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(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

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(c, c) -> F2(a, a)
F2(s1(X), c) -> F2(X, c)
F2(a, b) -> F2(s1(a), c)
F2(a, a) -> F2(a, b)

The TRS R consists of the following rules:

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

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(c, c) -> F2(a, a)
F2(s1(X), c) -> F2(X, c)
F2(a, b) -> F2(s1(a), c)
F2(a, a) -> F2(a, b)

The TRS R consists of the following rules:

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

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(c, c) -> F2(a, a)
The remaining pairs can at least be oriented weakly.

F2(s1(X), c) -> F2(X, c)
F2(a, b) -> F2(s1(a), c)
F2(a, a) -> F2(a, b)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1


POL( b ) = 1


POL( F2(x1, x2) ) = x1 + 1


POL( a ) = max{0, -1}


POL( c ) = 1



The following usable rules [14] were oriented: none



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

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

F2(s1(X), c) -> F2(X, c)
F2(a, b) -> F2(s1(a), c)
F2(a, a) -> F2(a, b)

The TRS R consists of the following rules:

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

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

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

F2(s1(X), c) -> F2(X, c)

The TRS R consists of the following rules:

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

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

POL( s1(x1) ) = x1 + 1


POL( F2(x1, x2) ) = max{0, x1 + x2 - 1}


POL( c ) = 1



The following usable rules [14] were oriented: none



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

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

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

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.