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(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(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:

A(a(f(x, y))) → A(b(a(x)))
A(a(f(x, y))) → A(y)
A(a(f(x, y))) → A(b(a(b(a(y)))))
A(a(f(x, y))) → A(b(a(b(a(x)))))
F(a(x), a(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → A(b(a(y)))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(b(x), b(y)) → F(x, y)
A(a(f(x, y))) → A(x)

The TRS R consists of the following rules:

a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(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:

A(a(f(x, y))) → A(b(a(x)))
A(a(f(x, y))) → A(y)
A(a(f(x, y))) → A(b(a(b(a(y)))))
A(a(f(x, y))) → A(b(a(b(a(x)))))
F(a(x), a(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → A(b(a(y)))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(b(x), b(y)) → F(x, y)
A(a(f(x, y))) → A(x)

The TRS R consists of the following rules:

a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(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:

A(a(f(x, y))) → A(y)
F(a(x), a(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, y))) → A(x)
F(b(x), b(y)) → F(x, y)

The TRS R consists of the following rules:

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

F(a(x), a(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(b(x), b(y)) → F(x, y)
Used ordering: Polynomial interpretation [25,35]:

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

a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))



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

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

F(a(x), a(y)) → A(f(x, y))
F(a(x), a(y)) → F(x, y)
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(b(x), b(y)) → F(x, y)

The TRS R consists of the following rules:

a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))

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