Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/RubioSLASHgcd.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, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(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:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(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:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

The set Q consists of the following terms:

minus₂(x0, s₁(x1))
minus₂(x0, 0)
pred₁(s₁(x0))
le₂(s₁(x0), s₁(x1))
le₂(s₁(x0), 0)
le₂(0, x0)
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if₃(true, s₁(x0), s₁(x1))
if₃(false, s₁(x0), s₁(x1))

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

IF₃(false, s₁(X), s₁(Y)) -> MINUS₂(Y, X)
IF₃(true, s₁(X), s₁(Y)) -> GCD₂(minus₂(X, Y), s₁(Y))
IF₃(false, s₁(X), s₁(Y)) -> GCD₂(minus₂(Y, X), s₁(X))
MINUS₂(X, s₁(Y)) -> PRED₁(minus₂(X, Y))
LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)
GCD₂(s₁(X), s₁(Y)) -> IF₃(le₂(Y, X), s₁(X), s₁(Y))
GCD₂(s₁(X), s₁(Y)) -> LE₂(Y, X)
IF₃(true, s₁(X), s₁(Y)) -> MINUS₂(X, Y)
MINUS₂(X, s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

The set Q consists of the following terms:

minus₂(x0, s₁(x1))
minus₂(x0, 0)
pred₁(s₁(x0))
le₂(s₁(x0), s₁(x1))
le₂(s₁(x0), 0)
le₂(0, x0)
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if₃(true, s₁(x0), s₁(x1))
if₃(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:

IF₃(false, s₁(X), s₁(Y)) -> MINUS₂(Y, X)
IF₃(true, s₁(X), s₁(Y)) -> GCD₂(minus₂(X, Y), s₁(Y))
IF₃(false, s₁(X), s₁(Y)) -> GCD₂(minus₂(Y, X), s₁(X))
MINUS₂(X, s₁(Y)) -> PRED₁(minus₂(X, Y))
LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)
GCD₂(s₁(X), s₁(Y)) -> IF₃(le₂(Y, X), s₁(X), s₁(Y))
GCD₂(s₁(X), s₁(Y)) -> LE₂(Y, X)
IF₃(true, s₁(X), s₁(Y)) -> MINUS₂(X, Y)
MINUS₂(X, s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

The set Q consists of the following terms:

minus₂(x0, s₁(x1))
minus₂(x0, 0)
pred₁(s₁(x0))
le₂(s₁(x0), s₁(x1))
le₂(s₁(x0), 0)
le₂(0, x0)
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if₃(true, s₁(x0), s₁(x1))
if₃(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:

LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

The set Q consists of the following terms:

minus₂(x0, s₁(x1))
minus₂(x0, 0)
pred₁(s₁(x0))
le₂(s₁(x0), s₁(x1))
le₂(s₁(x0), 0)
le₂(0, x0)
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if₃(true, s₁(x0), s₁(x1))
if₃(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

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

The set Q consists of the following terms:

minus₂(x0, s₁(x1))
minus₂(x0, 0)
pred₁(s₁(x0))
le₂(s₁(x0), s₁(x1))
le₂(s₁(x0), 0)
le₂(0, x0)
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if₃(true, s₁(x0), s₁(x1))
if₃(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:

MINUS₂(X, s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

The set Q consists of the following terms:

minus₂(x0, s₁(x1))
minus₂(x0, 0)
pred₁(s₁(x0))
le₂(s₁(x0), s₁(x1))
le₂(s₁(x0), 0)
le₂(0, x0)
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if₃(true, s₁(x0), s₁(x1))
if₃(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

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

The set Q consists of the following terms:

minus₂(x0, s₁(x1))
minus₂(x0, 0)
pred₁(s₁(x0))
le₂(s₁(x0), s₁(x1))
le₂(s₁(x0), 0)
le₂(0, x0)
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if₃(true, s₁(x0), s₁(x1))
if₃(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:

IF₃(true, s₁(X), s₁(Y)) -> GCD₂(minus₂(X, Y), s₁(Y))
IF₃(false, s₁(X), s₁(Y)) -> GCD₂(minus₂(Y, X), s₁(X))
GCD₂(s₁(X), s₁(Y)) -> IF₃(le₂(Y, X), s₁(X), s₁(Y))

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

The set Q consists of the following terms:

minus₂(x0, s₁(x1))
minus₂(x0, 0)
pred₁(s₁(x0))
le₂(s₁(x0), s₁(x1))
le₂(s₁(x0), 0)
le₂(0, x0)
gcd₂(0, x0)
gcd₂(s₁(x0), 0)
gcd₂(s₁(x0), s₁(x1))
if₃(true, s₁(x0), s₁(x1))
if₃(false, s₁(x0), s₁(x1))

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