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

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

c(a(a(b(a(x1))))) → a(a(b(a(c(c(a(a(b(x1)))))))))

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence

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

c(a(a(b(a(x1))))) → a(a(b(a(c(c(a(a(b(x1)))))))))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

c(a(a(b(a(x1))))) → a(a(b(a(c(c(a(a(b(x1)))))))))

The set Q is empty.
We have obtained the following QTRS:

a(b(a(a(c(x))))) → b(a(a(c(c(a(b(a(a(x)))))))))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence

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

a(b(a(a(c(x))))) → b(a(a(c(c(a(b(a(a(x)))))))))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

c(a(a(b(a(x1))))) → a(a(b(a(c(c(a(a(b(x1)))))))))

The set Q is empty.
We have obtained the following QTRS:

a(b(a(a(c(x))))) → b(a(a(c(c(a(b(a(a(x)))))))))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ Overlay + Local Confluence

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

a(b(a(a(c(x))))) → b(a(a(c(c(a(b(a(a(x)))))))))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

c(a(a(b(a(x1))))) → a(a(b(a(c(c(a(a(b(x1)))))))))

The set Q consists of the following terms:

c(a(a(b(a(x0)))))


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

C(a(a(b(a(x1))))) → C(a(a(b(x1))))
C(a(a(b(a(x1))))) → C(c(a(a(b(x1)))))

The TRS R consists of the following rules:

c(a(a(b(a(x1))))) → a(a(b(a(c(c(a(a(b(x1)))))))))

The set Q consists of the following terms:

c(a(a(b(a(x0)))))

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof

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

C(a(a(b(a(x1))))) → C(a(a(b(x1))))
C(a(a(b(a(x1))))) → C(c(a(a(b(x1)))))

The TRS R consists of the following rules:

c(a(a(b(a(x1))))) → a(a(b(a(c(c(a(a(b(x1)))))))))

The set Q consists of the following terms:

c(a(a(b(a(x0)))))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
            ↳ QDP
              ↳ NonTerminationProof

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

C(a(a(b(a(x1))))) → C(a(a(b(x1))))
C(a(a(b(a(x1))))) → C(c(a(a(b(x1)))))

The TRS R consists of the following rules:

c(a(a(b(a(x1))))) → a(a(b(a(c(c(a(a(b(x1)))))))))

Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

C(a(a(b(a(x1))))) → C(a(a(b(x1))))
C(a(a(b(a(x1))))) → C(c(a(a(b(x1)))))

The TRS R consists of the following rules:

c(a(a(b(a(x1))))) → a(a(b(a(c(c(a(a(b(x1)))))))))


s = C(c(a(a(b(c(c(a(a(b(a(x1'))))))))))) evaluates to t =C(c(a(a(b(c(c(a(a(b(a(b(a(c(c(a(a(b(c(c(a(a(b(x1')))))))))))))))))))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

C(c(a(a(b(c(c(a(a(b(a(x1'))))))))))) → C(c(a(a(b(c(a(a(b(a(c(c(a(a(b(x1')))))))))))))))
with rule c(a(a(b(a(x1''))))) → a(a(b(a(c(c(a(a(b(x1''))))))))) at position [0,0,0,0,0,0] and matcher [x1'' / x1']

C(c(a(a(b(c(a(a(b(a(c(c(a(a(b(x1'))))))))))))))) → C(c(a(a(b(a(a(b(a(c(c(a(a(b(c(c(a(a(b(x1')))))))))))))))))))
with rule c(a(a(b(a(x1))))) → a(a(b(a(c(c(a(a(b(x1))))))))) at position [0,0,0,0,0] and matcher [x1 / c(c(a(a(b(x1')))))]

C(c(a(a(b(a(a(b(a(c(c(a(a(b(c(c(a(a(b(x1'))))))))))))))))))) → C(a(a(b(a(c(c(a(a(b(a(b(a(c(c(a(a(b(c(c(a(a(b(x1')))))))))))))))))))))))
with rule c(a(a(b(a(x1''))))) → a(a(b(a(c(c(a(a(b(x1''))))))))) at position [0] and matcher [x1'' / a(b(a(c(c(a(a(b(c(c(a(a(b(x1')))))))))))))]

C(a(a(b(a(c(c(a(a(b(a(b(a(c(c(a(a(b(c(c(a(a(b(x1'))))))))))))))))))))))) → C(c(a(a(b(c(c(a(a(b(a(b(a(c(c(a(a(b(c(c(a(a(b(x1')))))))))))))))))))))))
with rule C(a(a(b(a(x1))))) → C(c(a(a(b(x1)))))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.