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

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

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

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

DOUBLE(s(x)) → DOUBLE(x)
SQUARE(s(x)) → SQUARE(x)
SQUARE(s(x)) → PLUS(square(x), double(x))
COND(false, x, y) → SQUARE(s(s(y)))
LE(s(u), s(v)) → LE(u, v)
COND(false, x, y) → LOG(x, square(s(s(y))))
SQUARE(s(x)) → DOUBLE(x)
PLUS(n, s(m)) → PLUS(n, m)
LOG(x, s(s(y))) → LE(x, s(s(y)))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)
COND(false, x, y) → DOUBLE(log(x, square(s(s(y)))))

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

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:

DOUBLE(s(x)) → DOUBLE(x)
SQUARE(s(x)) → SQUARE(x)
SQUARE(s(x)) → PLUS(square(x), double(x))
COND(false, x, y) → SQUARE(s(s(y)))
LE(s(u), s(v)) → LE(u, v)
COND(false, x, y) → LOG(x, square(s(s(y))))
SQUARE(s(x)) → DOUBLE(x)
PLUS(n, s(m)) → PLUS(n, m)
LOG(x, s(s(y))) → LE(x, s(s(y)))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)
COND(false, x, y) → DOUBLE(log(x, square(s(s(y)))))

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

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 [13] we can delete all non-usable rules [14] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(n, s(m)) → PLUS(n, m)

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

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

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.

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(n, s(m)) → PLUS(n, m)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [16] together with the size-change analysis [27] 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:

PLUS(n, s(m)) → PLUS(n, m)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

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

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

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.

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

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

DOUBLE(s(x)) → DOUBLE(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

SQUARE(s(x)) → SQUARE(x)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

SQUARE(s(x)) → SQUARE(x)

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

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

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.

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

SQUARE(s(x)) → SQUARE(x)

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

SQUARE(s(x)) → SQUARE(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

LE(s(u), s(v)) → LE(u, v)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

LE(s(u), s(v)) → LE(u, v)

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

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

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.

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

LE(s(u), s(v)) → LE(u, v)

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

LE(s(u), s(v)) → LE(u, v)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

COND(false, x, y) → LOG(x, square(s(s(y))))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

COND(false, x, y) → LOG(x, square(s(s(y))))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
cond(true, x0, x1)
log(x0, s(s(x1)))
square(s(x0))
double(s(x0))
cond(false, x0, x1)

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.

cond(true, x0, x1)
log(x0, s(s(x1)))
cond(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

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

COND(false, x, y) → LOG(x, square(s(s(y))))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, x, y) → LOG(x, square(s(s(y)))) at position [1] we obtained the following new rules:

COND(false, x, y) → LOG(x, s(plus(square(s(y)), double(s(y)))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

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

COND(false, x, y) → LOG(x, s(plus(square(s(y)), double(s(y)))))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, x, y) → LOG(x, s(plus(square(s(y)), double(s(y))))) at position [1,0,0] we obtained the following new rules:

COND(false, x, y) → LOG(x, s(plus(s(plus(square(y), double(y))), double(s(y)))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

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

COND(false, x, y) → LOG(x, s(plus(s(plus(square(y), double(y))), double(s(y)))))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, x, y) → LOG(x, s(plus(s(plus(square(y), double(y))), double(s(y))))) at position [1,0,1] we obtained the following new rules:

COND(false, x, y) → LOG(x, s(plus(s(plus(square(y), double(y))), s(s(double(y))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

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

COND(false, x, y) → LOG(x, s(plus(s(plus(square(y), double(y))), s(s(double(y))))))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, x, y) → LOG(x, s(plus(s(plus(square(y), double(y))), s(s(double(y)))))) at position [1,0] we obtained the following new rules:

COND(false, x, y) → LOG(x, s(s(plus(s(plus(square(y), double(y))), s(double(y))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

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

COND(false, x, y) → LOG(x, s(s(plus(s(plus(square(y), double(y))), s(double(y))))))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, x, y) → LOG(x, s(s(plus(s(plus(square(y), double(y))), s(double(y)))))) at position [1,0,0] we obtained the following new rules:

COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                        ↳ NonInfProof

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

COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y))))))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y) at position [0] we obtained the following new rules:

LOG(s(x0), s(s(y1))) → COND(le(x0, s(y1)), s(x0), y1)
LOG(0, s(s(y1))) → COND(true, 0, y1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                        ↳ NonInfProof

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

LOG(s(x0), s(s(y1))) → COND(le(x0, s(y1)), s(x0), y1)
LOG(0, s(s(y1))) → COND(true, 0, y1)
COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y))))))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                        ↳ NonInfProof

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

LOG(s(x0), s(s(y1))) → COND(le(x0, s(y1)), s(x0), y1)
COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y))))))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule LOG(s(x0), s(s(y1))) → COND(le(x0, s(y1)), s(x0), y1) we obtained the following new rules:

LOG(s(x0), s(s(s(y_4)))) → COND(le(x0, s(s(y_4))), s(x0), s(y_4))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                        ↳ NonInfProof

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

COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y))))))
LOG(s(x0), s(s(s(y_4)))) → COND(le(x0, s(s(y_4))), s(x0), s(y_4))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y)))))) we obtained the following new rules:

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(plus(square(s(z1)), double(s(z1)))), double(s(z1)))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Rewriting

                                        ↳ NonInfProof

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

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(plus(square(s(z1)), double(s(z1)))), double(s(z1)))))))
LOG(s(x0), s(s(s(y_4)))) → COND(le(x0, s(s(y_4))), s(x0), s(y_4))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(plus(square(s(z1)), double(s(z1)))), double(s(z1))))))) at position [1,0,0,0,0,0,0] we obtained the following new rules:

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(plus(s(plus(square(z1), double(z1))), double(s(z1)))), double(s(z1)))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ Rewriting

                                        ↳ NonInfProof

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

LOG(s(x0), s(s(s(y_4)))) → COND(le(x0, s(s(y_4))), s(x0), s(y_4))
COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(plus(s(plus(square(z1), double(z1))), double(s(z1)))), double(s(z1)))))))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(plus(s(plus(square(z1), double(z1))), double(s(z1)))), double(s(z1))))))) at position [1,0,0,0,0,0,1] we obtained the following new rules:

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(plus(s(plus(square(z1), double(z1))), s(s(double(z1))))), double(s(z1)))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ Rewriting

                                        ↳ NonInfProof

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

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(plus(s(plus(square(z1), double(z1))), s(s(double(z1))))), double(s(z1)))))))
LOG(s(x0), s(s(s(y_4)))) → COND(le(x0, s(s(y_4))), s(x0), s(y_4))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(plus(s(plus(square(z1), double(z1))), s(s(double(z1))))), double(s(z1))))))) at position [1,0,0,0,0,0] we obtained the following new rules:

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(s(plus(s(plus(square(z1), double(z1))), s(double(z1))))), double(s(z1)))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                        ↳ NonInfProof

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

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(s(plus(s(plus(square(z1), double(z1))), s(double(z1))))), double(s(z1)))))))
LOG(s(x0), s(s(s(y_4)))) → COND(le(x0, s(s(y_4))), s(x0), s(y_4))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(s(plus(s(plus(square(z1), double(z1))), s(double(z1))))), double(s(z1))))))) at position [1,0,0,0,0,0,0] we obtained the following new rules:

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), double(s(z1)))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Rewriting

                                        ↳ NonInfProof

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

LOG(s(x0), s(s(s(y_4)))) → COND(le(x0, s(s(y_4))), s(x0), s(y_4))
COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), double(s(z1)))))))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), double(s(z1))))))) at position [1,0,0,0,1] we obtained the following new rules:

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), s(s(double(z1))))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Rewriting

                                                                          ↳ QDP

                                                                            ↳ Rewriting

                                        ↳ NonInfProof

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

LOG(s(x0), s(s(s(y_4)))) → COND(le(x0, s(s(y_4))), s(x0), s(y_4))
COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), s(s(double(z1))))))))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), s(s(double(z1)))))))) at position [1,0,0,0] we obtained the following new rules:

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), s(double(z1))))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Rewriting

                                                                          ↳ QDP

                                                                            ↳ Rewriting

                                                                              ↳ QDP

                                                                                ↳ Rewriting

                                        ↳ NonInfProof

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

LOG(s(x0), s(s(s(y_4)))) → COND(le(x0, s(s(y_4))), s(x0), s(y_4))
COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), s(double(z1))))))))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), s(double(z1)))))))) at position [1,0,0,0,0] we obtained the following new rules:

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), double(z1))))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Rewriting

                                                                          ↳ QDP

                                                                            ↳ Rewriting

                                                                              ↳ QDP

                                                                                ↳ Rewriting

                                                                                  ↳ QDP

                                        ↳ NonInfProof

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

COND(false, s(z0), s(z1)) → LOG(s(z0), s(s(s(s(s(plus(s(s(s(plus(s(plus(square(z1), double(z1))), double(z1))))), double(z1))))))))
LOG(s(x0), s(s(s(y_4)))) → COND(le(x0, s(s(y_4))), s(x0), s(y_4))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus with the following steps:
Note that final constraints are written in bold face.

For Pair LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y) the following chains were created:

We consider the chain LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y), COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y)))))) which results in the following constraint:
- (1) (COND(le(x2, s(s(x3))), x2, x3)=COND(false, x4, x5) ⇒ LOG(x2, s(s(x3)))≥COND(le(x2, s(s(x3))), x2, x3))
We simplified constraint (1) using rules (I), (II), (IV), (VII) which results in the following new constraint:
- (2) (s(s(x3))=x12∧le(x2, x12)=false ⇒ LOG(x2, s(s(x3)))≥COND(le(x2, s(s(x3))), x2, x3))
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x2, x12)=false which results in the following new constraints:
- (3) (le(x14, x15)=false∧s(s(x3))=s(x15)∧(∀x16:le(x14, x15)=false∧s(s(x16))=x15 ⇒ LOG(x14, s(s(x16)))≥COND(le(x14, s(s(x16))), x14, x16)) ⇒ LOG(s(x14), s(s(x3)))≥COND(le(s(x14), s(s(x3))), s(x14), x3))
- (4) (false=false∧s(s(x3))=0 ⇒ LOG(s(x17), s(s(x3)))≥COND(le(s(x17), s(s(x3))), s(x17), x3))
We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:
- (5) (le(x14, x15)=false∧s(x3)=x15 ⇒ LOG(s(x14), s(s(x3)))≥COND(le(s(x14), s(s(x3))), s(x14), x3))
We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on le(x14, x15)=false which results in the following new constraints:
- (6) (le(x19, x20)=false∧s(x3)=s(x20)∧(∀x21:le(x19, x20)=false∧s(x21)=x20 ⇒ LOG(s(x19), s(s(x21)))≥COND(le(s(x19), s(s(x21))), s(x19), x21)) ⇒ LOG(s(s(x19)), s(s(x3)))≥COND(le(s(s(x19)), s(s(x3))), s(s(x19)), x3))
- (7) (false=false∧s(x3)=0 ⇒ LOG(s(s(x22)), s(s(x3)))≥COND(le(s(s(x22)), s(s(x3))), s(s(x22)), x3))
We simplified constraint (6) using rules (I), (II), (III), (IV) which results in the following new constraint:
- (8) (le(x19, x20)=false ⇒ LOG(s(s(x19)), s(s(x20)))≥COND(le(s(s(x19)), s(s(x20))), s(s(x19)), x20))
We solved constraint (7) using rules (I), (II).We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on le(x19, x20)=false which results in the following new constraints:
- (9) (le(x24, x25)=false∧(le(x24, x25)=false ⇒ LOG(s(s(x24)), s(s(x25)))≥COND(le(s(s(x24)), s(s(x25))), s(s(x24)), x25)) ⇒ LOG(s(s(s(x24))), s(s(s(x25))))≥COND(le(s(s(s(x24))), s(s(s(x25)))), s(s(s(x24))), s(x25)))
- (10) (false=false ⇒ LOG(s(s(s(x26))), s(s(0)))≥COND(le(s(s(s(x26))), s(s(0))), s(s(s(x26))), 0))
We simplified constraint (9) using rule (VI) where we applied the induction hypothesis (le(x24, x25)=false ⇒ LOG(s(s(x24)), s(s(x25)))≥COND(le(s(s(x24)), s(s(x25))), s(s(x24)), x25)) with σ = [ ] which results in the following new constraint:
- (11) (LOG(s(s(x24)), s(s(x25)))≥COND(le(s(s(x24)), s(s(x25))), s(s(x24)), x25) ⇒ LOG(s(s(s(x24))), s(s(s(x25))))≥COND(le(s(s(s(x24))), s(s(s(x25)))), s(s(s(x24))), s(x25)))
We simplified constraint (10) using rules (I), (II) which results in the following new constraint:
- (12) (LOG(s(s(s(x26))), s(s(0)))≥COND(le(s(s(s(x26))), s(s(0))), s(s(s(x26))), 0))

For Pair COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y)))))) the following chains were created:

We consider the chain COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y)))))), LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y) which results in the following constraint:
- (13) (LOG(x6, s(s(s(plus(s(plus(square(x7), double(x7))), double(x7))))))=LOG(x8, s(s(x9))) ⇒ COND(false, x6, x7)≥LOG(x6, s(s(s(plus(s(plus(square(x7), double(x7))), double(x7)))))))
We simplified constraint (13) using rules (I), (II), (IV) which results in the following new constraint:
- (14) (COND(false, x6, x7)≥LOG(x6, s(s(s(plus(s(plus(square(x7), double(x7))), double(x7)))))))

To summarize, we get the following constraints P_≥ for the following pairs.

LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)
- (LOG(s(s(x24)), s(s(x25)))≥COND(le(s(s(x24)), s(s(x25))), s(s(x24)), x25) ⇒ LOG(s(s(s(x24))), s(s(s(x25))))≥COND(le(s(s(s(x24))), s(s(s(x25)))), s(s(s(x24))), s(x25)))
- (LOG(s(s(s(x26))), s(s(0)))≥COND(le(s(s(s(x26))), s(s(0))), s(s(s(x26))), 0))
COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y))))))
- (COND(false, x6, x7)≥LOG(x6, s(s(s(plus(s(plus(square(x7), double(x7))), double(x7)))))))

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved.
Polynomial interpretation [21]:

POL(0) = 0
POL(COND(x₁, x₂, x₃)) = -1 - x₁ + 2·x₂ - x₃
POL(LOG(x₁, x₂)) = 2 + 2·x₁ - x₂
POL(c) = -2
POL(double(x₁)) = 2·x₁
POL(false) = 1
POL(le(x₁, x₂)) = 1
POL(plus(x₁, x₂)) = x₂
POL(s(x₁)) = 2 + x₁
POL(square(x₁)) = 0
POL(true) = 1

The following pairs are in P_>:

COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y))))))

The following pairs are in P_bound:

LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The following rules are usable:

false → le(s(u), 0)
s(s(double(x))) → double(s(x))
le(u, v) → le(s(u), s(v))
s(plus(n, m)) → plus(n, s(m))
n → plus(n, 0)
true → le(0, v)
0 → double(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                        ↳ NonInfProof

                                          ↳ AND

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                            ↳ QDP

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

LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                        ↳ NonInfProof

                                          ↳ AND

                                            ↳ QDP

                                            ↳ QDP

                                              ↳ DependencyGraphProof

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

COND(false, x, y) → LOG(x, s(s(s(plus(s(plus(square(y), double(y))), double(y))))))

The TRS R consists of the following rules:

le(0, v) → true
le(s(u), s(v)) → le(u, v)
le(s(u), 0) → false
square(s(x)) → s(plus(square(x), double(x)))
square(0) → 0
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

double(0)
square(0)
le(s(x0), s(x1))
plus(x0, s(x1))
le(s(x0), 0)
plus(x0, 0)
le(0, x0)
square(s(x0))
double(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.