Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/nonterminSLASHAG01SLASHHASH4.23.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

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:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
plus₂(0, x0)
plus₂(s₁(x0), x1)
quot₃(x0, 0, 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:

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
QUOT₃(x, 0, s₁(z)) -> PLUS₂(z, s₁(0))
QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
plus₂(0, x0)
plus₂(s₁(x0), x1)
quot₃(x0, 0, 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:

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
QUOT₃(x, 0, s₁(z)) -> PLUS₂(z, s₁(0))
QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
plus₂(0, x0)
plus₂(s₁(x0), x1)
quot₃(x0, 0, s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ 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:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
plus₂(0, x0)
plus₂(s₁(x0), x1)
quot₃(x0, 0, 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.

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

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

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
plus₂(0, x0)
plus₂(s₁(x0), x1)
quot₃(x0, 0, 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

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

QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
plus₂(0, x0)
plus₂(s₁(x0), x1)
quot₃(x0, 0, 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.

QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)

The remaining pairs can at least by weakly be oriented.

QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

Used ordering: Combined order from the following AFS and order.
QUOT₃(x1, x2, x3) = x1
s₁(x1) = s₁(x1)
0 = 0
plus₂(x1, x2) = plus₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

0 > plus₂ > s₁

The following usable rules [14] were oriented:

plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
plus₂(0, x0)
plus₂(s₁(x0), x1)
quot₃(x0, 0, 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.

QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOT₃(x1, x2, x3) = x2
0 = 0
s₁(x1) = s
plus₂(x1, x2) = x2

Lexicographic Path Order [19].
Precedence:

0 > s

The following usable rules [14] were oriented:

plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
plus₂(0, x0)
plus₂(s₁(x0), x1)
quot₃(x0, 0, 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.