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(s(x), y) → f(x, s(s(x)))
f(x, s(s(y))) → f(y, x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(s(x), y) → f(x, s(s(x)))
f(x, s(s(y))) → f(y, 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(x, s(s(y))) → F(y, x)
F(s(x), y) → F(x, s(s(x)))

The TRS R consists of the following rules:

f(s(x), y) → f(x, s(s(x)))
f(x, s(s(y))) → f(y, 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(x, s(s(y))) → F(y, x)
F(s(x), y) → F(x, s(s(x)))

The TRS R consists of the following rules:

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

POL(s(x1)) = 1 + (2)x_1   
POL(F(x1, x2)) = (2)x_1 + (1/2)x_2   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



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

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

f(s(x), y) → f(x, s(s(x)))
f(x, s(s(y))) → f(y, 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.