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:

f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 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:

F3(g1(x), s1(0), y) -> G1(s1(0))
F3(g1(x), s1(0), y) -> F3(g1(s1(0)), y, g1(x))
G1(s1(x)) -> G1(x)

The TRS R consists of the following rules:

f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 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:

F3(g1(x), s1(0), y) -> G1(s1(0))
F3(g1(x), s1(0), y) -> F3(g1(s1(0)), y, g1(x))
G1(s1(x)) -> G1(x)

The TRS R consists of the following rules:

f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

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

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

G1(s1(x)) -> G1(x)

The TRS R consists of the following rules:

f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 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.


G1(s1(x)) -> G1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( G1(x1) ) = x1



The following usable rules [14] were oriented: none



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

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

f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(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.

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

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

F3(g1(x), s1(0), y) -> F3(g1(s1(0)), y, g1(x))

The TRS R consists of the following rules:

f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0

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