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:

a1(a1(f2(b, a1(x)))) -> f2(a1(a1(a1(x))), b)
a1(a1(x)) -> f2(b, a1(f2(a1(x), b)))
f2(a1(x), b) -> f2(b, a1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a1(a1(f2(b, a1(x)))) -> f2(a1(a1(a1(x))), b)
a1(a1(x)) -> f2(b, a1(f2(a1(x), b)))
f2(a1(x), b) -> f2(b, a1(x))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A1(a1(f2(b, a1(x)))) -> A1(a1(x))
A1(a1(x)) -> F2(a1(x), b)
A1(a1(x)) -> A1(f2(a1(x), b))
A1(a1(x)) -> F2(b, a1(f2(a1(x), b)))
A1(a1(f2(b, a1(x)))) -> A1(a1(a1(x)))
A1(a1(f2(b, a1(x)))) -> F2(a1(a1(a1(x))), b)
F2(a1(x), b) -> F2(b, a1(x))

The TRS R consists of the following rules:

a1(a1(f2(b, a1(x)))) -> f2(a1(a1(a1(x))), b)
a1(a1(x)) -> f2(b, a1(f2(a1(x), b)))
f2(a1(x), b) -> f2(b, a1(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:

A1(a1(f2(b, a1(x)))) -> A1(a1(x))
A1(a1(x)) -> F2(a1(x), b)
A1(a1(x)) -> A1(f2(a1(x), b))
A1(a1(x)) -> F2(b, a1(f2(a1(x), b)))
A1(a1(f2(b, a1(x)))) -> A1(a1(a1(x)))
A1(a1(f2(b, a1(x)))) -> F2(a1(a1(a1(x))), b)
F2(a1(x), b) -> F2(b, a1(x))

The TRS R consists of the following rules:

a1(a1(f2(b, a1(x)))) -> f2(a1(a1(a1(x))), b)
a1(a1(x)) -> f2(b, a1(f2(a1(x), b)))
f2(a1(x), b) -> f2(b, a1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 5 less nodes.

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

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

A1(a1(f2(b, a1(x)))) -> A1(a1(x))
A1(a1(f2(b, a1(x)))) -> A1(a1(a1(x)))

The TRS R consists of the following rules:

a1(a1(f2(b, a1(x)))) -> f2(a1(a1(a1(x))), b)
a1(a1(x)) -> f2(b, a1(f2(a1(x), b)))
f2(a1(x), b) -> f2(b, a1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


A1(a1(f2(b, a1(x)))) -> A1(a1(x))
The remaining pairs can at least be oriented weakly.

A1(a1(f2(b, a1(x)))) -> A1(a1(a1(x)))
Used ordering: Polynomial interpretation [21]:

POL(A1(x1)) = 2·x1   
POL(a1(x1)) = 2 + x1   
POL(b) = 0   
POL(f2(x1, x2)) = x1 + x2   

The following usable rules [14] were oriented:

a1(a1(x)) -> f2(b, a1(f2(a1(x), b)))
a1(a1(f2(b, a1(x)))) -> f2(a1(a1(a1(x))), b)
f2(a1(x), b) -> f2(b, a1(x))



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

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

A1(a1(f2(b, a1(x)))) -> A1(a1(a1(x)))

The TRS R consists of the following rules:

a1(a1(f2(b, a1(x)))) -> f2(a1(a1(a1(x))), b)
a1(a1(x)) -> f2(b, a1(f2(a1(x), b)))
f2(a1(x), b) -> f2(b, a1(x))

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