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) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)



↳ QTRS
  ↳ DependencyPairsProof

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

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)


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)
G(s(X)) → G(X)
F(X) → F(g(X))
SEL(s(X), cons(Y, Z)) → SEL(X, Z)

The TRS R consists of the following rules:

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)

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)
G(s(X)) → G(X)
F(X) → F(g(X))
SEL(s(X), cons(Y, Z)) → SEL(X, Z)

The TRS R consists of the following rules:

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 1 less node.

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

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

The TRS R consists of the following rules:

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)

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
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
          ↳ QDP
          ↳ QDP

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

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

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)

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.

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

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
            ↳ 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(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)

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
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
          ↳ QDP

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

G(s(X)) → G(X)

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

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)

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.

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)



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

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

F(X) → F(g(X))

The TRS R consists of the following rules:

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)

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
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof

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

F(X) → F(g(X))

The TRS R consists of the following rules:

g(0) → s(0)
g(s(X)) → s(s(g(X)))

The set Q consists of the following terms:

g(s(x0))
sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))
g(0)

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.

sel(0, cons(x0, x1))
f(x0)
sel(s(x0), cons(x1, x2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ MNOCProof

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

F(X) → F(g(X))

The TRS R consists of the following rules:

g(0) → s(0)
g(s(X)) → s(s(g(X)))

The set Q consists of the following terms:

g(s(x0))
g(0)

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ NonTerminationProof

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

F(X) → F(g(X))

The TRS R consists of the following rules:

g(0) → s(0)
g(s(X)) → s(s(g(X)))

Q is empty.
We have to consider all (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(X) → F(g(X))

The TRS R consists of the following rules:

g(0) → s(0)
g(s(X)) → s(s(g(X)))


s = F(X) evaluates to t =F(g(X))

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(X) to F(g(X)).