Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

a(b(a(x))) → b(a(b(x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(b(a(x))) → b(a(b(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:

A(b(a(x))) → A(b(x))

The TRS R consists of the following rules:

a(b(a(x))) → b(a(b(x)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

A(b(a(x))) → A(b(x))

The TRS R consists of the following rules:

a(b(a(x))) → b(a(b(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.


A(b(a(x))) → A(b(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A(x1)  =  A(x1)
b(x1)  =  b(x1)
a(x1)  =  a(x1)

Recursive path order with status [2].
Quasi-Precedence:
[A1, b1, a1]

Status:
a1: multiset
b1: multiset
A1: multiset


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a(b(a(x))) → b(a(b(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.