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))


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(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
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


A1(b1(x)) -> A1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
A1(x1)  =  A1(x1)
b1(x1)  =  b1(x1)

Lexicographic Path Order [19].
Precedence:
b1 > A1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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.