Termination w.r.t. Q proof of TRS/TRCSREx49_GM04

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:

a__minus₂(0, Y) -> 0
a__minus₂(s₁(X), s₁(Y)) -> a__minus₂(X, Y)
a__geq₂(X, 0) -> true
a__geq₂(0, s₁(Y)) -> false
a__geq₂(s₁(X), s₁(Y)) -> a__geq₂(X, Y)
a__div₂(0, s₁(Y)) -> 0
a__div₂(s₁(X), s₁(Y)) -> a__if₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(minus₂(X1, X2)) -> a__minus₂(X1, X2)
mark₁(geq₂(X1, X2)) -> a__geq₂(X1, X2)
mark₁(div₂(X1, X2)) -> a__div₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__minus₂(X1, X2) -> minus₂(X1, X2)
a__geq₂(X1, X2) -> geq₂(X1, X2)
a__div₂(X1, X2) -> div₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__minus₂(0, Y) -> 0
a__minus₂(s₁(X), s₁(Y)) -> a__minus₂(X, Y)
a__geq₂(X, 0) -> true
a__geq₂(0, s₁(Y)) -> false
a__geq₂(s₁(X), s₁(Y)) -> a__geq₂(X, Y)
a__div₂(0, s₁(Y)) -> 0
a__div₂(s₁(X), s₁(Y)) -> a__if₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(minus₂(X1, X2)) -> a__minus₂(X1, X2)
mark₁(geq₂(X1, X2)) -> a__geq₂(X1, X2)
mark₁(div₂(X1, X2)) -> a__div₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__minus₂(X1, X2) -> minus₂(X1, X2)
a__geq₂(X1, X2) -> geq₂(X1, X2)
a__div₂(X1, X2) -> div₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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:

MARK₁(geq₂(X1, X2)) -> A__GEQ₂(X1, X2)
A__GEQ₂(s₁(X), s₁(Y)) -> A__GEQ₂(X, Y)
A__IF₃(false, X, Y) -> MARK₁(Y)
MARK₁(minus₂(X1, X2)) -> A__MINUS₂(X1, X2)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(s₁(X)) -> MARK₁(X)
A__DIV₂(s₁(X), s₁(Y)) -> A__IF₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
A__DIV₂(s₁(X), s₁(Y)) -> A__GEQ₂(X, Y)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(div₂(X1, X2)) -> A__DIV₂(mark₁(X1), X2)
MARK₁(div₂(X1, X2)) -> MARK₁(X1)
A__MINUS₂(s₁(X), s₁(Y)) -> A__MINUS₂(X, Y)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)

The TRS R consists of the following rules:

a__minus₂(0, Y) -> 0
a__minus₂(s₁(X), s₁(Y)) -> a__minus₂(X, Y)
a__geq₂(X, 0) -> true
a__geq₂(0, s₁(Y)) -> false
a__geq₂(s₁(X), s₁(Y)) -> a__geq₂(X, Y)
a__div₂(0, s₁(Y)) -> 0
a__div₂(s₁(X), s₁(Y)) -> a__if₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(minus₂(X1, X2)) -> a__minus₂(X1, X2)
mark₁(geq₂(X1, X2)) -> a__geq₂(X1, X2)
mark₁(div₂(X1, X2)) -> a__div₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__minus₂(X1, X2) -> minus₂(X1, X2)
a__geq₂(X1, X2) -> geq₂(X1, X2)
a__div₂(X1, X2) -> div₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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:

MARK₁(geq₂(X1, X2)) -> A__GEQ₂(X1, X2)
A__GEQ₂(s₁(X), s₁(Y)) -> A__GEQ₂(X, Y)
A__IF₃(false, X, Y) -> MARK₁(Y)
MARK₁(minus₂(X1, X2)) -> A__MINUS₂(X1, X2)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(s₁(X)) -> MARK₁(X)
A__DIV₂(s₁(X), s₁(Y)) -> A__IF₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
A__DIV₂(s₁(X), s₁(Y)) -> A__GEQ₂(X, Y)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(div₂(X1, X2)) -> A__DIV₂(mark₁(X1), X2)
MARK₁(div₂(X1, X2)) -> MARK₁(X1)
A__MINUS₂(s₁(X), s₁(Y)) -> A__MINUS₂(X, Y)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)

The TRS R consists of the following rules:

a__minus₂(0, Y) -> 0
a__minus₂(s₁(X), s₁(Y)) -> a__minus₂(X, Y)
a__geq₂(X, 0) -> true
a__geq₂(0, s₁(Y)) -> false
a__geq₂(s₁(X), s₁(Y)) -> a__geq₂(X, Y)
a__div₂(0, s₁(Y)) -> 0
a__div₂(s₁(X), s₁(Y)) -> a__if₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(minus₂(X1, X2)) -> a__minus₂(X1, X2)
mark₁(geq₂(X1, X2)) -> a__geq₂(X1, X2)
mark₁(div₂(X1, X2)) -> a__div₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__minus₂(X1, X2) -> minus₂(X1, X2)
a__geq₂(X1, X2) -> geq₂(X1, X2)
a__div₂(X1, X2) -> div₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

A__GEQ₂(s₁(X), s₁(Y)) -> A__GEQ₂(X, Y)

The TRS R consists of the following rules:

a__minus₂(0, Y) -> 0
a__minus₂(s₁(X), s₁(Y)) -> a__minus₂(X, Y)
a__geq₂(X, 0) -> true
a__geq₂(0, s₁(Y)) -> false
a__geq₂(s₁(X), s₁(Y)) -> a__geq₂(X, Y)
a__div₂(0, s₁(Y)) -> 0
a__div₂(s₁(X), s₁(Y)) -> a__if₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(minus₂(X1, X2)) -> a__minus₂(X1, X2)
mark₁(geq₂(X1, X2)) -> a__geq₂(X1, X2)
mark₁(div₂(X1, X2)) -> a__div₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__minus₂(X1, X2) -> minus₂(X1, X2)
a__geq₂(X1, X2) -> geq₂(X1, X2)
a__div₂(X1, X2) -> div₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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.

A__GEQ₂(s₁(X), s₁(Y)) -> A__GEQ₂(X, Y)

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

POL( A__GEQ₂(x₁, x₂) ) = max{0, x₂ - 2}

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

a__minus₂(0, Y) -> 0
a__minus₂(s₁(X), s₁(Y)) -> a__minus₂(X, Y)
a__geq₂(X, 0) -> true
a__geq₂(0, s₁(Y)) -> false
a__geq₂(s₁(X), s₁(Y)) -> a__geq₂(X, Y)
a__div₂(0, s₁(Y)) -> 0
a__div₂(s₁(X), s₁(Y)) -> a__if₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(minus₂(X1, X2)) -> a__minus₂(X1, X2)
mark₁(geq₂(X1, X2)) -> a__geq₂(X1, X2)
mark₁(div₂(X1, X2)) -> a__div₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__minus₂(X1, X2) -> minus₂(X1, X2)
a__geq₂(X1, X2) -> geq₂(X1, X2)
a__div₂(X1, X2) -> div₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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:

A__MINUS₂(s₁(X), s₁(Y)) -> A__MINUS₂(X, Y)

The TRS R consists of the following rules:

a__minus₂(0, Y) -> 0
a__minus₂(s₁(X), s₁(Y)) -> a__minus₂(X, Y)
a__geq₂(X, 0) -> true
a__geq₂(0, s₁(Y)) -> false
a__geq₂(s₁(X), s₁(Y)) -> a__geq₂(X, Y)
a__div₂(0, s₁(Y)) -> 0
a__div₂(s₁(X), s₁(Y)) -> a__if₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(minus₂(X1, X2)) -> a__minus₂(X1, X2)
mark₁(geq₂(X1, X2)) -> a__geq₂(X1, X2)
mark₁(div₂(X1, X2)) -> a__div₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__minus₂(X1, X2) -> minus₂(X1, X2)
a__geq₂(X1, X2) -> geq₂(X1, X2)
a__div₂(X1, X2) -> div₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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.

A__MINUS₂(s₁(X), s₁(Y)) -> A__MINUS₂(X, Y)

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

POL( A__MINUS₂(x₁, x₂) ) = max{0, x₂ - 2}

POL( s₁(x₁) ) = x₁ + 3

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:

a__minus₂(0, Y) -> 0
a__minus₂(s₁(X), s₁(Y)) -> a__minus₂(X, Y)
a__geq₂(X, 0) -> true
a__geq₂(0, s₁(Y)) -> false
a__geq₂(s₁(X), s₁(Y)) -> a__geq₂(X, Y)
a__div₂(0, s₁(Y)) -> 0
a__div₂(s₁(X), s₁(Y)) -> a__if₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(minus₂(X1, X2)) -> a__minus₂(X1, X2)
mark₁(geq₂(X1, X2)) -> a__geq₂(X1, X2)
mark₁(div₂(X1, X2)) -> a__div₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__minus₂(X1, X2) -> minus₂(X1, X2)
a__geq₂(X1, X2) -> geq₂(X1, X2)
a__div₂(X1, X2) -> div₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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

            ↳ QDPOrderProof

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

MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(s₁(X)) -> MARK₁(X)
A__DIV₂(s₁(X), s₁(Y)) -> A__IF₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
A__IF₃(false, X, Y) -> MARK₁(Y)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(div₂(X1, X2)) -> A__DIV₂(mark₁(X1), X2)
MARK₁(div₂(X1, X2)) -> MARK₁(X1)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)

The TRS R consists of the following rules:

a__minus₂(0, Y) -> 0
a__minus₂(s₁(X), s₁(Y)) -> a__minus₂(X, Y)
a__geq₂(X, 0) -> true
a__geq₂(0, s₁(Y)) -> false
a__geq₂(s₁(X), s₁(Y)) -> a__geq₂(X, Y)
a__div₂(0, s₁(Y)) -> 0
a__div₂(s₁(X), s₁(Y)) -> a__if₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(minus₂(X1, X2)) -> a__minus₂(X1, X2)
mark₁(geq₂(X1, X2)) -> a__geq₂(X1, X2)
mark₁(div₂(X1, X2)) -> a__div₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__minus₂(X1, X2) -> minus₂(X1, X2)
a__geq₂(X1, X2) -> geq₂(X1, X2)
a__div₂(X1, X2) -> div₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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.

MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)

The remaining pairs can at least be oriented weakly.

MARK₁(s₁(X)) -> MARK₁(X)
A__DIV₂(s₁(X), s₁(Y)) -> A__IF₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
A__IF₃(false, X, Y) -> MARK₁(Y)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(div₂(X1, X2)) -> A__DIV₂(mark₁(X1), X2)
MARK₁(div₂(X1, X2)) -> MARK₁(X1)

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

POL( MARK₁(x₁) ) = max{0, x₁ - 2}

POL( if₃(x₁, ..., x₃) ) = x₁ + x₂ + x₃ + 3

POL( s₁(x₁) ) = x₁

POL( A__DIV₂(x₁, x₂) ) = 0

POL( A__IF₃(x₁, ..., x₃) ) = x₂ + x₃

POL( div₂(x₁, x₂) ) = x₁

POL( minus₂(x₁, x₂) ) = 0

POL( 0 ) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

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

MARK₁(s₁(X)) -> MARK₁(X)
A__DIV₂(s₁(X), s₁(Y)) -> A__IF₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
A__IF₃(false, X, Y) -> MARK₁(Y)
MARK₁(div₂(X1, X2)) -> A__DIV₂(mark₁(X1), X2)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(div₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__minus₂(0, Y) -> 0
a__minus₂(s₁(X), s₁(Y)) -> a__minus₂(X, Y)
a__geq₂(X, 0) -> true
a__geq₂(0, s₁(Y)) -> false
a__geq₂(s₁(X), s₁(Y)) -> a__geq₂(X, Y)
a__div₂(0, s₁(Y)) -> 0
a__div₂(s₁(X), s₁(Y)) -> a__if₃(a__geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(minus₂(X1, X2)) -> a__minus₂(X1, X2)
mark₁(geq₂(X1, X2)) -> a__geq₂(X1, X2)
mark₁(div₂(X1, X2)) -> a__div₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__minus₂(X1, X2) -> minus₂(X1, X2)
a__geq₂(X1, X2) -> geq₂(X1, X2)
a__div₂(X1, X2) -> div₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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