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(X, g(X)) → f(1, g(X))
g(1) → g(0)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(X, g(X)) → f(1, g(X))
g(1) → g(0)

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:

F(X, g(X)) → F(1, g(X))
G(1) → G(0)

The TRS R consists of the following rules:

f(X, g(X)) → f(1, g(X))
g(1) → g(0)

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:

F(X, g(X)) → F(1, g(X))
G(1) → G(0)

The TRS R consists of the following rules:

f(X, g(X)) → f(1, g(X))
g(1) → g(0)

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 1 less node.

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

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

F(X, g(X)) → F(1, g(X))

The TRS R consists of the following rules:

f(X, g(X)) → f(1, g(X))
g(1) → g(0)

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
              ↳ Instantiation

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

F(X, g(X)) → F(1, g(X))

The TRS R consists of the following rules:

g(1) → g(0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(X, g(X)) → F(1, g(X)) we obtained the following new rules:

F(1, g(1)) → F(1, g(1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ RuleRemovalProof

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

F(1, g(1)) → F(1, g(1))

The TRS R consists of the following rules:

g(1) → g(0)

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(1) → g(0)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1   
POL(1) = 2   
POL(F(x1, x2)) = x1 + x2   
POL(g(x1)) = 2·x1   



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

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

F(1, g(1)) → F(1, g(1))

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(1, g(1)) → F(1, g(1))

The TRS R consists of the following rules:none


s = F(1, g(1)) evaluates to t =F(1, g(1))

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(1, g(1)) to F(1, g(1)).