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:

f1(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(X)

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:

F1(c1(a)) -> F1(d1(b))
F1(c1(b)) -> F1(d1(a))
F1(a) -> F1(d1(a))
E1(g1(X)) -> E1(X)
F1(a) -> F1(c1(a))

The TRS R consists of the following rules:

f1(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(X)

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:

F1(c1(a)) -> F1(d1(b))
F1(c1(b)) -> F1(d1(a))
F1(a) -> F1(d1(a))
E1(g1(X)) -> E1(X)
F1(a) -> F1(c1(a))

The TRS R consists of the following rules:

f1(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(X)

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 4 less nodes.

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

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

E1(g1(X)) -> E1(X)

The TRS R consists of the following rules:

f1(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(X)

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.


E1(g1(X)) -> E1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(E1(x1)) = 2·x1   
POL(g1(x1)) = 2 + x1   

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:

f1(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(X)

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.