Termination w.r.t. Q proof of TRS/secret05aprove4.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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

LE₂(p₁(s₁(x)), x) -> LE₂(x, x)

The remaining pairs can at least be oriented weakly.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( LE₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( p₁(x₁) ) = x₁ + 1

POL( s₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ 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:

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( LE₂(x₁, x₂) ) = x₂

POL( s₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

P₁(p₁(s₁(x))) -> P₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( P₁(x₁) ) = max{0, x₁ - 1}

POL( p₁(x₁) ) = x₁ + 1

POL( s₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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.