Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/AG01SLASHHASH3.5a.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)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

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)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), 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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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

MINUS₂(x, s₁(y)) -> PRED₁(minus₂(x, y))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
IF_MOD₃(true, s₁(x), s₁(y)) -> MINUS₂(x, y)
MOD₂(s₁(x), s₁(y)) -> IF_MOD₃(le₂(y, x), s₁(x), s₁(y))
MINUS₂(x, s₁(y)) -> MINUS₂(x, y)
IF_MOD₃(true, s₁(x), s₁(y)) -> MOD₂(minus₂(x, y), s₁(y))
MOD₂(s₁(x), s₁(y)) -> LE₂(y, x)

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)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, 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:

MINUS₂(x, s₁(y)) -> PRED₁(minus₂(x, y))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
IF_MOD₃(true, s₁(x), s₁(y)) -> MINUS₂(x, y)
MOD₂(s₁(x), s₁(y)) -> IF_MOD₃(le₂(y, x), s₁(x), s₁(y))
MINUS₂(x, s₁(y)) -> MINUS₂(x, y)
IF_MOD₃(true, s₁(x), s₁(y)) -> MOD₂(minus₂(x, y), s₁(y))
MOD₂(s₁(x), s₁(y)) -> LE₂(y, x)

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)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

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

MINUS₂(x, s₁(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)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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.

MINUS₂(x, s₁(y)) -> MINUS₂(x, y)

Used argument filtering: MINUS₂(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

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)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, 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

                ↳ QDPAfsSolverProof

              ↳ 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)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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

              ↳ 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)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, 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

                ↳ QDPAfsSolverProof

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

MOD₂(s₁(x), s₁(y)) -> IF_MOD₃(le₂(y, x), s₁(x), s₁(y))
IF_MOD₃(true, s₁(x), s₁(y)) -> MOD₂(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)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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_MOD₃(true, s₁(x), s₁(y)) -> MOD₂(minus₂(x, y), s₁(y))

Used argument filtering: MOD₂(x1, x2) = x1
s₁(x1) = s₁(x1)
IF_MOD₃(x1, x2, x3) = x2
minus₂(x1, x2) = x1
pred₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

MOD₂(s₁(x), s₁(y)) -> IF_MOD₃(le₂(y, x), s₁(x), 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)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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