Termination w.r.t. Q proof of /tmp/tmpR5dHF2/jw3.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:

a(b(b(x1))) → P(a(b(x1)))
a(P(x1)) → P(a(x(x1)))
a(x(x1)) → x(a(x1))
b(P(x1)) → b(Q(x1))
Q(x(x1)) → a(Q(x1))
Q(a(x1)) → b(b(a(x1)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a(b(b(x1))) → P(a(b(x1)))
a(P(x1)) → P(a(x(x1)))
a(x(x1)) → x(a(x1))
b(P(x1)) → b(Q(x1))
Q(x(x1)) → a(Q(x1))
Q(a(x1)) → b(b(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:

Q¹(x(x1)) → A(Q(x1))
B(P(x1)) → B(Q(x1))
Q¹(x(x1)) → Q¹(x1)
A(x(x1)) → A(x1)
A(P(x1)) → A(x(x1))
Q¹(a(x1)) → B(a(x1))
B(P(x1)) → Q¹(x1)
Q¹(a(x1)) → B(b(a(x1)))
A(b(b(x1))) → A(b(x1))

The TRS R consists of the following rules:

a(b(b(x1))) → P(a(b(x1)))
a(P(x1)) → P(a(x(x1)))
a(x(x1)) → x(a(x1))
b(P(x1)) → b(Q(x1))
Q(x(x1)) → a(Q(x1))
Q(a(x1)) → b(b(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:

Q¹(x(x1)) → A(Q(x1))
B(P(x1)) → B(Q(x1))
Q¹(x(x1)) → Q¹(x1)
A(x(x1)) → A(x1)
A(P(x1)) → A(x(x1))
Q¹(a(x1)) → B(a(x1))
B(P(x1)) → Q¹(x1)
Q¹(a(x1)) → B(b(a(x1)))
A(b(b(x1))) → A(b(x1))

The TRS R consists of the following rules:

a(b(b(x1))) → P(a(b(x1)))
a(P(x1)) → P(a(x(x1)))
a(x(x1)) → x(a(x1))
b(P(x1)) → b(Q(x1))
Q(x(x1)) → a(Q(x1))
Q(a(x1)) → b(b(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 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

A(x(x1)) → A(x1)
A(P(x1)) → A(x(x1))
A(b(b(x1))) → A(b(x1))

The TRS R consists of the following rules:

a(b(b(x1))) → P(a(b(x1)))
a(P(x1)) → P(a(x(x1)))
a(x(x1)) → x(a(x1))
b(P(x1)) → b(Q(x1))
Q(x(x1)) → a(Q(x1))
Q(a(x1)) → b(b(a(x1)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

B(P(x1)) → B(Q(x1))
Q¹(x(x1)) → Q¹(x1)
Q¹(a(x1)) → B(a(x1))
B(P(x1)) → Q¹(x1)
Q¹(a(x1)) → B(b(a(x1)))

The TRS R consists of the following rules:

a(b(b(x1))) → P(a(b(x1)))
a(P(x1)) → P(a(x(x1)))
a(x(x1)) → x(a(x1))
b(P(x1)) → b(Q(x1))
Q(x(x1)) → a(Q(x1))
Q(a(x1)) → b(b(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.

Q¹(a(x1)) → B(b(a(x1)))

The remaining pairs can at least be oriented weakly.

B(P(x1)) → B(Q(x1))
Q¹(x(x1)) → Q¹(x1)
Q¹(a(x1)) → B(a(x1))
B(P(x1)) → Q¹(x1)

Used ordering: Polynomial interpretation [25,35]:

POL(Q(x₁)) = 1
POL(P(x₁)) = 1
POL(x(x₁)) = 0
POL(B(x₁)) = (2)x_1
POL(a(x₁)) = 1
POL(b(x₁)) = 0
POL(Q¹(x₁)) = 2

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

a(b(b(x1))) → P(a(b(x1)))
a(P(x1)) → P(a(x(x1)))
a(x(x1)) → x(a(x1))
b(P(x1)) → b(Q(x1))
Q(x(x1)) → a(Q(x1))
Q(a(x1)) → b(b(a(x1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

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

B(P(x1)) → B(Q(x1))
Q¹(x(x1)) → Q¹(x1)
B(P(x1)) → Q¹(x1)
Q¹(a(x1)) → B(a(x1))

The TRS R consists of the following rules:

a(b(b(x1))) → P(a(b(x1)))
a(P(x1)) → P(a(x(x1)))
a(x(x1)) → x(a(x1))
b(P(x1)) → b(Q(x1))
Q(x(x1)) → a(Q(x1))
Q(a(x1)) → b(b(a(x1)))

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