Termination w.r.t. Q proof of /tmp/tmp8plmnf/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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

      ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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.
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:

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

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 2 + x₁
POL(B(x₁)) = x₁
POL(C(x₁)) = 2 + x₁
POL(a(x₁)) = 2 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 2 + x₁
POL(d(x₁)) = 2 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(b(x1)) → A(x1)
A(a(x1)) → A(b(a(x1)))
B(a(a(x1))) → A(b(c(x1)))
B(d(x1)) → A(b(x1))
C(b(x1)) → B(a(x1))
C(a(x1)) → A(c(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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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.
We have reversed the following QTRS:
The set of rules R is

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)))

The set Q is empty.
We have obtained the following QTRS:

a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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

a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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)))

The set Q is empty.
We have obtained the following QTRS:

a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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

a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))

Q is empty.