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

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(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:

MARK(geq(X1, X2)) → A__GEQ(X1, X2)
A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)
A__IF(false, X, Y) → MARK(Y)
MARK(minus(X1, X2)) → A__MINUS(X1, X2)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__DIV(s(X), s(Y)) → A__GEQ(X, Y)
A__IF(true, X, Y) → MARK(X)
MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
MARK(div(X1, X2)) → MARK(X1)
A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

MARK(geq(X1, X2)) → A__GEQ(X1, X2)
A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)
A__IF(false, X, Y) → MARK(Y)
MARK(minus(X1, X2)) → A__MINUS(X1, X2)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__DIV(s(X), s(Y)) → A__GEQ(X, Y)
A__IF(true, X, Y) → MARK(X)
MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
MARK(div(X1, X2)) → MARK(X1)
A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK(geq(X1, X2)) → A__GEQ(X1, X2)
A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)
A__IF(false, X, Y) → MARK(Y)
MARK(minus(X1, X2)) → A__MINUS(X1, X2)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__DIV(s(X), s(Y)) → A__GEQ(X, Y)
MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
A__IF(true, X, Y) → MARK(X)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(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__GEQ(s(X), s(Y)) → A__GEQ(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A__GEQ(x1, x2) = A__GEQ(x1)
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(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__MINUS(s(X), s(Y)) → A__MINUS(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A__MINUS(x1, x2) = A__MINUS(x1)
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__IF(false, X, Y) → MARK(Y)
A__IF(true, X, Y) → MARK(X)
MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(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.

MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__IF(false, X, Y) → MARK(Y)
A__IF(true, X, Y) → MARK(X)
MARK(div(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.

MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)

Used ordering: Combined order from the following AFS and order.
MARK(x1) = x1
if(x1, x2, x3) = if(x1, x2, x3)
s(x1) = s(x1)
A__DIV(x1, x2) = A__DIV(x1, x2)
A__IF(x1, x2, x3) = A__IF(x1, x2, x3)
a__geq(x1, x2) = a__geq(x1)
div(x1, x2) = div(x1, x2)
minus(x1, x2) = x1
0 = 0
false = false
true = true
mark(x1) = x1
a__if(x1, x2, x3) = a__if(x1, x2, x3)
a__div(x1, x2) = a__div(x1, x2)
a__minus(x1, x2) = x1
geq(x1, x2) = geq(x1)

Lexicographic Path Order [19].
Precedence:

[A_{_DIV₂}, div₂, a_{_{div_2]}} > [s₁, a_{_geq₁}, geq_1] > 0 > true > [if₃, A_{_IF₃}, a_{_{if_3]}}
[A_{_DIV₂}, div₂, a_{_{div_2]}} > [s₁, a_{_geq₁}, geq_1] > false > [if₃, A_{_IF₃}, a_{_{if_3]}}

The following usable rules [14] were oriented:

a__if(X1, X2, X3) → if(X1, X2, X3)
a__div(0, s(Y)) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
mark(false) → false
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(true) → true
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__minus(0, Y) → 0
a__if(true, X, Y) → mark(X)
mark(div(X1, X2)) → a__div(mark(X1), X2)
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
a__if(false, X, Y) → mark(Y)
a__div(X1, X2) → div(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
mark(0) → 0
a__geq(0, s(Y)) → false
a__geq(X, 0) → true
a__minus(X1, X2) → minus(X1, X2)
mark(s(X)) → s(mark(X))
mark(minus(X1, X2)) → a__minus(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(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 0 SCCs with 2 less nodes.