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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a(b(a(x1))) → a(a(b(b(a(a(x1))))))
b(a(a(b(x1)))) → b(a(b(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 2 less nodes.

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

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

B(a(a(b(x1)))) → B(a(b(x1)))

The TRS R consists of the following rules:

a(b(a(x1))) → a(a(b(b(a(a(x1))))))
b(a(a(b(x1)))) → b(a(b(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:

A(b(a(x1))) → A(a(b(b(a(a(x1))))))
A(b(a(x1))) → A(b(b(a(a(x1)))))
A(b(a(x1))) → A(a(x1))

The TRS R consists of the following rules:

a(b(a(x1))) → a(a(b(b(a(a(x1))))))
b(a(a(b(x1)))) → b(a(b(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(b(a(x1))) → A(a(b(b(a(a(x1))))))
A(b(a(x1))) → A(a(x1))
The remaining pairs can at least be oriented weakly.

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

POL(a(x1)) = 0   
POL(A(x1)) = (1/2)x_1   
POL(b(x1)) = 1/2   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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



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

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

A(b(a(x1))) → A(b(b(a(a(x1)))))

The TRS R consists of the following rules:

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

POL(a(x1)) = 2   
POL(A(x1)) = (4)x_1   
POL(b(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(a(x1))) → a(a(b(b(a(a(x1))))))
b(a(a(b(x1)))) → b(a(b(x1)))



↳ 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(b(a(x1))) → a(a(b(b(a(a(x1))))))
b(a(a(b(x1)))) → b(a(b(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.