Termination w.r.t. Q proof of /tmp/tmpxQoHQB/#4.29.trs

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:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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:

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
HALF(s(s(x))) → HALF(x)
EVEN(s(s(x))) → EVEN(x)
IF_TIMES(true, s(x), y) → PLUS(times(half(s(x)), y), times(half(s(x)), y))
TIMES(s(x), y) → EVEN(s(x))
IF_TIMES(false, s(x), y) → TIMES(x, y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(true, s(x), y) → HALF(s(x))
IF_TIMES(false, s(x), y) → PLUS(y, times(x, y))
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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:

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
HALF(s(s(x))) → HALF(x)
EVEN(s(s(x))) → EVEN(x)
IF_TIMES(true, s(x), y) → PLUS(times(half(s(x)), y), times(half(s(x)), y))
TIMES(s(x), y) → EVEN(s(x))
IF_TIMES(false, s(x), y) → TIMES(x, y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(true, s(x), y) → HALF(s(x))
IF_TIMES(false, s(x), y) → PLUS(y, times(x, y))
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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(s(x), y) → PLUS(x, y)

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

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

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ 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:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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:

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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(s(s(x))) → HALF(x)

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

POL(HALF(x₁)) = (2)x_1
POL(s(x₁)) = 1 + x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ 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:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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:

EVEN(s(s(x))) → EVEN(x)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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.

EVEN(s(s(x))) → EVEN(x)

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

POL(EVEN(x₁)) = (2)x_1
POL(s(x₁)) = 1 + x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(false, s(x), y) → TIMES(x, y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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.

IF_TIMES(false, s(x), y) → TIMES(x, y)

The remaining pairs can at least be oriented weakly.

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)

Used ordering: Polynomial interpretation [25,35]:

POL(TIMES(x₁, x₂)) = (4)x_1
POL(half(x₁)) = x_1
POL(IF_TIMES(x₁, x₂, x₃)) = (4)x_2
POL(even(x₁)) = 0
POL(true) = 0
POL(false) = 0
POL(s(x₁)) = 4 + x_1
POL(0) = 0

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

half(0) → 0
half(s(s(x))) → s(half(x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

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