Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/TRCSRSLASHEx49_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__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 strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
A__GEQ₂(x1, x2) = A__GEQ₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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 strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
A__MINUS₂(x1, x2) = A__MINUS₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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 strictly oriented and are deleted.

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)) -> MARK₁(X1)

The remaining pairs can at least by weakly be oriented.

MARK₁(div₂(X1, X2)) -> A__DIV₂(mark₁(X1), X2)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)

Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = x1
if₃(x1, x2, x3) = if₃(x1, x2, x3)
s₁(x1) = s₁(x1)
A__DIV₂(x1, x2) = A__DIV₁(x1)
A__IF₃(x1, x2, x3) = A__IF₃(x1, x2, x3)
a__geq₂(x1, x2) = x1
div₂(x1, x2) = div₁(x1)
minus₂(x1, x2) = minus₁(x1)
0 = 0
false = false
true = true
mark₁(x1) = x1
geq₂(x1, x2) = x1
a__minus₂(x1, x2) = a__minus₁(x1)
a__if₃(x1, x2, x3) = a__if₃(x1, x2, x3)
a__div₂(x1, x2) = a__div₁(x1)

Lexicographic Path Order [19].
Precedence:

[A_{_DIV₁}, div₁, a_{_{div_1]}} > s₁ > [if₃, A_{_IF₃}, minus₁, true, a_{_minus₁}, a_{_{if_3]}}
[A_{_DIV₁}, div₁, a_{_{div_1]}} > [0, false] > [if₃, A_{_IF₃}, minus₁, true, a_{_minus₁}, a_{_{if_3]}}

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

MARK₁(div₂(X1, X2)) -> A__DIV₂(mark₁(X1), X2)
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 0 SCCs with 2 less nodes.