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
  ↳ AAECC Innermost

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.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

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

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
    ↳ 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)

The set Q consists of the following terms:

f(s(x0), x1)
f(x0, s(s(x1)))


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)

The set Q consists of the following terms:

f(s(x0), x1)
f(x0, s(s(x1)))

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof

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)

The set Q consists of the following terms:

f(s(x0), x1)
f(x0, s(s(x1)))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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)))

R is empty.
The set Q consists of the following terms:

f(s(x0), x1)
f(x0, s(s(x1)))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(s(x0), x1)
f(x0, s(s(x1)))



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ 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)))

R is empty.
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)) = 1/4 + (4)x_1   
POL(F(x1, x2)) = (3/2)x_1 + (1/4)x_2   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
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.