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(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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:

B(0, a(1, a(x, y))) → B(1, a(0, a(x, y)))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
A(0, a(x, y)) → A(1, a(1, a(x, y)))
A(0, b(0, x)) → A(0, x)
A(0, a(1, a(x, y))) → A(1, a(0, a(x, y)))
A(0, a(x, y)) → A(1, a(x, y))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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:

B(0, a(1, a(x, y))) → B(1, a(0, a(x, y)))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
A(0, a(x, y)) → A(1, a(1, a(x, y)))
A(0, b(0, x)) → A(0, x)
A(0, a(1, a(x, y))) → A(1, a(0, a(x, y)))
A(0, a(x, y)) → A(1, a(x, y))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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 4 less nodes.

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

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
A(0, b(0, x)) → A(0, x)

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, b(0, x)) → A(0, x)
The remaining pairs can at least be oriented weakly.

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
Used ordering: Polynomial interpretation [25,35]:

POL(b(x1, x2)) = x_1 + x_2   
POL(a(x1, x2)) = 1/2 + x_1 + x_2   
POL(B(x1, x2)) = (1/4)x_2   
POL(A(x1, x2)) = (1/4)x_1 + (1/4)x_2   
POL(0) = 1/2   
POL(1) = 0   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))



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

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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(0, x) → B(0, b(0, x))
The remaining pairs can at least be oriented weakly.

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, b(0, x)) → B(0, a(0, x))
Used ordering: Polynomial interpretation [25,35]:

POL(b(x1, x2)) = 0   
POL(a(x1, x2)) = 1   
POL(B(x1, x2)) = (4)x_2   
POL(A(x1, x2)) = 4   
POL(0) = 0   
POL(1) = 0   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))



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

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, b(0, x)) → B(0, a(0, x))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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