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:

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

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

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

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(0)
f(s(0))
f(s(s(x0)))


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

F(s(s(x))) → F(f(s(x)))
F(s(s(x))) → F(s(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(0)
f(s(0))
f(s(s(x0)))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof

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

F(s(s(x))) → F(f(s(x)))
F(s(s(x))) → F(s(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(0)
f(s(0))
f(s(s(x0)))

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(s(x))) → F(s(x))
The remaining pairs can at least be oriented weakly.

F(s(s(x))) → F(f(s(x)))
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
s(x1)  =  s(x1)
f(x1)  =  f(x1)
0  =  0

Lexicographic Path Order [19].
Precedence:
F1 > [s1, f1, 0]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP

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

F(s(s(x))) → F(f(s(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(0)
f(s(0))
f(s(s(x0)))

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