Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/secret2005SLASHaprove4.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) -> s₁(s₁(0))
p₁(s₁(x)) -> x
p₁(p₁(s₁(x))) -> p₁(x)
le₂(p₁(s₁(x)), x) -> le₂(x, x)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(x, y) -> if₃(le₂(x, y), x, y)
if₃(true, x, y) -> 0
if₃(false, x, y) -> s₁(minus₂(p₁(x), y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

LE₂(p₁(s₁(x)), x) -> LE₂(x, x)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The TRS R consists of the following rules:

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

Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

LE₂(p₁(s₁(x)), x) -> LE₂(x, x)

The TRS R consists of the following rules:

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

Q is empty.
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(p₁(s₁(x)), x) -> LE₂(x, x)

Used argument filtering: LE₂(x1, x2) = x1
p₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

P₁(p₁(s₁(x))) -> P₁(x)

The TRS R consists of the following rules:

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

Q is empty.
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.

P₁(p₁(s₁(x))) -> P₁(x)

Used argument filtering: P₁(x1) = x1
p₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ 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:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS₂(x, y) -> IF₃(le₂(x, y), x, y)
IF₃(false, x, y) -> MINUS₂(p₁(x), y)

The TRS R consists of the following rules:

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

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