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:

f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(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:

F(c(a)) → F(d(b))
F(c(b)) → F(d(a))
F(a) → F(d(a))
E(g(X)) → E(X)
F(a) → F(c(a))

The TRS R consists of the following rules:

f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(X)

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:

F(c(a)) → F(d(b))
F(c(b)) → F(d(a))
F(a) → F(d(a))
E(g(X)) → E(X)
F(a) → F(c(a))

The TRS R consists of the following rules:

f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(X)

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:

F(c(a)) → F(d(b))
F(c(b)) → F(d(a))
F(a) → F(d(a))
F(a) → F(c(a))
E(g(X)) → E(X)

The TRS R consists of the following rules:

f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

E(g(X)) → E(X)

The TRS R consists of the following rules:

f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(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.


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

Recursive Path Order [2].
Precedence:
trivial

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:

f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(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.