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:

2nd(cons(X, cons(Y, Z))) → Y
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
2nd(cons(x0, cons(x1, x2)))



↳ QTRS
  ↳ DependencyPairsProof

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

2nd(cons(X, cons(Y, Z))) → Y
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
2nd(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:

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

2nd(cons(X, cons(Y, Z))) → Y
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
2nd(cons(x0, cons(x1, x2)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

2nd(cons(X, cons(Y, Z))) → Y
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

from(x0)
2nd(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
      ↳ UsableRulesProof
        ↳ QDP
          ↳ QReductionProof

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

FROM(X) → FROM(s(X))

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

from(x0)
2nd(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.

from(x0)
2nd(cons(x0, cons(x1, x2)))



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

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

FROM(X) → FROM(s(X))

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

FROM(s(z0)) → FROM(s(s(z0)))



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

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

FROM(s(z0)) → FROM(s(s(z0)))

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

FROM(s(s(z0))) → FROM(s(s(s(z0))))



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

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

FROM(s(s(z0))) → FROM(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:

FROM(s(s(z0))) → FROM(s(s(s(z0))))

The TRS R consists of the following rules:none


s = FROM(s(s(z0))) evaluates to t =FROM(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 FROM(s(s(z0))) to FROM(s(s(s(z0)))).