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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(f(x)) → F(g(f(x), x))
F(f(x)) → H(f(x), f(x))
H(x, x) → G(x, 0)
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
      ↳ EdgeDeletionProof

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

F(f(x)) → F(g(f(x), x))
F(f(x)) → H(f(x), f(x))
H(x, x) → G(x, 0)
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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(f(x)) → F(g(f(x), x))
F(f(x)) → H(f(x), f(x))
H(x, x) → G(x, 0)
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 [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
F(x1)  =  F(x1)
f(x1)  =  f(x1)
g(x1, x2)  =  g(x2)
h(x1, x2)  =  h
0  =  0

Recursive Path Order [2].
Precedence:
F1 > h > g1 > 0
f1 > h > g1 > 0

The following usable rules [14] were oriented:

g(x, y) → y
h(x, x) → g(x, 0)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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.