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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

a(c(x1)) → a(x1)
d(a(x1)) → a(c(b(c(d(x1)))))
a(c(b(c(x1)))) → c(b(c(c(x1))))
c(x1) → b(a(a(x1)))
d(c(x1)) → a(c(d(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 6 less nodes.

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

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

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

The TRS R consists of the following rules:

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

D(a(x1)) → D(x1)
D(c(x1)) → D(a(x1))

The TRS R consists of the following rules:

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


D(c(x1)) → D(a(x1))
The remaining pairs can at least be oriented weakly.

D(a(x1)) → D(x1)
Used ordering: Polynomial interpretation [25,35]:

POL(c(x1)) = 3 + (5/4)x_1   
POL(D(x1)) = x_1   
POL(a(x1)) = (5/4)x_1   
POL(b(x1)) = 1/4   
The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented:

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



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

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

D(a(x1)) → D(x1)

The TRS R consists of the following rules:

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


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

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



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

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

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