Termination w.r.t. Q proof of /tmp/tmpqjhuIu/ExSec11_1

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:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

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

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(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:

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
HALF(s(s(X))) → HALF(X)
TERMS(N) → SQR(N)
SQR(s(X)) → ADD(sqr(X), dbl(X))
SQR(s(X)) → SQR(X)
ADD(s(X), Y) → ADD(X, Y)
DBL(s(X)) → DBL(X)
SQR(s(X)) → DBL(X)
TERMS(N) → TERMS(s(N))

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(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:

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
HALF(s(s(X))) → HALF(X)
TERMS(N) → SQR(N)
SQR(s(X)) → ADD(sqr(X), dbl(X))
SQR(s(X)) → SQR(X)
ADD(s(X), Y) → ADD(X, Y)
DBL(s(X)) → DBL(X)
SQR(s(X)) → DBL(X)
TERMS(N) → TERMS(s(N))

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

HALF(s(s(X))) → HALF(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(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

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

HALF(s(s(X))) → HALF(X)

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

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(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.

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

HALF(s(s(X))) → HALF(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:

HALF(s(s(X))) → HALF(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

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

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(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.

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

From the DPs we obtained the following set of size-change graphs:

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

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

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(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.

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

From the DPs we obtained the following set of size-change graphs:

ADD(s(X), Y) → ADD(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

DBL(s(X)) → DBL(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

DBL(s(X)) → DBL(X)

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

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(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.

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

DBL(s(X)) → DBL(X)

From the DPs we obtained the following set of size-change graphs:

DBL(s(X)) → DBL(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

SQR(s(X)) → SQR(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

SQR(s(X)) → SQR(X)

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

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(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.

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

SQR(s(X)) → SQR(X)

From the DPs we obtained the following set of size-change graphs:

SQR(s(X)) → SQR(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

TERMS(N) → TERMS(s(N))

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

TERMS(N) → TERMS(s(N))

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

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(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.

terms(x0)
half(dbl(x0))
half(0)
half(s(s(x0)))
sqr(s(x0))
half(s(0))
first(0, x0)
add(s(x0), x1)
add(0, x0)
first(s(x0), cons(x1, x2))
dbl(0)
dbl(s(x0))
sqr(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

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

TERMS(N) → TERMS(s(N))

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

TERMS(s(z0)) → TERMS(s(s(z0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

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

TERMS(s(z0)) → TERMS(s(s(z0)))

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

TERMS(s(s(z0))) → TERMS(s(s(s(z0))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

                          ↳ QDP

                            ↳ NonTerminationProof

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

TERMS(s(s(z0))) → TERMS(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:

TERMS(s(s(z0))) → TERMS(s(s(s(z0))))

The TRS R consists of the following rules:none

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

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [z0 / s(z0)]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from TERMS(s(s(z0))) to TERMS(s(s(s(z0)))).