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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

The set Q consists of the following terms:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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


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

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



Rewriting sequence

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

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

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

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