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:

cons(x, cons(y, z)) → big
inf(x) → cons(x, inf(s(x)))

The set Q consists of the following terms:

inf(x0)
cons(x0, cons(x1, x2))



↳ QTRS
  ↳ DependencyPairsProof

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

cons(x, cons(y, z)) → big
inf(x) → cons(x, inf(s(x)))

The set Q consists of the following terms:

inf(x0)
cons(x0, cons(x1, x2))


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

INF(x) → CONS(x, inf(s(x)))
INF(x) → INF(s(x))

The TRS R consists of the following rules:

cons(x, cons(y, z)) → big
inf(x) → cons(x, inf(s(x)))

The set Q consists of the following terms:

inf(x0)
cons(x0, cons(x1, x2))

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:

INF(x) → CONS(x, inf(s(x)))
INF(x) → INF(s(x))

The TRS R consists of the following rules:

cons(x, cons(y, z)) → big
inf(x) → cons(x, inf(s(x)))

The set Q consists of the following terms:

inf(x0)
cons(x0, cons(x1, x2))

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:

INF(x) → INF(s(x))

The TRS R consists of the following rules:

cons(x, cons(y, z)) → big
inf(x) → cons(x, inf(s(x)))

The set Q consists of the following terms:

inf(x0)
cons(x0, cons(x1, x2))

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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

INF(x) → INF(s(x))

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

inf(x0)
cons(x0, cons(x1, x2))

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.

inf(x0)
cons(x0, cons(x1, x2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Instantiation

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

INF(x) → INF(s(x))

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

INF(s(z0)) → INF(s(s(z0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation

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

INF(s(z0)) → INF(s(s(z0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule INF(s(z0)) → INF(s(s(z0))) we obtained the following new rules:

INF(s(s(z0))) → INF(s(s(s(z0))))



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

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

INF(s(s(z0))) → INF(s(s(s(z0))))

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:

INF(s(s(z0))) → INF(s(s(s(z0))))

The TRS R consists of the following rules:none


s = INF(s(s(z0))) evaluates to t =INF(s(s(s(z0))))

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 INF(s(s(z0))) to INF(s(s(s(z0)))).