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

Q is empty.


↳ 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)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(s(s(x))) → F(s(x))
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)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

F(s(s(x))) → F(s(x))
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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


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: Polynomial interpretation [25,35]:

POL(f(x1)) = 2   
POL(s(x1)) = 1 + x_1   
POL(0) = 1   
POL(F(x1)) = x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof

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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(s(s(x))) → F(f(s(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = 4 + (2)x_1   
POL(s(x1)) = 4 + (3)x_1   
POL(0) = 0   
POL(F(x1)) = (2)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

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



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

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.