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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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:

G(s(0)) → F(s(0))
G(s(0)) → G(f(s(0)))
F(s(x)) → F(x)

The TRS R consists of the following rules:

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

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:

G(s(0)) → F(s(0))
G(s(0)) → G(f(s(0)))
F(s(x)) → F(x)

The TRS R consists of the following rules:

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

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:

G(s(0)) → F(s(0))
F(s(x)) → F(x)
G(s(0)) → G(f(s(0)))

The TRS R consists of the following rules:

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

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 2 less nodes.

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

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

F(s(x)) → F(x)

The TRS R consists of the following rules:

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

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.


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

Recursive path order with status [2].
Quasi-Precedence:
trivial

Status:
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(f(x)) → f(x)
f(s(x)) → f(x)
g(s(0)) → g(f(s(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.