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:

f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(x)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(x)

Q is empty.

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

F1(f1(x)) -> F1(c1(f1(x)))
F1(f1(x)) -> F1(d1(f1(x)))
G1(c1(h1(0))) -> G1(d1(1))
G1(h1(x)) -> G1(x)
G1(c1(1)) -> G1(d1(h1(0)))

The TRS R consists of the following rules:

f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(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(c1(f1(x)))
F1(f1(x)) -> F1(d1(f1(x)))
G1(c1(h1(0))) -> G1(d1(1))
G1(h1(x)) -> G1(x)
G1(c1(1)) -> G1(d1(h1(0)))

The TRS R consists of the following rules:

f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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

G1(h1(x)) -> G1(x)

The TRS R consists of the following rules:

f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(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(h1(x)) -> G1(x)
Used argument filtering: G1(x1)  =  x1
h1(x1)  =  h1(x1)
Used ordering: Quasi Precedence: trivial


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

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

f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(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.