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:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> x

Q is empty.

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

F1(f1(x)) -> F1(d1(f1(x)))
F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
G1(c2(x, s1(y))) -> G1(c2(s1(x), y))

The TRS R consists of the following rules:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> x

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:

F1(f1(x)) -> F1(d1(f1(x)))
F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
G1(c2(x, s1(y))) -> G1(c2(s1(x), y))

The TRS R consists of the following rules:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> x

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:

F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))

The TRS R consists of the following rules:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> x

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.

F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
Used argument filtering: F1(x1)  =  x1
c2(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:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> 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.

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

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

G1(c2(x, s1(y))) -> G1(c2(s1(x), y))

The TRS R consists of the following rules:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> x

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.

G1(c2(x, s1(y))) -> G1(c2(s1(x), y))
Used argument filtering: G1(x1)  =  x1
c2(x1, x2)  =  x2
s1(x1)  =  s1(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:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> 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.