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:

c(b(a(X))) → a(a(b(b(c(c(X))))))
a(X) → e
b(X) → e
c(X) → e

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

c(b(a(X))) → a(a(b(b(c(c(X))))))
a(X) → e
b(X) → e
c(X) → e

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

c(b(a(X))) → a(a(b(b(c(c(X))))))
a(X) → e
b(X) → e
c(X) → e

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

c(b(a(X))) → a(a(b(b(c(c(X))))))
a(X) → e
b(X) → e
c(X) → e

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

c(b(a(X))) → a(a(b(b(c(c(X))))))
a(X) → e
b(X) → e
c(X) → e

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 5 less nodes.

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

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

C(b(a(X))) → C(X)

The TRS R consists of the following rules:

c(b(a(X))) → a(a(b(b(c(c(X))))))
a(X) → e
b(X) → e
c(X) → e

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


C(b(a(X))) → C(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
C(x1)  =  C(x1)
b(x1)  =  b(x1)
a(x1)  =  x1

Recursive Path Order [2].
Precedence:
b1 > C1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

c(b(a(X))) → a(a(b(b(c(c(X))))))
a(X) → e
b(X) → e
c(X) → e

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.