Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, Y)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, Y)

Q is empty.

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

H2(X, Z) -> F3(X, s1(X), Z)
G2(X, s1(Y)) -> G2(X, Y)
F3(X, Y, g2(X, Y)) -> H2(0, g2(X, Y))

The TRS R consists of the following rules:

h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, 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:

H2(X, Z) -> F3(X, s1(X), Z)
G2(X, s1(Y)) -> G2(X, Y)
F3(X, Y, g2(X, Y)) -> H2(0, g2(X, Y))

The TRS R consists of the following rules:

h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, Y)

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

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

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

G2(X, s1(Y)) -> G2(X, Y)

The TRS R consists of the following rules:

h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, 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.

G2(X, s1(Y)) -> G2(X, Y)
Used argument filtering: G2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi 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:

h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, 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

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

H2(X, Z) -> F3(X, s1(X), Z)
F3(X, Y, g2(X, Y)) -> H2(0, g2(X, Y))

The TRS R consists of the following rules:

h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, Y)

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