NO Termination w.r.t. Q proof of ../tpdb/TRS/nontermin/HM/n001.trs
Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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(f(f(x))) → H(f(g(f(x))))
H(f(f(x))) → F(g(f(x)))
F(g(f(x))) → F(f(x))

The TRS R consists of the following rules:

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

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(f(f(x))) → H(f(g(f(x))))
H(f(f(x))) → F(g(f(x)))
F(g(f(x))) → F(f(x))

The TRS R consists of the following rules:

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

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 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof

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

H(f(f(x))) → H(f(g(f(x))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ Narrowing

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

H(f(f(x))) → H(f(g(f(x))))

The TRS R consists of the following rules:

f(g(f(x))) → f(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule H(f(f(x))) → H(f(g(f(x)))) at position [0] we obtained the following new rules:

H(f(f(x0))) → H(f(f(x0)))
H(f(f(g(f(x0))))) → H(f(g(f(f(x0)))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ RuleRemovalProof

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

H(f(f(x0))) → H(f(f(x0)))
H(f(f(g(f(x0))))) → H(f(g(f(f(x0)))))

The TRS R consists of the following rules:

f(g(f(x))) → f(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

H(f(f(g(f(x0))))) → H(f(g(f(f(x0)))))

Strictly oriented rules of the TRS R:

f(g(f(x))) → f(f(x))

Used ordering: POLO with Polynomial interpretation [25]:

POL(H(x1)) = 2·x1   
POL(f(x1)) = 2·x1   
POL(g(x1)) = 2 + 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ RuleRemovalProof
                    ↳ QDP
                      ↳ NonTerminationProof

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

H(f(f(x0))) → H(f(f(x0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

H(f(f(x0))) → H(f(f(x0)))

The TRS R consists of the following rules:none


s = H(f(f(x0))) evaluates to t =H(f(f(x0)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from H(f(f(x0))) to H(f(f(x0))).