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:

c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(X) -> e

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(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:

C1(b1(a1(X))) -> A1(b1(b1(c1(c1(X)))))
C1(b1(a1(X))) -> C1(X)
C1(b1(a1(X))) -> B1(b1(c1(c1(X))))
C1(b1(a1(X))) -> A1(a1(b1(b1(c1(c1(X))))))
C1(b1(a1(X))) -> B1(c1(c1(X)))
C1(b1(a1(X))) -> C1(c1(X))

The TRS R consists of the following rules:

c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(X) -> e

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:

C1(b1(a1(X))) -> A1(b1(b1(c1(c1(X)))))
C1(b1(a1(X))) -> C1(X)
C1(b1(a1(X))) -> B1(b1(c1(c1(X))))
C1(b1(a1(X))) -> A1(a1(b1(b1(c1(c1(X))))))
C1(b1(a1(X))) -> B1(c1(c1(X)))
C1(b1(a1(X))) -> C1(c1(X))

The TRS R consists of the following rules:

c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

C1(b1(a1(X))) -> C1(X)

The TRS R consists of the following rules:

c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(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.


C1(b1(a1(X))) -> C1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( b1(x1) ) = x1 + 1


POL( C1(x1) ) = max{0, x1 - 1}


POL( a1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(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.