Termination w.r.t. Q proof of TRS/nontermin/AG01#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:

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(x, y))

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:

EVEN₁(s₁(s₁(x))) -> EVEN₁(x)
TIMES₂(s₁(x), y) -> IF_TIMES₃(even₁(s₁(x)), s₁(x), y)
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
IF_TIMES₃(true, s₁(x), y) -> TIMES₂(half₁(s₁(x)), y)
IF_TIMES₃(true, s₁(x), y) -> HALF₁(s₁(x))
IF_TIMES₃(true, s₁(x), y) -> PLUS₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
IF_TIMES₃(false, s₁(x), y) -> PLUS₂(y, times₂(x, y))
TIMES₂(s₁(x), y) -> EVEN₁(s₁(x))
IF_TIMES₃(false, s₁(x), y) -> TIMES₂(x, y)
HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(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:

EVEN₁(s₁(s₁(x))) -> EVEN₁(x)
TIMES₂(s₁(x), y) -> IF_TIMES₃(even₁(s₁(x)), s₁(x), y)
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
IF_TIMES₃(true, s₁(x), y) -> TIMES₂(half₁(s₁(x)), y)
IF_TIMES₃(true, s₁(x), y) -> HALF₁(s₁(x))
IF_TIMES₃(true, s₁(x), y) -> PLUS₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
IF_TIMES₃(false, s₁(x), y) -> PLUS₂(y, times₂(x, y))
TIMES₂(s₁(x), y) -> EVEN₁(s₁(x))
IF_TIMES₃(false, s₁(x), y) -> TIMES₂(x, y)
HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] 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:

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The TRS R consists of the following rules:

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(x, y))

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.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

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

POL(PLUS₂(x₁, x₂)) = 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

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

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(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:

HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(x, y))

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

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(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:

EVEN₁(s₁(s₁(x))) -> EVEN₁(x)

The TRS R consists of the following rules:

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(x, y))

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.

EVEN₁(s₁(s₁(x))) -> EVEN₁(x)

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

POL(EVEN₁(x₁)) = 2·x₁
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

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

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(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:

TIMES₂(s₁(x), y) -> IF_TIMES₃(even₁(s₁(x)), s₁(x), y)
IF_TIMES₃(true, s₁(x), y) -> TIMES₂(half₁(s₁(x)), y)
IF_TIMES₃(false, s₁(x), y) -> TIMES₂(x, y)

The TRS R consists of the following rules:

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(x, y))

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.

IF_TIMES₃(false, s₁(x), y) -> TIMES₂(x, y)

The remaining pairs can at least be oriented weakly.

TIMES₂(s₁(x), y) -> IF_TIMES₃(even₁(s₁(x)), s₁(x), y)
IF_TIMES₃(true, s₁(x), y) -> TIMES₂(half₁(s₁(x)), y)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(IF_TIMES₃(x₁, x₂, x₃)) = 2·x₂
POL(TIMES₂(x₁, x₂)) = 2·x₁
POL(even₁(x₁)) = 0
POL(false) = 0
POL(half₁(x₁)) = x₁
POL(s₁(x₁)) = 2 + x₁
POL(true) = 0

The following usable rules [14] were oriented:

half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

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

TIMES₂(s₁(x), y) -> IF_TIMES₃(even₁(s₁(x)), s₁(x), y)
IF_TIMES₃(true, s₁(x), y) -> TIMES₂(half₁(s₁(x)), y)

The TRS R consists of the following rules:

even₁(0) -> true
even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
half₁(0) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> if_times₃(even₁(s₁(x)), s₁(x), y)
if_times₃(true, s₁(x), y) -> plus₂(times₂(half₁(s₁(x)), y), times₂(half₁(s₁(x)), y))
if_times₃(false, s₁(x), y) -> plus₂(y, times₂(x, y))

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