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:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(x))

The set Q consists of the following terms:

g(x0)
f(f(x0))



↳ QTRS
  ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse

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

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(x))

The set Q consists of the following terms:

g(x0)
f(f(x0))


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

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(x))

The set Q consists of the following terms:

g(x0)
f(f(x0))

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

f(g(a)) → f(s(g(b)))
Used ordering:
Polynomial interpretation [25]:

POL(a) = 2   
POL(b) = 0   
POL(f(x1)) = x1   
POL(g(x1)) = x1   
POL(s(x1)) = 1 + x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

f(f(x)) → b
g(x) → f(g(x))

The set Q consists of the following terms:

g(x0)
f(f(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:

G(x) → F(g(x))
G(x) → G(x)

The TRS R consists of the following rules:

f(f(x)) → b
g(x) → f(g(x))

The set Q consists of the following terms:

g(x0)
f(f(x0))

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

G(x) → F(g(x))
G(x) → G(x)

The TRS R consists of the following rules:

f(f(x)) → b
g(x) → f(g(x))

The set Q consists of the following terms:

g(x0)
f(f(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
              ↳ UsableRulesProof
  ↳ QTRS Reverse

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

G(x) → G(x)

The TRS R consists of the following rules:

f(f(x)) → b
g(x) → f(g(x))

The set Q consists of the following terms:

g(x0)
f(f(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
              ↳ UsableRulesProof
  ↳ QTRS Reverse

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

G(x) → G(x)

R is empty.
The set Q consists of the following terms:

g(x0)
f(f(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

g(x0)
f(f(x0))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ NonTerminationProof
              ↳ UsableRulesProof
  ↳ QTRS Reverse

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

G(x) → G(x)

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:

G(x) → G(x)

The TRS R consists of the following rules:none


s = G(x) evaluates to t =G(x)

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 G(x) to G(x).




As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
              ↳ UsableRulesProof
                ↳ QDP
  ↳ QTRS Reverse

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

G(x) → G(x)

R is empty.
The set Q consists of the following terms:

g(x0)
f(f(x0))

We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(x))

The set Q is {g(x0), f(f(x0))}.
We have obtained the following QTRS:

a'(g(f(x))) → b'(g(s(f(x))))
f(f(x)) → b'(x)
g(x) → g(f(x))

The set Q is empty.

↳ QTRS
  ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse
    ↳ QTRS

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

a'(g(f(x))) → b'(g(s(f(x))))
f(f(x)) → b'(x)
g(x) → g(f(x))

Q is empty.