Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a(s(x1)) → s(a(x1))
b(a(b(s(x1)))) → a(b(s(a(x1))))
b(a(b(b(x1)))) → c(s(x1))
c(s(x1)) → a(b(a(b(x1))))
a(b(a(a(x1)))) → b(a(b(a(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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

a(s(x1)) → s(a(x1))
b(a(b(s(x1)))) → a(b(s(a(x1))))
b(a(b(b(x1)))) → c(s(x1))
c(s(x1)) → a(b(a(b(x1))))
a(b(a(a(x1)))) → b(a(b(a(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 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

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

The TRS R consists of the following rules:

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

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