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:

f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)

Q is empty.

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

F2(s1(x), y) -> F2(x, s1(x))
F2(x, s1(y)) -> F2(y, x)

The TRS R consists of the following rules:

f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(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:

F2(s1(x), y) -> F2(x, s1(x))
F2(x, s1(y)) -> F2(y, x)

The TRS R consists of the following rules:

f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)

Q is empty.
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.


F2(s1(x), y) -> F2(x, s1(x))
The remaining pairs can at least be oriented weakly.

F2(x, s1(y)) -> F2(y, x)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F2(x1, x2) ) = max{0, 2x1 + x2 - 3}


POL( s1(x1) ) = 2x1 + 3



The following usable rules [14] were oriented: none



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

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

F2(x, s1(y)) -> F2(y, x)

The TRS R consists of the following rules:

f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)

Q is empty.
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.


F2(x, s1(y)) -> F2(y, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F2(x1, x2) ) = max{0, x1 + x2 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



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

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

f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(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.