Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/nonterminSLASHAG01SLASHHASH4.29.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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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

The following pairs can be strictly oriented and are deleted.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PLUS₂(x1, x2) = PLUS₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > PLUS₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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

The following pairs can be strictly oriented and are deleted.

HALF₁(s₁(s₁(x))) -> HALF₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
HALF₁(x1) = HALF₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > HALF₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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

The following pairs can be strictly oriented and are deleted.

EVEN₁(s₁(s₁(x))) -> EVEN₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
EVEN₁(x1) = EVEN₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > EVEN₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

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: Combined order from the following AFS and order.
TIMES₂(x1, x2) = x1
s₁(x1) = s₁(x1)
IF_TIMES₃(x1, x2, x3) = x2
even₁(x1) = even₁(x1)
true = true
half₁(x1) = x1
false = false
0 = 0

Lexicographic Path Order [19].
Precedence:

even₁ > true
0 > true
0 > false

The following usable rules [14] were oriented:

even₁(s₁(0)) -> false
even₁(s₁(s₁(x))) -> even₁(x)
even₁(0) -> true
half₁(s₁(s₁(x))) -> s₁(half₁(x))
half₁(0) -> 0

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

even₁(0)
even₁(s₁(0))
even₁(s₁(s₁(x0)))
half₁(0)
half₁(s₁(s₁(x0)))
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
if_times₃(true, s₁(x0), x1)
if_times₃(false, s₁(x0), x1)

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