Termination w.r.t. Q proof of /tmp/tmpBGi5m

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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))

Q is empty.

↳ 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))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

Q is empty.
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:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

PLUS(n, s(m)) → PLUS(n, m)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(PLUS(x₁, x₂)) = (1/4)x_2
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

DOUBLE(s(x)) → DOUBLE(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(DOUBLE(x₁)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

SQUARE(s(x)) → SQUARE(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(SQUARE(x₁)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

LE(s(u), s(v)) → LE(u, v)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x₁)) = 1 + x_1
POL(LE(x₁, x₂)) = x_2

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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)
COND(false, x, y) → 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))

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