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(x, y))) -> f2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))
f2(a1(x), a1(y)) -> a1(f2(x, y))
f2(b1(x), b1(y)) -> b1(f2(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a1(a1(f2(x, y))) -> f2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))
f2(a1(x), a1(y)) -> a1(f2(x, y))
f2(b1(x), b1(y)) -> b1(f2(x, y))

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(x, y))) -> A1(y)
F2(a1(x), a1(y)) -> F2(x, y)
A1(a1(f2(x, y))) -> A1(b1(a1(y)))
A1(a1(f2(x, y))) -> A1(b1(a1(b1(a1(x)))))
A1(a1(f2(x, y))) -> A1(x)
F2(b1(x), b1(y)) -> F2(x, y)
F2(a1(x), a1(y)) -> A1(f2(x, y))
A1(a1(f2(x, y))) -> A1(b1(a1(b1(a1(y)))))
A1(a1(f2(x, y))) -> F2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))
A1(a1(f2(x, y))) -> A1(b1(a1(x)))

The TRS R consists of the following rules:

a1(a1(f2(x, y))) -> f2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))
f2(a1(x), a1(y)) -> a1(f2(x, y))
f2(b1(x), b1(y)) -> b1(f2(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:

A1(a1(f2(x, y))) -> A1(y)
F2(a1(x), a1(y)) -> F2(x, y)
A1(a1(f2(x, y))) -> A1(b1(a1(y)))
A1(a1(f2(x, y))) -> A1(b1(a1(b1(a1(x)))))
A1(a1(f2(x, y))) -> A1(x)
F2(b1(x), b1(y)) -> F2(x, y)
F2(a1(x), a1(y)) -> A1(f2(x, y))
A1(a1(f2(x, y))) -> A1(b1(a1(b1(a1(y)))))
A1(a1(f2(x, y))) -> F2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))
A1(a1(f2(x, y))) -> A1(b1(a1(x)))

The TRS R consists of the following rules:

a1(a1(f2(x, y))) -> f2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))
f2(a1(x), a1(y)) -> a1(f2(x, y))
f2(b1(x), b1(y)) -> b1(f2(x, y))

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 4 less nodes.

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

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

A1(a1(f2(x, y))) -> A1(y)
F2(a1(x), a1(y)) -> F2(x, y)
A1(a1(f2(x, y))) -> A1(x)
F2(b1(x), b1(y)) -> F2(x, y)
F2(a1(x), a1(y)) -> A1(f2(x, y))
A1(a1(f2(x, y))) -> F2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))

The TRS R consists of the following rules:

a1(a1(f2(x, y))) -> f2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))
f2(a1(x), a1(y)) -> a1(f2(x, y))
f2(b1(x), b1(y)) -> b1(f2(x, y))

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

F2(a1(x), a1(y)) -> F2(x, y)
F2(b1(x), b1(y)) -> F2(x, y)
F2(a1(x), a1(y)) -> A1(f2(x, y))
A1(a1(f2(x, y))) -> F2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))
Used ordering: Polynomial interpretation [21]:

POL(A1(x1)) = x1   
POL(F2(x1, x2)) = 1 + x1 + x2   
POL(a1(x1)) = x1   
POL(b1(x1)) = x1   
POL(f2(x1, x2)) = 1 + x1 + x2   

The following usable rules [14] were oriented:

f2(a1(x), a1(y)) -> a1(f2(x, y))
a1(a1(f2(x, y))) -> f2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))
f2(b1(x), b1(y)) -> b1(f2(x, y))



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

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

F2(a1(x), a1(y)) -> F2(x, y)
F2(b1(x), b1(y)) -> F2(x, y)
F2(a1(x), a1(y)) -> A1(f2(x, y))
A1(a1(f2(x, y))) -> F2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))

The TRS R consists of the following rules:

a1(a1(f2(x, y))) -> f2(a1(b1(a1(b1(a1(x))))), a1(b1(a1(b1(a1(y))))))
f2(a1(x), a1(y)) -> a1(f2(x, y))
f2(b1(x), b1(y)) -> b1(f2(x, y))

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