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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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 5 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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



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

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

A(a(f(b, a(x)))) → A(a(a(x)))

The TRS R consists of the following rules:

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

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