Termination w.r.t. Q proof of /tmp/tmpOb-V0W/beans6.srs

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:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(a(x1)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(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:

A(d(x1)) → A(x1)
C(a(x1)) → C(x1)
A(a(b(x1))) → A(x1)
A(a(x1)) → A(b(a(x1)))
B(d(x1)) → A(b(x1))
A(a(x1)) → B(a(x1))
C(b(x1)) → A(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
B(a(a(x1))) → A(b(c(x1)))
B(d(x1)) → B(x1)
C(b(x1)) → B(a(x1))
C(a(x1)) → A(c(x1))
A(a(b(x1))) → B(a(x1))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(a(x1)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

A(d(x1)) → A(x1)
C(a(x1)) → C(x1)
A(a(b(x1))) → A(x1)
A(a(x1)) → A(b(a(x1)))
B(d(x1)) → A(b(x1))
A(a(x1)) → B(a(x1))
C(b(x1)) → A(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
B(a(a(x1))) → A(b(c(x1)))
B(d(x1)) → B(x1)
C(b(x1)) → B(a(x1))
C(a(x1)) → A(c(x1))
A(a(b(x1))) → B(a(x1))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(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.

A(d(x1)) → A(x1)
C(a(x1)) → C(x1)
A(a(b(x1))) → A(x1)
B(d(x1)) → A(b(x1))
A(a(x1)) → B(a(x1))
C(b(x1)) → A(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
B(a(a(x1))) → A(b(c(x1)))
B(d(x1)) → B(x1)
C(a(x1)) → A(c(x1))
A(a(b(x1))) → B(a(x1))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = 1/2 + x_1
POL(c(x₁)) = 1 + (2)x_1
POL(B(x₁)) = (1/2)x_1
POL(a(x₁)) = 1 + (2)x_1
POL(A(x₁)) = 1/4 + x_1
POL(d(x₁)) = 1 + (2)x_1
POL(b(x₁)) = x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

b(a(a(x1))) → a(b(c(x1)))
c(b(x1)) → b(a(x1))
c(a(x1)) → a(c(x1))
a(d(x1)) → d(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(a(x1)) → a(b(a(x1)))
b(d(x1)) → a(b(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(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 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

A(a(x1)) → A(b(a(x1)))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(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.

A(a(x1)) → A(b(a(x1)))

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

POL(c(x₁)) = 3/4 + (13/4)x_1
POL(a(x₁)) = 1 + (13/4)x_1
POL(A(x₁)) = (5/4)x_1
POL(b(x₁)) = (3/4)x_1
POL(d(x₁)) = 3/2 + (13/4)x_1

The value of delta used in the strict ordering is 5/16.
The following usable rules [17] were oriented:

c(b(x1)) → b(a(x1))
c(a(x1)) → a(c(x1))
a(d(x1)) → d(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(a(x1)) → a(b(a(x1)))
b(d(x1)) → a(b(x1))
b(a(a(x1))) → a(b(c(x1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(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.