Termination w.r.t. Q proof of TRS/HMt001.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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

Q is empty.

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

F₃(x, s₁(y), z) -> ODD₁(s₁(y))
POW₂(x, y) -> F₃(x, y, s₁(0))
F₃(x, s₁(y), z) -> *¹₂(x, x)
F₃(x, s₁(y), z) -> F₃(*₂(x, x), half₁(s₁(y)), z)
*¹₂(x, s₁(y)) -> *¹₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
F₃(x, s₁(y), z) -> F₃(x, y, *₂(x, z))
F₃(x, s₁(y), z) -> HALF₁(s₁(y))
F₃(x, s₁(y), z) -> *¹₂(x, z)
ODD₁(s₁(s₁(x))) -> ODD₁(x)
F₃(x, s₁(y), z) -> IF₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))
HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

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:

F₃(x, s₁(y), z) -> ODD₁(s₁(y))
POW₂(x, y) -> F₃(x, y, s₁(0))
F₃(x, s₁(y), z) -> *¹₂(x, x)
F₃(x, s₁(y), z) -> F₃(*₂(x, x), half₁(s₁(y)), z)
*¹₂(x, s₁(y)) -> *¹₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
F₃(x, s₁(y), z) -> F₃(x, y, *₂(x, z))
F₃(x, s₁(y), z) -> HALF₁(s₁(y))
F₃(x, s₁(y), z) -> *¹₂(x, z)
ODD₁(s₁(s₁(x))) -> ODD₁(x)
F₃(x, s₁(y), z) -> IF₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))
HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

HALF₁(s₁(s₁(x))) -> HALF₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(HALF₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] 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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

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:

ODD₁(s₁(s₁(x))) -> ODD₁(x)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ODD₁(s₁(s₁(x))) -> ODD₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ODD₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] 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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

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:

*¹₂(x, s₁(y)) -> *¹₂(x, y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

*¹₂(x, s₁(y)) -> *¹₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*¹₂(x₁, x₂)) = 2·x₂
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] 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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

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:

F₃(x, s₁(y), z) -> F₃(*₂(x, x), half₁(s₁(y)), z)
F₃(x, s₁(y), z) -> F₃(x, y, *₂(x, z))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

F₃(x, s₁(y), z) -> F₃(x, y, *₂(x, z))

The remaining pairs can at least be oriented weakly.

F₃(x, s₁(y), z) -> F₃(*₂(x, x), half₁(s₁(y)), z)

Used ordering: Polynomial interpretation [21]:

POL(*₂(x₁, x₂)) = x₂
POL(+₂(x₁, x₂)) = 1
POL(0) = 0
POL(F₃(x₁, x₂, x₃)) = 2·x₁ + x₂ + x₃
POL(half₁(x₁)) = x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented:

*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
*₂(x, 0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

          ↳ QDP

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

F₃(x, s₁(y), z) -> F₃(*₂(x, x), half₁(s₁(y)), z)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(-¹₂(x₁, x₂)) = 2·x₁ + x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(*₂(x, y), x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
odd₁(0) -> false
odd₁(s₁(0)) -> true
odd₁(s₁(s₁(x))) -> odd₁(x)
half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
if₃(true, x, y) -> true
if₃(false, x, y) -> false
pow₂(x, y) -> f₃(x, y, s₁(0))
f₃(x, 0, z) -> z
f₃(x, s₁(y), z) -> if₃(odd₁(s₁(y)), f₃(x, y, *₂(x, z)), f₃(*₂(x, x), half₁(s₁(y)), z))

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.