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:

g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0

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:

F(s(x)) → G(x)
G(s(x)) → F(x)

The TRS R consists of the following rules:

g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0

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:

F(s(x)) → G(x)
G(s(x)) → F(x)

The TRS R consists of the following rules:

g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0

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

POL(s(x1)) = 1/4 + x_1   
POL(G(x1)) = (1/2)x_1   
POL(F(x1)) = (1/2)x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented: none



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

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

g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0

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.