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))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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))

The set Q consists of the following terms:

f(g(x0))
f(h(x0))


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(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))

The set Q consists of the following terms:

f(g(x0))
f(h(x0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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))

The set Q consists of the following terms:

f(g(x0))
f(h(x0))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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))

The set Q consists of the following terms:

f(g(x0))
f(h(x0))

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