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:

f1(0) -> cons2(0, f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

f1(0) -> cons2(0, f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

f1(0) -> cons2(0, f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0

The set Q consists of the following terms:

f1(0)
f1(s1(0))
p1(s1(0))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F1(s1(0)) -> P1(s1(0))
F1(s1(0)) -> F1(p1(s1(0)))
F1(0) -> F1(s1(0))

The TRS R consists of the following rules:

f1(0) -> cons2(0, f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0

The set Q consists of the following terms:

f1(0)
f1(s1(0))
p1(s1(0))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

F1(s1(0)) -> P1(s1(0))
F1(s1(0)) -> F1(p1(s1(0)))
F1(0) -> F1(s1(0))

The TRS R consists of the following rules:

f1(0) -> cons2(0, f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0

The set Q consists of the following terms:

f1(0)
f1(s1(0))
p1(s1(0))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP

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

F1(s1(0)) -> F1(p1(s1(0)))
F1(0) -> F1(s1(0))

The TRS R consists of the following rules:

f1(0) -> cons2(0, f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0

The set Q consists of the following terms:

f1(0)
f1(s1(0))
p1(s1(0))

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