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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

A(b(x1)) → A(c(b(x1)))
A(b(x1)) → A(a(c(b(x1))))
B(x1) → A(x1)
A(b(x1)) → B(a(a(c(b(x1)))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → A(a(c(b(x1)))) at position [0] we obtained the following new rules:

A(b(y0)) → A(c(b(y0)))
A(b(x0)) → A(a(c(a(x0))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

A(b(x1)) → A(c(b(x1)))
A(b(x0)) → A(a(c(a(x0))))
B(x1) → A(x1)
A(b(x1)) → B(a(a(c(b(x1)))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → A(c(b(x1))) at position [0] we obtained the following new rules:

A(b(x0)) → A(c(a(x0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
          ↳ NonTerminationProof
  ↳ QTRS Reverse

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

A(b(x0)) → A(c(a(x0)))
B(x1) → A(x1)
A(b(x0)) → A(a(c(a(x0))))
A(b(x1)) → B(a(a(c(b(x1)))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (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:

A(b(x1)) → A(c(b(x1)))
A(b(x1)) → A(a(c(b(x1))))
B(x1) → A(x1)
A(b(x1)) → B(a(a(c(b(x1)))))

The TRS R consists of the following rules:

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


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

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



Rewriting sequence

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

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

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

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

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

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

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




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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS

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

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

Q is empty.