Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(4) Obligation:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x1, x2)
a(x1)  =  x1
A(x1)  =  x1
f(x1, x2)  =  f(x1, x2)
b(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

[F₂, f₂]

The following usable rules [FROCOS05] 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))

(6) Obligation:

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

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(a(x), a(y)) → F(x, 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.