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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
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)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)


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)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ QDPAfsSolverProof

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)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F2(c, c) -> F2(a, a)
Used argument filtering: F2(x1, x2)  =  x1
c  =  c
a  =  a
s1(x1)  =  x1
Used ordering: Quasi Precedence: c > a


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof
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)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ DependencyGraphProof
QDP
                  ↳ QDPAfsSolverProof

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)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F2(s1(X), c) -> F2(X, c)
Used argument filtering: F2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPAfsSolverProof
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)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)

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