Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/AG01SLASHHASH3.6.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:

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))
gcd₂(0, y) -> y
gcd₂(s₁(x), 0) -> s₁(x)
gcd₂(s₁(x), s₁(y)) -> if_gcd₃(le₂(y, x), s₁(x), s₁(y))
if_gcd₃(true, s₁(x), s₁(y)) -> gcd₂(minus₂(x, y), s₁(y))
if_gcd₃(false, s₁(x), s₁(y)) -> gcd₂(minus₂(y, x), 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))
gcd₂(0, y) -> y
gcd₂(s₁(x), 0) -> s₁(x)
gcd₂(s₁(x), s₁(y)) -> if_gcd₃(le₂(y, x), s₁(x), s₁(y))
if_gcd₃(true, s₁(x), s₁(y)) -> gcd₂(minus₂(x, y), s₁(y))
if_gcd₃(false, s₁(x), s₁(y)) -> gcd₂(minus₂(y, x), 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))
gcd₂(0, y) -> y
gcd₂(s₁(x), 0) -> s₁(x)
gcd₂(s₁(x), s₁(y)) -> if_gcd₃(le₂(y, x), s₁(x), s₁(y))
if_gcd₃(true, s₁(x), s₁(y)) -> gcd₂(minus₂(x, y), s₁(y))
if_gcd₃(false, s₁(x), s₁(y)) -> gcd₂(minus₂(y, x), 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))
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if_gcd₃(true, s₁(x0), s₁(x1))
if_gcd₃(false, s₁(x0), s₁(x1))

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

IF_GCD₃(false, s₁(x), s₁(y)) -> GCD₂(minus₂(y, x), s₁(x))
IF_GCD₃(true, s₁(x), s₁(y)) -> MINUS₂(x, y)
MINUS₂(x, s₁(y)) -> PRED₁(minus₂(x, y))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
GCD₂(s₁(x), s₁(y)) -> LE₂(y, x)
IF_GCD₃(true, s₁(x), s₁(y)) -> GCD₂(minus₂(x, y), s₁(y))
GCD₂(s₁(x), s₁(y)) -> IF_GCD₃(le₂(y, x), s₁(x), s₁(y))
MINUS₂(x, s₁(y)) -> MINUS₂(x, y)
IF_GCD₃(false, s₁(x), s₁(y)) -> MINUS₂(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))
gcd₂(0, y) -> y
gcd₂(s₁(x), 0) -> s₁(x)
gcd₂(s₁(x), s₁(y)) -> if_gcd₃(le₂(y, x), s₁(x), s₁(y))
if_gcd₃(true, s₁(x), s₁(y)) -> gcd₂(minus₂(x, y), s₁(y))
if_gcd₃(false, s₁(x), s₁(y)) -> gcd₂(minus₂(y, x), 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))
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if_gcd₃(true, s₁(x0), s₁(x1))
if_gcd₃(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:

IF_GCD₃(false, s₁(x), s₁(y)) -> GCD₂(minus₂(y, x), s₁(x))
IF_GCD₃(true, s₁(x), s₁(y)) -> MINUS₂(x, y)
MINUS₂(x, s₁(y)) -> PRED₁(minus₂(x, y))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
GCD₂(s₁(x), s₁(y)) -> LE₂(y, x)
IF_GCD₃(true, s₁(x), s₁(y)) -> GCD₂(minus₂(x, y), s₁(y))
GCD₂(s₁(x), s₁(y)) -> IF_GCD₃(le₂(y, x), s₁(x), s₁(y))
MINUS₂(x, s₁(y)) -> MINUS₂(x, y)
IF_GCD₃(false, s₁(x), s₁(y)) -> MINUS₂(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))
gcd₂(0, y) -> y
gcd₂(s₁(x), 0) -> s₁(x)
gcd₂(s₁(x), s₁(y)) -> if_gcd₃(le₂(y, x), s₁(x), s₁(y))
if_gcd₃(true, s₁(x), s₁(y)) -> gcd₂(minus₂(x, y), s₁(y))
if_gcd₃(false, s₁(x), s₁(y)) -> gcd₂(minus₂(y, x), 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))
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if_gcd₃(true, s₁(x0), s₁(x1))
if_gcd₃(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 4 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))
gcd₂(0, y) -> y
gcd₂(s₁(x), 0) -> s₁(x)
gcd₂(s₁(x), s₁(y)) -> if_gcd₃(le₂(y, x), s₁(x), s₁(y))
if_gcd₃(true, s₁(x), s₁(y)) -> gcd₂(minus₂(x, y), s₁(y))
if_gcd₃(false, s₁(x), s₁(y)) -> gcd₂(minus₂(y, x), 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))
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if_gcd₃(true, s₁(x0), s₁(x1))
if_gcd₃(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: 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))
gcd₂(0, y) -> y
gcd₂(s₁(x), 0) -> s₁(x)
gcd₂(s₁(x), s₁(y)) -> if_gcd₃(le₂(y, x), s₁(x), s₁(y))
if_gcd₃(true, s₁(x), s₁(y)) -> gcd₂(minus₂(x, y), s₁(y))
if_gcd₃(false, s₁(x), s₁(y)) -> gcd₂(minus₂(y, x), 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))
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if_gcd₃(true, s₁(x0), s₁(x1))
if_gcd₃(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))
gcd₂(0, y) -> y
gcd₂(s₁(x), 0) -> s₁(x)
gcd₂(s₁(x), s₁(y)) -> if_gcd₃(le₂(y, x), s₁(x), s₁(y))
if_gcd₃(true, s₁(x), s₁(y)) -> gcd₂(minus₂(x, y), s₁(y))
if_gcd₃(false, s₁(x), s₁(y)) -> gcd₂(minus₂(y, x), 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))
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if_gcd₃(true, s₁(x0), s₁(x1))
if_gcd₃(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: 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))
gcd₂(0, y) -> y
gcd₂(s₁(x), 0) -> s₁(x)
gcd₂(s₁(x), s₁(y)) -> if_gcd₃(le₂(y, x), s₁(x), s₁(y))
if_gcd₃(true, s₁(x), s₁(y)) -> gcd₂(minus₂(x, y), s₁(y))
if_gcd₃(false, s₁(x), s₁(y)) -> gcd₂(minus₂(y, x), 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))
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if_gcd₃(true, s₁(x0), s₁(x1))
if_gcd₃(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

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

IF_GCD₃(false, s₁(x), s₁(y)) -> GCD₂(minus₂(y, x), s₁(x))
IF_GCD₃(true, s₁(x), s₁(y)) -> GCD₂(minus₂(x, y), s₁(y))
GCD₂(s₁(x), s₁(y)) -> IF_GCD₃(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))
gcd₂(0, y) -> y
gcd₂(s₁(x), 0) -> s₁(x)
gcd₂(s₁(x), s₁(y)) -> if_gcd₃(le₂(y, x), s₁(x), s₁(y))
if_gcd₃(true, s₁(x), s₁(y)) -> gcd₂(minus₂(x, y), s₁(y))
if_gcd₃(false, s₁(x), s₁(y)) -> gcd₂(minus₂(y, x), 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))
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if_gcd₃(true, s₁(x0), s₁(x1))
if_gcd₃(false, s₁(x0), s₁(x1))

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