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(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), y) -> f2(x, s1(c1(y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), y) -> f2(x, s1(c1(y)))

Q is empty.

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

F2(x, c1(y)) -> F2(y, y)
F2(s1(x), y) -> F2(x, s1(c1(y)))
F2(x, c1(y)) -> F2(x, s1(f2(y, y)))

The TRS R consists of the following rules:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), y) -> f2(x, s1(c1(y)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

F2(x, c1(y)) -> F2(y, y)
F2(s1(x), y) -> F2(x, s1(c1(y)))
F2(x, c1(y)) -> F2(x, s1(f2(y, y)))

The TRS R consists of the following rules:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), y) -> f2(x, s1(c1(y)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

F2(s1(x), y) -> F2(x, s1(c1(y)))

The TRS R consists of the following rules:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), y) -> f2(x, s1(c1(y)))

Q is empty.
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), y) -> F2(x, s1(c1(y)))
Used argument filtering: F2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), y) -> f2(x, s1(c1(y)))

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof

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

F2(x, c1(y)) -> F2(y, y)

The TRS R consists of the following rules:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), y) -> f2(x, s1(c1(y)))

Q is empty.
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(x, c1(y)) -> F2(y, y)
Used argument filtering: F2(x1, x2)  =  x2
c1(x1)  =  c1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof

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

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), y) -> f2(x, s1(c1(y)))

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.