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:

f(f(x)) → f(g(f(x), x))
f(f(x)) → f(h(f(x), f(x)))
g(x, y) → y
h(x, x) → g(x, 0)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(f(x)) → f(g(f(x), x))
f(f(x)) → f(h(f(x), f(x)))
g(x, y) → y
h(x, x) → g(x, 0)

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:

H(x, x) → G(x, 0)
F(f(x)) → F(g(f(x), x))
F(f(x)) → H(f(x), f(x))
F(f(x)) → G(f(x), x)
F(f(x)) → F(h(f(x), f(x)))

The TRS R consists of the following rules:

f(f(x)) → f(g(f(x), x))
f(f(x)) → f(h(f(x), f(x)))
g(x, y) → y
h(x, x) → g(x, 0)

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:

H(x, x) → G(x, 0)
F(f(x)) → F(g(f(x), x))
F(f(x)) → H(f(x), f(x))
F(f(x)) → G(f(x), x)
F(f(x)) → F(h(f(x), f(x)))

The TRS R consists of the following rules:

f(f(x)) → f(g(f(x), x))
f(f(x)) → f(h(f(x), f(x)))
g(x, y) → y
h(x, x) → g(x, 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

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

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

F(f(x)) → F(g(f(x), x))
F(f(x)) → F(h(f(x), f(x)))

The TRS R consists of the following rules:

f(f(x)) → f(g(f(x), x))
f(f(x)) → f(h(f(x), f(x)))
g(x, y) → y
h(x, x) → g(x, 0)

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.


F(f(x)) → F(g(f(x), x))
F(f(x)) → F(h(f(x), f(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(F(x1)) = x1   
POL(f(x1)) = 1 + x1   
POL(g(x1, x2)) = x2   
POL(h(x1, x2)) = 0   

The following usable rules [17] were oriented:

h(x, x) → g(x, 0)
g(x, y) → y



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

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

f(f(x)) → f(g(f(x), x))
f(f(x)) → f(h(f(x), f(x)))
g(x, y) → y
h(x, x) → g(x, 0)

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.