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(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

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:

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

The set Q consists of the following terms:

f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(cons2(x0, x1))


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

H1(cons2(X, Y)) -> G1(cons2(X, Y))
G1(cons2(0, Y)) -> G1(Y)
H1(cons2(X, Y)) -> H1(g1(cons2(X, Y)))
F1(s1(X)) -> F1(X)

The TRS R consists of the following rules:

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

The set Q consists of the following terms:

f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(cons2(x0, x1))

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

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

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

H1(cons2(X, Y)) -> G1(cons2(X, Y))
G1(cons2(0, Y)) -> G1(Y)
H1(cons2(X, Y)) -> H1(g1(cons2(X, Y)))
F1(s1(X)) -> F1(X)

The TRS R consists of the following rules:

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

The set Q consists of the following terms:

f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(cons2(x0, x1))

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

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

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

G1(cons2(0, Y)) -> G1(Y)

The TRS R consists of the following rules:

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

The set Q consists of the following terms:

f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(cons2(x0, x1))

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(cons2(0, Y)) -> G1(Y)
Used argument filtering: G1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
0  =  0
Used ordering: Quasi Precedence: trivial 


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

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

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

The set Q consists of the following terms:

f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(cons2(x0, x1))

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

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

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

F1(s1(X)) -> F1(X)

The TRS R consists of the following rules:

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

The set Q consists of the following terms:

f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(cons2(x0, x1))

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(s1(X)) -> F1(X)
Used argument filtering: F1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


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

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

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

The set Q consists of the following terms:

f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(cons2(x0, x1))

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