Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/nonterminSLASHAG01SLASHHASH4.27.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:

p₁(0) -> 0
p₁(s₁(x)) -> x
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> if₃(le₂(x, s₁(y)), 0, p₁(minus₂(x, p₁(s₁(y)))))
if₃(true, x, y) -> x
if₃(false, x, y) -> y

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

p₁(0) -> 0
p₁(s₁(x)) -> x
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> if₃(le₂(x, s₁(y)), 0, p₁(minus₂(x, p₁(s₁(y)))))
if₃(true, x, y) -> x
if₃(false, x, y) -> y

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:

p₁(0) -> 0
p₁(s₁(x)) -> x
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> if₃(le₂(x, s₁(y)), 0, p₁(minus₂(x, p₁(s₁(y)))))
if₃(true, x, y) -> x
if₃(false, x, y) -> y

The set Q consists of the following terms:

p₁(0)
p₁(s₁(x0))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)

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

MINUS₂(x, s₁(y)) -> MINUS₂(x, p₁(s₁(y)))
MINUS₂(x, s₁(y)) -> P₁(minus₂(x, p₁(s₁(y))))
MINUS₂(x, s₁(y)) -> LE₂(x, s₁(y))
MINUS₂(x, s₁(y)) -> IF₃(le₂(x, s₁(y)), 0, p₁(minus₂(x, p₁(s₁(y)))))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
MINUS₂(x, s₁(y)) -> P₁(s₁(y))

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> if₃(le₂(x, s₁(y)), 0, p₁(minus₂(x, p₁(s₁(y)))))
if₃(true, x, y) -> x
if₃(false, x, y) -> y

The set Q consists of the following terms:

p₁(0)
p₁(s₁(x0))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, 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)) -> MINUS₂(x, p₁(s₁(y)))
MINUS₂(x, s₁(y)) -> P₁(minus₂(x, p₁(s₁(y))))
MINUS₂(x, s₁(y)) -> LE₂(x, s₁(y))
MINUS₂(x, s₁(y)) -> IF₃(le₂(x, s₁(y)), 0, p₁(minus₂(x, p₁(s₁(y)))))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
MINUS₂(x, s₁(y)) -> P₁(s₁(y))

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> if₃(le₂(x, s₁(y)), 0, p₁(minus₂(x, p₁(s₁(y)))))
if₃(true, x, y) -> x
if₃(false, x, y) -> y

The set Q consists of the following terms:

p₁(0)
p₁(s₁(x0))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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:

p₁(0) -> 0
p₁(s₁(x)) -> x
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> if₃(le₂(x, s₁(y)), 0, p₁(minus₂(x, p₁(s₁(y)))))
if₃(true, x, y) -> x
if₃(false, x, y) -> y

The set Q consists of the following terms:

p₁(0)
p₁(s₁(x0))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, 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

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

p₁(0) -> 0
p₁(s₁(x)) -> x
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> if₃(le₂(x, s₁(y)), 0, p₁(minus₂(x, p₁(s₁(y)))))
if₃(true, x, y) -> x
if₃(false, x, y) -> y

The set Q consists of the following terms:

p₁(0)
p₁(s₁(x0))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, 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

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

MINUS₂(x, s₁(y)) -> MINUS₂(x, p₁(s₁(y)))

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> if₃(le₂(x, s₁(y)), 0, p₁(minus₂(x, p₁(s₁(y)))))
if₃(true, x, y) -> x
if₃(false, x, y) -> y

The set Q consists of the following terms:

p₁(0)
p₁(s₁(x0))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)

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