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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a(c(a(x1))) → c(a(c(x1)))
a(a(b(x1))) → a(d(b(x1)))
a(b(x1)) → b(a(a(x1)))
d(d(x1)) → a(d(b(x1)))
b(b(x1)) → b(c(x1))
a(d(c(x1))) → c(a(x1))
b(c(x1)) → a(a(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 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

A(c(a(x1))) → A(c(x1))

The TRS R consists of the following rules:

a(c(a(x1))) → c(a(c(x1)))
a(a(b(x1))) → a(d(b(x1)))
a(b(x1)) → b(a(a(x1)))
d(d(x1)) → a(d(b(x1)))
b(b(x1)) → b(c(x1))
a(d(c(x1))) → c(a(x1))
b(c(x1)) → a(a(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(c(a(x1))) → A(c(x1))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(c(x1)) = (2)x_1   
POL(a(x1)) = 1/4 + (11/4)x_1   
POL(A(x1)) = (2)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

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

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

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

The TRS R consists of the following rules:

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

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