → A(b(x1))
F(s(s(s(x1)))) → F(s(h(x1)))
F(s(s(s(x1)))) → H(x1)
B(b(b(b(x1)))) → A(x1)
G(h(x1)) → G(f(s(x1)))
G(h(x1)) → F(s(x1))
F(s(s(s(x1)))) → H(f(s(h(x1))))
B(a(x1)) → B(x1)
F(h(x1)) → H(f(s(h(x1))))
A(a(a(x1))) → B(a(a(b(x1))))
F(f(s(s(x1)))) → F(f(x1))
A(a(a(x1))) → A(a(b(x1)))
A(a(a(x1))) → A(b(x1))
F(f(s(s(x1)))) → F(x1)

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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:

A(a(a(x1))) → B(x1)
F(h(x1)) → F(s(h(x1)))
B(a(x1)) → A(b(x1))
F(s(s(s(x1)))) → F(s(h(x1)))
F(s(s(s(x1)))) → H(x1)
B(b(b(b(x1)))) → A(x1)
G(h(x1)) → G(f(s(x1)))
G(h(x1)) → F(s(x1))
F(s(s(s(x1)))) → H(f(s(h(x1))))
B(a(x1)) → B(x1)
F(h(x1)) → H(f(s(h(x1))))
A(a(a(x1))) → B(a(a(b(x1))))
F(f(s(s(x1)))) → F(f(x1))
A(a(a(x1))) → A(a(b(x1)))
A(a(a(x1))) → A(b(x1))
F(f(s(s(x1)))) → F(x1)

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

A(a(a(x1))) → B(x1)
B(a(x1)) → A(b(x1))
A(a(a(x1))) → B(a(a(b(x1))))
A(a(a(x1))) → A(a(b(x1)))
B(b(b(b(x1)))) → A(x1)
A(a(a(x1))) → A(b(x1))
B(a(x1)) → B(x1)

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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.

A(a(a(x1))) → B(x1)
B(a(x1)) → A(b(x1))
A(a(a(x1))) → A(a(b(x1)))
B(b(b(b(x1)))) → A(x1)
A(a(a(x1))) → A(b(x1))
B(a(x1)) → B(x1)

The remaining pairs can at least be oriented weakly.

A(a(a(x1))) → B(a(a(b(x1))))

Used ordering: Polynomial interpretation [25,35]:

POL(B(x₁)) = (4)x_1
POL(a(x₁)) = 4 + x_1
POL(A(x₁)) = 4 + (4)x_1
POL(b(x₁)) = 1 + x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

b(a(x1)) → a(b(x1))
b(b(b(b(x1)))) → a(x1)
a(a(a(x1))) → b(a(a(b(x1))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

A(a(a(x1))) → B(a(a(b(x1))))

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

F(s(s(s(x1)))) → F(s(h(x1)))

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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(s(s(s(x1)))) → F(s(h(x1)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(h(x₁)) = 3 + (2)x_1
POL(s(x₁)) = 2 + (3)x_1
POL(F(x₁)) = (3)x_1

The value of delta used in the strict ordering is 45.
The following usable rules [17] were oriented:

h(x1) → x1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

F(f(s(s(x1)))) → F(f(x1))
F(f(s(s(x1)))) → F(x1)

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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(f(s(s(x1)))) → F(x1)

The remaining pairs can at least be oriented weakly.

F(f(s(s(x1)))) → F(f(x1))

Used ordering: Polynomial interpretation [25,35]:

POL(f(x₁)) = 4 + (4)x_1
POL(h(x₁)) = x_1
POL(s(x₁)) = x_1
POL(F(x₁)) = (1/2)x_1

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

f(h(x1)) → h(f(s(h(x1))))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(x1) → x1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

          ↳ QDP

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

F(f(s(s(x1)))) → F(f(x1))

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

G(h(x1)) → G(f(s(x1)))

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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