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(f(x)) → g(f(x))
g(g(x)) → f(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(f(x)) → g(f(x))
g(g(x)) → f(x)

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(x)) → F(x)
F(f(x)) → G(f(x))

The TRS R consists of the following rules:

f(f(x)) → g(f(x))
g(g(x)) → f(x)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

G(g(x)) → F(x)
F(f(x)) → G(f(x))

The TRS R consists of the following rules:

f(f(x)) → g(f(x))
g(g(x)) → f(x)

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.


G(g(x)) → F(x)
The remaining pairs can at least be oriented weakly.

F(f(x)) → G(f(x))
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = 1/2 + (4)x_1   
POL(g(x1)) = 3/4 + (3)x_1   
POL(G(x1)) = 2 + (1/4)x_1   
POL(F(x1)) = 7/4 + (3/4)x_1   
The value of delta used in the strict ordering is 7/16.
The following usable rules [17] were oriented:

g(g(x)) → f(x)
f(f(x)) → g(f(x))



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

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

F(f(x)) → G(f(x))

The TRS R consists of the following rules:

f(f(x)) → g(f(x))
g(g(x)) → f(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.