Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

g(c(x1)) → g(f(c(x1)))
g(f(c(x1))) → g(f(f(c(x1))))
g(g(x1)) → g(f(g(x1)))
f(f(g(x1))) → g(f(x1))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g(c(x1)) → g(f(c(x1)))
g(f(c(x1))) → g(f(f(c(x1))))
g(g(x1)) → g(f(g(x1)))
f(f(g(x1))) → g(f(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:

G(g(x1)) → G(f(g(x1)))
G(c(x1)) → G(f(c(x1)))
G(g(x1)) → F(g(x1))
F(f(g(x1))) → F(x1)
G(f(c(x1))) → F(f(c(x1)))
F(f(g(x1))) → G(f(x1))
G(f(c(x1))) → G(f(f(c(x1))))
G(c(x1)) → F(c(x1))

The TRS R consists of the following rules:

g(c(x1)) → g(f(c(x1)))
g(f(c(x1))) → g(f(f(c(x1))))
g(g(x1)) → g(f(g(x1)))
f(f(g(x1))) → g(f(x1))

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:

G(g(x1)) → G(f(g(x1)))
G(c(x1)) → G(f(c(x1)))
G(g(x1)) → F(g(x1))
F(f(g(x1))) → F(x1)
G(f(c(x1))) → F(f(c(x1)))
F(f(g(x1))) → G(f(x1))
G(f(c(x1))) → G(f(f(c(x1))))
G(c(x1)) → F(c(x1))

The TRS R consists of the following rules:

g(c(x1)) → g(f(c(x1)))
g(f(c(x1))) → g(f(f(c(x1))))
g(g(x1)) → g(f(g(x1)))
f(f(g(x1))) → g(f(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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

G(g(x1)) → G(f(g(x1)))

The TRS R consists of the following rules:

g(c(x1)) → g(f(c(x1)))
g(f(c(x1))) → g(f(f(c(x1))))
g(g(x1)) → g(f(g(x1)))
f(f(g(x1))) → g(f(x1))

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

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

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

F(f(g(x1))) → F(x1)

The TRS R consists of the following rules:

g(c(x1)) → g(f(c(x1)))
g(f(c(x1))) → g(f(f(c(x1))))
g(g(x1)) → g(f(g(x1)))
f(f(g(x1))) → g(f(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.


F(f(g(x1))) → F(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = (1/4)x_1   
POL(g(x1)) = 1/4 + (4)x_1   
POL(F(x1)) = (1/2)x_1   
The value of delta used in the strict ordering is 1/32.
The following usable rules [17] were oriented: none



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

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

g(c(x1)) → g(f(c(x1)))
g(f(c(x1))) → g(f(f(c(x1))))
g(g(x1)) → g(f(g(x1)))
f(f(g(x1))) → g(f(x1))

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.