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:

a1(b1(x)) -> b1(a1(x))
a1(c1(x)) -> x

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

a1(b1(x)) -> b1(a1(x))
a1(c1(x)) -> x

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

a1(b1(x)) -> b1(a1(x))
a1(c1(x)) -> x

The set Q consists of the following terms:

a1(b1(x0))
a1(c1(x0))


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

A1(b1(x)) -> A1(x)

The TRS R consists of the following rules:

a1(b1(x)) -> b1(a1(x))
a1(c1(x)) -> x

The set Q consists of the following terms:

a1(b1(x0))
a1(c1(x0))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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

A1(b1(x)) -> A1(x)

The TRS R consists of the following rules:

a1(b1(x)) -> b1(a1(x))
a1(c1(x)) -> x

The set Q consists of the following terms:

a1(b1(x0))
a1(c1(x0))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

A1(b1(x)) -> A1(x)
Used argument filtering: A1(x1)  =  x1
b1(x1)  =  b1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ PisEmptyProof

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

a1(b1(x)) -> b1(a1(x))
a1(c1(x)) -> x

The set Q consists of the following terms:

a1(b1(x0))
a1(c1(x0))

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