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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a(b(x1)) → d(x1)
b(a(x1)) → a(b(x1))
d(c(x1)) → f(a(b(b(c(x1)))))
d(f(x1)) → f(a(b(x1)))
a(f(x1)) → 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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

a(b(x1)) → d(x1)
b(a(x1)) → a(b(x1))
d(c(x1)) → f(a(b(b(c(x1)))))
d(f(x1)) → f(a(b(x1)))
a(f(x1)) → 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(f(x1)) → B(x1)
B(a(x1)) → B(x1)
The remaining pairs can at least be oriented weakly.

A(f(x1)) → A(x1)
B(a(x1)) → A(b(x1))
D(c(x1)) → A(b(b(c(x1))))
D(f(x1)) → A(b(x1))
A(b(x1)) → D(x1)
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = x_1   
POL(c(x1)) = 0   
POL(D(x1)) = 2 + (2)x_1   
POL(B(x1)) = (2)x_1   
POL(a(x1)) = 1 + x_1   
POL(A(x1)) = 2 + x_1   
POL(b(x1)) = (2)x_1   
POL(d(x1)) = 1 + (2)x_1   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

a(b(x1)) → d(x1)
b(a(x1)) → a(b(x1))
d(c(x1)) → f(a(b(b(c(x1)))))
d(f(x1)) → f(a(b(x1)))
a(f(x1)) → 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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP

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

A(f(x1)) → A(x1)
D(c(x1)) → A(b(b(c(x1))))
D(f(x1)) → A(b(x1))
A(b(x1)) → D(x1)

The TRS R consists of the following rules:

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

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