Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/AG01SLASHHASH3.8b.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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))
MINUS₂(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)
MINUS₂(s₁(x), y) -> LE₂(s₁(x), y)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
LOG₁(s₁(s₁(x))) -> QUOT₂(x, s₁(s₁(0)))
LOG₁(s₁(s₁(x))) -> LOG₁(s₁(quot₂(x, s₁(s₁(0)))))
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
IF_MINUS₃(false, s₁(x), y) -> MINUS₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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:

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))
MINUS₂(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)
MINUS₂(s₁(x), y) -> LE₂(s₁(x), y)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
LOG₁(s₁(s₁(x))) -> QUOT₂(x, s₁(s₁(0)))
LOG₁(s₁(s₁(x))) -> LOG₁(s₁(quot₂(x, s₁(s₁(0)))))
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
IF_MINUS₃(false, s₁(x), y) -> MINUS₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 3 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

Used argument filtering: LE₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

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

MINUS₂(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)
IF_MINUS₃(false, s₁(x), y) -> MINUS₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IF_MINUS₃(false, s₁(x), y) -> MINUS₂(x, y)

Used argument filtering: MINUS₂(x1, x2) = x1
s₁(x1) = s₁(x1)
IF_MINUS₃(x1, x2, x3) = x2
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

              ↳ QDP

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

MINUS₂(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: QUOT₂(x1, x2) = x1
s₁(x1) = s₁(x1)
minus₂(x1, x2) = x1
0 = 0
if_minus₃(x1, x2, x3) = x2
Used ordering: Quasi Precedence: s_1 > 0

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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

                ↳ QDPAfsSolverProof

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

LOG₁(s₁(s₁(x))) -> LOG₁(s₁(quot₂(x, s₁(s₁(0)))))

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LOG₁(s₁(s₁(x))) -> LOG₁(s₁(quot₂(x, s₁(s₁(0)))))

Used argument filtering: LOG₁(x1) = x1
s₁(x1) = s₁(x1)
quot₂(x1, x2) = x1
0 = 0
minus₂(x1, x2) = x1
if_minus₃(x1, x2, x3) = x2
Used ordering: Quasi Precedence: s_1 > 0

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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