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(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(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:

G(s(x)) → G(x)
F(s(0), g(x)) → F(x, g(x))

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(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:

G(s(x)) → G(x)
F(s(0), g(x)) → F(x, g(x))

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

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

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

G(s(x)) → G(x)

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(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
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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

G(s(x)) → G(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ ForwardInstantiation

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

F(s(0), g(x)) → F(x, g(x))

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(s(0), g(x)) → F(x, g(x)) we obtained the following new rules:

F(s(0), g(s(0))) → F(s(0), g(s(0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ ForwardInstantiation
              ↳ QDP
                ↳ UsableRulesProof

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

F(s(0), g(s(0))) → F(s(0), g(s(0)))

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(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
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ ForwardInstantiation
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ RuleRemovalProof

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

F(s(0), g(s(0))) → F(s(0), g(s(0)))

The TRS R consists of the following rules:

g(s(x)) → g(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 rules of the TRS R:

g(s(x)) → g(x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(F(x1, x2)) = x1 + x2   
POL(g(x1)) = 2·x1   
POL(s(x1)) = 2 + 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ ForwardInstantiation
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ RuleRemovalProof
                      ↳ QDP
                        ↳ NonTerminationProof

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

F(s(0), g(s(0))) → F(s(0), g(s(0)))

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:

F(s(0), g(s(0))) → F(s(0), g(s(0)))

The TRS R consists of the following rules:none


s = F(s(0), g(s(0))) evaluates to t =F(s(0), g(s(0)))

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 F(s(0), g(s(0))) to F(s(0), g(s(0))).