Termination w.r.t. Q proof of TRS/secret05aprove5.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, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(x, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

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:

DIV₂(plus₂(x, y), z) -> PLUS₂(div₂(x, z), div₂(y, z))
DIV₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
MINUS₂(s₁(x), s₁(y)) -> P₁(s₁(x))
MINUS₂(s₁(x), s₁(y)) -> P₁(s₁(y))
MINUS₂(x, plus₂(y, z)) -> MINUS₂(minus₂(x, y), z)
PLUS₂(s₁(x), y) -> PLUS₂(y, minus₂(s₁(x), s₁(0)))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(p₁(s₁(x)), p₁(s₁(y)))
P₁(s₁(s₁(x))) -> P₁(s₁(x))
DIV₂(s₁(x), s₁(y)) -> DIV₂(minus₂(x, y), s₁(y))
MINUS₂(x, plus₂(y, z)) -> MINUS₂(x, y)
PLUS₂(s₁(x), y) -> MINUS₂(s₁(x), s₁(0))
DIV₂(plus₂(x, y), z) -> DIV₂(x, z)
DIV₂(plus₂(x, y), z) -> DIV₂(y, z)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

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:

DIV₂(plus₂(x, y), z) -> PLUS₂(div₂(x, z), div₂(y, z))
DIV₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
MINUS₂(s₁(x), s₁(y)) -> P₁(s₁(x))
MINUS₂(s₁(x), s₁(y)) -> P₁(s₁(y))
MINUS₂(x, plus₂(y, z)) -> MINUS₂(minus₂(x, y), z)
PLUS₂(s₁(x), y) -> PLUS₂(y, minus₂(s₁(x), s₁(0)))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(p₁(s₁(x)), p₁(s₁(y)))
P₁(s₁(s₁(x))) -> P₁(s₁(x))
DIV₂(s₁(x), s₁(y)) -> DIV₂(minus₂(x, y), s₁(y))
MINUS₂(x, plus₂(y, z)) -> MINUS₂(x, y)
PLUS₂(s₁(x), y) -> MINUS₂(s₁(x), s₁(0))
DIV₂(plus₂(x, y), z) -> DIV₂(x, z)
DIV₂(plus₂(x, y), z) -> DIV₂(y, z)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

P₁(s₁(s₁(x))) -> P₁(s₁(x))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

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(s₁(s₁(x))) -> P₁(s₁(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(P₁(x₁)) = 2·x₁ + 3·x₁²
POL(s₁(x₁)) = 3 + 3·x₁ + 3·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus₂(x, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

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

          ↳ QDP

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

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(p₁(s₁(x)), p₁(s₁(y)))
MINUS₂(x, plus₂(y, z)) -> MINUS₂(x, y)
MINUS₂(x, plus₂(y, z)) -> MINUS₂(minus₂(x, y), z)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

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.

MINUS₂(x, plus₂(y, z)) -> MINUS₂(x, y)
MINUS₂(x, plus₂(y, z)) -> MINUS₂(minus₂(x, y), z)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(0) = 2
POL(MINUS₂(x₁, x₂)) = x₁ + 3·x₂
POL(minus₂(x₁, x₂)) = 2 + x₁ + 3·x₂
POL(p₁(x₁)) = 0
POL(plus₂(x₁, x₂)) = 3 + 2·x₁ + 3·x₁·x₂ + 2·x₂
POL(s₁(x₁)) = 3·x₁ + 3·x₁²

The following usable rules [14] were oriented:

minus₂(0, y) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, 0) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(y, minus₂(s₁(x), s₁(0)))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

DIV₂(s₁(x), s₁(y)) -> DIV₂(minus₂(x, y), s₁(y))
DIV₂(plus₂(x, y), z) -> DIV₂(x, z)
DIV₂(plus₂(x, y), z) -> DIV₂(y, z)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

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.

DIV₂(s₁(x), s₁(y)) -> DIV₂(minus₂(x, y), s₁(y))
DIV₂(plus₂(x, y), z) -> DIV₂(x, z)
DIV₂(plus₂(x, y), z) -> DIV₂(y, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 1
POL(DIV₂(x₁, x₂)) = 2·x₁ + x₁·x₂
POL(minus₂(x₁, x₂)) = x₁
POL(p₁(x₁)) = x₁
POL(plus₂(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented:

minus₂(0, y) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, 0) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

minus₂(x, 0) -> x
minus₂(0, y) -> 0
minus₂(s₁(x), s₁(y)) -> minus₂(p₁(s₁(x)), p₁(s₁(y)))
minus₂(x, plus₂(y, z)) -> minus₂(minus₂(x, y), z)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(0) -> s₁(s₁(0))
div₂(s₁(x), s₁(y)) -> s₁(div₂(minus₂(x, y), s₁(y)))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(y, minus₂(s₁(x), s₁(0))))

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.