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:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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:

MINUS₂(ok₁(X1), ok₁(X2)) -> MINUS₂(X1, X2)
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(minus₂(X1, X2)) -> MINUS₂(proper₁(X1), proper₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
ACTIVE₁(div₂(X1, X2)) -> ACTIVE₁(X1)
S₁(mark₁(X)) -> S₁(X)
GEQ₂(ok₁(X1), ok₁(X2)) -> GEQ₂(X1, X2)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)
DIV₂(ok₁(X1), ok₁(X2)) -> DIV₂(X1, X2)
PROPER₁(minus₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> IF₃(proper₁(X1), proper₁(X2), proper₁(X3))
ACTIVE₁(if₃(X1, X2, X3)) -> IF₃(active₁(X1), X2, X3)
PROPER₁(geq₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(geq₂(s₁(X), s₁(Y))) -> GEQ₂(X, Y)
PROPER₁(minus₂(X1, X2)) -> PROPER₁(X1)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
PROPER₁(geq₂(X1, X2)) -> GEQ₂(proper₁(X1), proper₁(X2))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
ACTIVE₁(div₂(X1, X2)) -> DIV₂(active₁(X1), X2)
PROPER₁(div₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(div₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(div₂(s₁(X), s₁(Y))) -> S₁(div₂(minus₂(X, Y), s₁(Y)))
PROPER₁(div₂(X1, X2)) -> DIV₂(proper₁(X1), proper₁(X2))
ACTIVE₁(div₂(s₁(X), s₁(Y))) -> DIV₂(minus₂(X, Y), s₁(Y))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
ACTIVE₁(div₂(s₁(X), s₁(Y))) -> GEQ₂(X, Y)
ACTIVE₁(div₂(s₁(X), s₁(Y))) -> MINUS₂(X, Y)
DIV₂(mark₁(X1), X2) -> DIV₂(X1, X2)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
ACTIVE₁(minus₂(s₁(X), s₁(Y))) -> MINUS₂(X, Y)
PROPER₁(s₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
PROPER₁(geq₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(div₂(s₁(X), s₁(Y))) -> IF₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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:

MINUS₂(ok₁(X1), ok₁(X2)) -> MINUS₂(X1, X2)
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(minus₂(X1, X2)) -> MINUS₂(proper₁(X1), proper₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
ACTIVE₁(div₂(X1, X2)) -> ACTIVE₁(X1)
S₁(mark₁(X)) -> S₁(X)
GEQ₂(ok₁(X1), ok₁(X2)) -> GEQ₂(X1, X2)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)
DIV₂(ok₁(X1), ok₁(X2)) -> DIV₂(X1, X2)
PROPER₁(minus₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> IF₃(proper₁(X1), proper₁(X2), proper₁(X3))
ACTIVE₁(if₃(X1, X2, X3)) -> IF₃(active₁(X1), X2, X3)
PROPER₁(geq₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(geq₂(s₁(X), s₁(Y))) -> GEQ₂(X, Y)
PROPER₁(minus₂(X1, X2)) -> PROPER₁(X1)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
PROPER₁(geq₂(X1, X2)) -> GEQ₂(proper₁(X1), proper₁(X2))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
ACTIVE₁(div₂(X1, X2)) -> DIV₂(active₁(X1), X2)
PROPER₁(div₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(div₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(div₂(s₁(X), s₁(Y))) -> S₁(div₂(minus₂(X, Y), s₁(Y)))
PROPER₁(div₂(X1, X2)) -> DIV₂(proper₁(X1), proper₁(X2))
ACTIVE₁(div₂(s₁(X), s₁(Y))) -> DIV₂(minus₂(X, Y), s₁(Y))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
ACTIVE₁(div₂(s₁(X), s₁(Y))) -> GEQ₂(X, Y)
ACTIVE₁(div₂(s₁(X), s₁(Y))) -> MINUS₂(X, Y)
DIV₂(mark₁(X1), X2) -> DIV₂(X1, X2)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
ACTIVE₁(minus₂(s₁(X), s₁(Y))) -> MINUS₂(X, Y)
PROPER₁(s₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
PROPER₁(geq₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(div₂(s₁(X), s₁(Y))) -> IF₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 8 SCCs with 17 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

GEQ₂(ok₁(X1), ok₁(X2)) -> GEQ₂(X1, X2)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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.

GEQ₂(ok₁(X1), ok₁(X2)) -> GEQ₂(X1, X2)

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

POL( ok₁(x₁) ) = x₁ + 3

POL( GEQ₂(x₁, x₂) ) = 3x₂ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS₂(ok₁(X1), ok₁(X2)) -> MINUS₂(X1, X2)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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(ok₁(X1), ok₁(X2)) -> MINUS₂(X1, X2)

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

POL( MINUS₂(x₁, x₂) ) = 3x₂ + 3

POL( ok₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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.

IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)

The remaining pairs can at least be oriented weakly.

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)

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

POL( IF₃(x₁, ..., x₃) ) = 2x₁ + 2x₃ + 1

POL( mark₁(x₁) ) = x₁

POL( ok₁(x₁) ) = 2x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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.

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)

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

POL( mark₁(x₁) ) = 2x₁ + 1

POL( IF₃(x₁, ..., x₃) ) = 2x₁ + 2x₂ + 3x₃ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DIV₂(mark₁(X1), X2) -> DIV₂(X1, X2)
DIV₂(ok₁(X1), ok₁(X2)) -> DIV₂(X1, X2)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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(mark₁(X1), X2) -> DIV₂(X1, X2)
DIV₂(ok₁(X1), ok₁(X2)) -> DIV₂(X1, X2)

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

POL( DIV₂(x₁, x₂) ) = max{0, 2x₁ + 2x₂ - 1}

POL( mark₁(x₁) ) = 2x₁ + 1

POL( ok₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S₁(ok₁(X)) -> S₁(X)
S₁(mark₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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.

S₁(ok₁(X)) -> S₁(X)
S₁(mark₁(X)) -> S₁(X)

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

POL( S₁(x₁) ) = 2x₁

POL( mark₁(x₁) ) = 2x₁ + 2

POL( ok₁(x₁) ) = 3x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(div₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(minus₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(div₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(geq₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(geq₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(minus₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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.

PROPER₁(div₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(minus₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(div₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(geq₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(geq₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(minus₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)

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

POL( s₁(x₁) ) = 2x₁ + 3

POL( minus₂(x₁, x₂) ) = x₁ + 2x₂ + 3

POL( if₃(x₁, ..., x₃) ) = 2x₁ + 2x₂ + 2x₃ + 3

POL( div₂(x₁, x₂) ) = 3x₁ + 2x₂ + 3

POL( geq₂(x₁, x₂) ) = 2x₁ + 2x₂ + 3

POL( PROPER₁(x₁) ) = x₁ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(div₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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.

ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)

The remaining pairs can at least be oriented weakly.

ACTIVE₁(div₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

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

POL( s₁(x₁) ) = 2x₁

POL( if₃(x₁, ..., x₃) ) = 2x₁ + x₂ + 2x₃ + 1

POL( ACTIVE₁(x₁) ) = 3x₁

POL( div₂(x₁, x₂) ) = 2x₁ + 3x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

ACTIVE₁(div₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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.

ACTIVE₁(div₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

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

POL( ACTIVE₁(x₁) ) = 2x₁

POL( div₂(x₁, x₂) ) = 2x₁ + 2x₂ + 3

POL( s₁(x₁) ) = 2x₁ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The TRS R consists of the following rules:

active₁(minus₂(0, Y)) -> mark₁(0)
active₁(minus₂(s₁(X), s₁(Y))) -> mark₁(minus₂(X, Y))
active₁(geq₂(X, 0)) -> mark₁(true)
active₁(geq₂(0, s₁(Y))) -> mark₁(false)
active₁(geq₂(s₁(X), s₁(Y))) -> mark₁(geq₂(X, Y))
active₁(div₂(0, s₁(Y))) -> mark₁(0)
active₁(div₂(s₁(X), s₁(Y))) -> mark₁(if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(div₂(X1, X2)) -> div₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
s₁(mark₁(X)) -> mark₁(s₁(X))
div₂(mark₁(X1), X2) -> mark₁(div₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
proper₁(minus₂(X1, X2)) -> minus₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(geq₂(X1, X2)) -> geq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(div₂(X1, X2)) -> div₂(proper₁(X1), proper₁(X2))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
minus₂(ok₁(X1), ok₁(X2)) -> ok₁(minus₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
geq₂(ok₁(X1), ok₁(X2)) -> ok₁(geq₂(X1, X2))
div₂(ok₁(X1), ok₁(X2)) -> ok₁(div₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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