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:

from(X) → cons(X, from(s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, XS)

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

from(X) → cons(X, from(s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, XS)

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

from(X) → cons(X, from(s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, XS)

The set Q consists of the following terms:

from(x0)
after(0, x0)
after(s(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:

AFTER(s(N), cons(X, XS)) → AFTER(N, XS)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, XS)

The set Q consists of the following terms:

from(x0)
after(0, x0)
after(s(x0), cons(x1, x2))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

AFTER(s(N), cons(X, XS)) → AFTER(N, XS)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, XS)

The set Q consists of the following terms:

from(x0)
after(0, x0)
after(s(x0), cons(x1, x2))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP

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

AFTER(s(N), cons(X, XS)) → AFTER(N, XS)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, XS)

The set Q consists of the following terms:

from(x0)
after(0, x0)
after(s(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP

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

AFTER(s(N), cons(X, XS)) → AFTER(N, XS)

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

from(x0)
after(0, x0)
after(s(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)
after(0, x0)
after(s(x0), cons(x1, x2))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

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

AFTER(s(N), cons(X, XS)) → AFTER(N, XS)

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ 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:

from(X) → cons(X, from(s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, XS)

The set Q consists of the following terms:

from(x0)
after(0, x0)
after(s(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ 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)
after(0, x0)
after(s(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)
after(0, x0)
after(s(x0), cons(x1, x2))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ 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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ 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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ 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)))).