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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B(b(x1)) → C(c(x1))
A(a(x1)) → B(b(x1))
C(c(c(c(x1)))) → B(x1)
C(c(c(c(x1)))) → A(b(x1))
B(b(x1)) → C(x1)
A(a(x1)) → B(x1)
The remaining pairs can at least be oriented weakly.

B(b(x1)) → C(c(c(x1)))
A(a(x1)) → B(b(b(x1)))
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = 5/4 + (1/2)x_1   
POL(c(x1)) = 5/4 + x_1   
POL(B(x1)) = 3/2 + (1/2)x_1   
POL(a(x1)) = 3 + x_1   
POL(A(x1)) = 2 + (1/2)x_1   
POL(b(x1)) = 2 + x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.