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(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is

f(h(x)) → h(g(x))
f(g(x)) → g(f(f(x)))

The TRS R 2 is

f'(s(x), y, y) → f'(y, x, s(x))

The signature Sigma is {f'}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))

The set Q consists of the following terms:

f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)


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(g(x)) → F(f(x))
F'(s(x), y, y) → F'(y, x, s(x))
F(g(x)) → F(x)

The TRS R consists of the following rules:

f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))

The set Q consists of the following terms:

f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

F(g(x)) → F(f(x))
F'(s(x), y, y) → F'(y, x, s(x))
F(g(x)) → F(x)

The TRS R consists of the following rules:

f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))

The set Q consists of the following terms:

f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

F(g(x)) → F(f(x))
F'(s(x), y, y) → F'(y, x, s(x))
F(g(x)) → F(x)

The TRS R consists of the following rules:

f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))

The set Q consists of the following terms:

f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

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

F(g(x)) → F(f(x))
F(g(x)) → F(x)

The TRS R consists of the following rules:

f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))

The set Q consists of the following terms:

f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)

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(g(x)) → F(f(x))
F(g(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)  =  x1
g(x1)  =  g(x1)
f(x1)  =  x1
h(x1)  =  h

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented:

f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))



↳ QTRS
  ↳ AAECC Innermost
    ↳ 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(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))

The set Q consists of the following terms:

f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)

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