Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/HMSLASHt002.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:

leq₂(0, y) -> true
leq₂(s₁(x), 0) -> false
leq₂(s₁(x), s₁(y)) -> leq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

leq₂(0, y) -> true
leq₂(s₁(x), 0) -> false
leq₂(s₁(x), s₁(y)) -> leq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))

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:

leq₂(0, y) -> true
leq₂(s₁(x), 0) -> false
leq₂(s₁(x), s₁(y)) -> leq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))

The set Q consists of the following terms:

leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(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:

MOD₂(s₁(x), s₁(y)) -> LEQ₂(y, x)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
MOD₂(s₁(x), s₁(y)) -> MOD₂(-₂(s₁(x), s₁(y)), s₁(y))
LEQ₂(s₁(x), s₁(y)) -> LEQ₂(x, y)
MOD₂(s₁(x), s₁(y)) -> IF₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))
MOD₂(s₁(x), s₁(y)) -> -¹₂(s₁(x), s₁(y))

The TRS R consists of the following rules:

leq₂(0, y) -> true
leq₂(s₁(x), 0) -> false
leq₂(s₁(x), s₁(y)) -> leq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))

The set Q consists of the following terms:

leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(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:

MOD₂(s₁(x), s₁(y)) -> LEQ₂(y, x)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
MOD₂(s₁(x), s₁(y)) -> MOD₂(-₂(s₁(x), s₁(y)), s₁(y))
LEQ₂(s₁(x), s₁(y)) -> LEQ₂(x, y)
MOD₂(s₁(x), s₁(y)) -> IF₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))
MOD₂(s₁(x), s₁(y)) -> -¹₂(s₁(x), s₁(y))

The TRS R consists of the following rules:

leq₂(0, y) -> true
leq₂(s₁(x), 0) -> false
leq₂(s₁(x), s₁(y)) -> leq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))

The set Q consists of the following terms:

leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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:

leq₂(0, y) -> true
leq₂(s₁(x), 0) -> false
leq₂(s₁(x), s₁(y)) -> leq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))

The set Q consists of the following terms:

leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(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.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
-¹₂(x1, x2) = x2
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

leq₂(0, y) -> true
leq₂(s₁(x), 0) -> false
leq₂(s₁(x), s₁(y)) -> leq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))

The set Q consists of the following terms:

leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(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

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

LEQ₂(s₁(x), s₁(y)) -> LEQ₂(x, y)

The TRS R consists of the following rules:

leq₂(0, y) -> true
leq₂(s₁(x), 0) -> false
leq₂(s₁(x), s₁(y)) -> leq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))

The set Q consists of the following terms:

leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(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.

LEQ₂(s₁(x), s₁(y)) -> LEQ₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LEQ₂(x1, x2) = x2
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

leq₂(0, y) -> true
leq₂(s₁(x), 0) -> false
leq₂(s₁(x), s₁(y)) -> leq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))

The set Q consists of the following terms:

leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(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

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

MOD₂(s₁(x), s₁(y)) -> MOD₂(-₂(s₁(x), s₁(y)), s₁(y))

The TRS R consists of the following rules:

leq₂(0, y) -> true
leq₂(s₁(x), 0) -> false
leq₂(s₁(x), s₁(y)) -> leq₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if₃(leq₂(y, x), mod₂(-₂(s₁(x), s₁(y)), s₁(y)), s₁(x))

The set Q consists of the following terms:

leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))

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