Termination w.r.t. Q proof of /tmp/tmpL5fEje/aprove4.srs

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(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

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:

HALF(s(s(x1))) → P(s(s(x1)))
HALF(0(x1)) → 0¹(s(s(half(x1))))
HALF(0(x1)) → S(half(x1))
HALF(0(x1)) → HALF(x1)
HALF(half(s(s(s(s(x1)))))) → HALF(x1)
HALF(s(s(x1))) → HALF(p(s(s(x1))))
HALF(half(s(s(s(s(x1)))))) → S(half(half(x1)))
P(s(s(s(x1)))) → S(p(s(s(x1))))
S(s(p(s(x1)))) → S(s(x1))
LOG(s(x1)) → HALF(s(x1))
P(s(s(s(x1)))) → P(s(s(x1)))
LOG(s(x1)) → LOG(half(s(x1)))
HALF(s(s(x1))) → S(half(p(s(s(x1)))))
HALF(half(s(s(s(s(x1)))))) → HALF(half(x1))
LOG(s(x1)) → S(log(half(s(x1))))
HALF(0(x1)) → S(s(half(x1)))
HALF(half(s(s(s(s(x1)))))) → S(s(half(half(x1))))

The TRS R consists of the following rules:

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

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:

HALF(s(s(x1))) → P(s(s(x1)))
HALF(0(x1)) → 0¹(s(s(half(x1))))
HALF(0(x1)) → S(half(x1))
HALF(0(x1)) → HALF(x1)
HALF(half(s(s(s(s(x1)))))) → HALF(x1)
HALF(s(s(x1))) → HALF(p(s(s(x1))))
HALF(half(s(s(s(s(x1)))))) → S(half(half(x1)))
P(s(s(s(x1)))) → S(p(s(s(x1))))
S(s(p(s(x1)))) → S(s(x1))
LOG(s(x1)) → HALF(s(x1))
P(s(s(s(x1)))) → P(s(s(x1)))
LOG(s(x1)) → LOG(half(s(x1)))
HALF(s(s(x1))) → S(half(p(s(s(x1)))))
HALF(half(s(s(s(s(x1)))))) → HALF(half(x1))
LOG(s(x1)) → S(log(half(s(x1))))
HALF(0(x1)) → S(s(half(x1)))
HALF(half(s(s(s(s(x1)))))) → S(s(half(half(x1))))

The TRS R consists of the following rules:

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(s(p(s(x1)))) → S(s(x1))

The TRS R consists of the following rules:

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

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.

S(s(p(s(x1)))) → S(s(x1))

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

POL(s(x₁)) = 1/4 + (4)x_1
POL(p(x₁)) = (1/4)x_1
POL(S(x₁)) = (4)x_1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

s(s(p(s(x1)))) → s(s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

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

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

P(s(s(s(x1)))) → P(s(s(x1)))

The TRS R consists of the following rules:

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

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.

P(s(s(s(x1)))) → P(s(s(x1)))

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

POL(P(x₁)) = (1/4)x_1
POL(s(x₁)) = 1/4 + x_1
POL(p(x₁)) = x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

s(s(p(s(x1)))) → s(s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

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

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

HALF(0(x1)) → HALF(x1)
HALF(half(s(s(s(s(x1)))))) → HALF(x1)
HALF(s(s(x1))) → HALF(p(s(s(x1))))
HALF(half(s(s(s(s(x1)))))) → HALF(half(x1))

The TRS R consists of the following rules:

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

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.

HALF(0(x1)) → HALF(x1)

The remaining pairs can at least be oriented weakly.

HALF(half(s(s(s(s(x1)))))) → HALF(x1)
HALF(s(s(x1))) → HALF(p(s(s(x1))))
HALF(half(s(s(s(s(x1)))))) → HALF(half(x1))

Used ordering: Polynomial interpretation [25,35]:

POL(HALF(x₁)) = (1/2)x_1
POL(half(x₁)) = (4)x_1
POL(s(x₁)) = x_1
POL(p(x₁)) = x_1
POL(0(x₁)) = 4 + (4)x_1

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(s(0(x1))) → 0(x1)
p(s(s(s(x1)))) → s(p(s(s(x1))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
0(x1) → x1
s(s(p(s(x1)))) → s(s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

HALF(half(s(s(s(s(x1)))))) → HALF(x1)
HALF(s(s(x1))) → HALF(p(s(s(x1))))
HALF(half(s(s(s(s(x1)))))) → HALF(half(x1))

The TRS R consists of the following rules:

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

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.

HALF(half(s(s(s(s(x1)))))) → HALF(x1)

The remaining pairs can at least be oriented weakly.

HALF(s(s(x1))) → HALF(p(s(s(x1))))
HALF(half(s(s(s(s(x1)))))) → HALF(half(x1))

Used ordering: Polynomial interpretation [25,35]:

POL(HALF(x₁)) = (1/4)x_1
POL(half(x₁)) = 4 + (4)x_1
POL(s(x₁)) = x_1
POL(p(x₁)) = x_1
POL(0(x₁)) = 4 + (4)x_1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(s(0(x1))) → 0(x1)
p(s(s(s(x1)))) → s(p(s(s(x1))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
0(x1) → x1
s(s(p(s(x1)))) → s(s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

          ↳ QDP

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

HALF(s(s(x1))) → HALF(p(s(s(x1))))
HALF(half(s(s(s(s(x1)))))) → HALF(half(x1))

The TRS R consists of the following rules:

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LOG(s(x1)) → LOG(half(s(x1)))

The TRS R consists of the following rules:

log(s(x1)) → s(log(half(s(x1))))
half(0(x1)) → 0(s(s(half(x1))))
half(s(0(x1))) → 0(x1)
half(s(s(x1))) → s(half(p(s(s(x1)))))
half(half(s(s(s(s(x1)))))) → s(s(half(half(x1))))
p(s(s(s(x1)))) → s(p(s(s(x1))))
s(s(p(s(x1)))) → s(s(x1))
0(x1) → x1

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