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:

f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(0)))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(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:

G1(s1(0)) -> F1(s1(0))
G1(s1(0)) -> G1(f1(s1(0)))
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(0)))

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:

G1(s1(0)) -> F1(s1(0))
G1(s1(0)) -> G1(f1(s1(0)))
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(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
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(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.


F1(s1(x)) -> F1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F1(x1) ) = x1


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(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.