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:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G(X, s(Y)) → G(X, Y)
H(X, Z) → F(X, s(X), Z)
F(X, Y, g(X, Y)) → H(0, g(X, Y))

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(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:

G(X, s(Y)) → G(X, Y)
H(X, Z) → F(X, s(X), Z)
F(X, Y, g(X, Y)) → H(0, g(X, Y))

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

G(X, s(Y)) → G(X, Y)

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


G(X, s(Y)) → G(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1/4 + (7/2)x_1   
POL(G(x1, x2)) = (2)x_2   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(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:

H(X, Z) → F(X, s(X), Z)
F(X, Y, g(X, Y)) → H(0, g(X, Y))

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

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