Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/AG01SLASHHASH3.40.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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> plus₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
plus₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> plus₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> plus₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
plus₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> plus₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))

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:

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))
PLUS₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> PLUS₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
PLUS₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> PLUS₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> plus₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
plus₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> plus₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))

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:

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))
PLUS₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> PLUS₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
PLUS₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> PLUS₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> plus₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
plus₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> plus₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
PLUS₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> PLUS₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
PLUS₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> PLUS₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> plus₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
plus₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> plus₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> plus₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
plus₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> plus₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))

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 strictly oriented and are deleted.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS₂(x1, x2) = MINUS₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ 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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> plus₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
plus₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> plus₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))

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

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

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> plus₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
plus₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> plus₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))

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 strictly oriented and are deleted.

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOT₂(x1, x2) = x1
s₁(x1) = s₁(x1)
minus₂(x1, x2) = x1
0 = 0

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(minus₂(x, s₁(0)), minus₂(y, s₁(s₁(z)))) -> plus₂(minus₂(y, s₁(s₁(z))), minus₂(x, s₁(0)))
plus₂(plus₂(x, s₁(0)), plus₂(y, s₁(s₁(z)))) -> plus₂(plus₂(y, s₁(s₁(z))), plus₂(x, s₁(0)))

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.