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
  ↳ RRRPoloQTRSProof

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.

The following Q TRS is given: 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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

f'(s(x), y, y) → f'(y, x, s(x))
Used ordering:
Polynomial interpretation [25]:

POL(f(x1)) = x1   
POL(f'(x1, x2, x3)) = 2·x1 + 2·x2 + x3   
POL(g(x1)) = x1   
POL(h(x1)) = 1 + 2·x1   
POL(s(x1)) = 2 + 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ 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 [19] we can switch to innermost.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ 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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(g(x)) → F(x)
F(g(x)) → F(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ QDPOrderProof

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

F(g(x)) → F(x)
F(g(x)) → F(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 [15].


The following pairs can be oriented strictly and are deleted.


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

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

The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ 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.