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 [1,13] 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(y)
F(a(x), a(y)) → F(x, y)
A(a(f(x, y))) → A(b(a(y)))
A(a(f(x, y))) → A(b(a(b(a(x)))))
A(a(f(x, y))) → A(x)
F(b(x), b(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → A(b(a(b(a(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(b(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
      ↳ EdgeDeletionProof

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)
A(a(f(x, y))) → A(b(a(y)))
A(a(f(x, y))) → A(b(a(b(a(x)))))
A(a(f(x, y))) → A(x)
F(b(x), b(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → A(b(a(b(a(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(b(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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

A(a(f(x, y))) → A(y)
A(a(f(x, y))) → A(b(a(b(a(x)))))
A(a(f(x, y))) → A(b(a(y)))
F(a(x), a(y)) → F(x, y)
A(a(f(x, y))) → A(x)
F(b(x), b(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(b(a(b(a(y)))))
A(a(f(x, y))) → A(b(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 [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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)
A(a(f(x, y))) → A(x)
F(b(x), b(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))))))

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 [13].


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(b(x), b(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))))))
Used ordering: Combined order from the following AFS and order.
A(x1)  =  x1
a(x1)  =  x1
f(x1, x2)  =  f(x1, x2)
F(x1, x2)  =  F(x1, x2)
b(x1)  =  x1

Recursive path order with status [2].
Quasi-Precedence:
[f2, F2]

Status:
f2: multiset
F2: multiset


The following usable rules [14] were oriented:

f(b(x), b(y)) → b(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))))))



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

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

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

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.