Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar2/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:

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.

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 contains 8 SCCs with 17 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: GEQ₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: MINUS₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: IF₃(x1, x2, x3) = x3
ok₁(x1) = ok₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: IF₃(x1, x2, x3) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: DIV₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DIV₂(mark₁(X1), 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DIV₂(mark₁(X1), X2) -> DIV₂(X1, X2)

Used argument filtering: DIV₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(mark₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S₁(ok₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(ok₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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(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)

Used argument filtering: PROPER₁(x1) = x1
div₂(x1, x2) = div₂(x1, x2)
minus₂(x1, x2) = minus₂(x1, x2)
geq₂(x1, x2) = geq₂(x1, x2)
s₁(x1) = x1
if₃(x1, x2, x3) = if₃(x1, x2, x3)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(s₁(X)) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(s₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
if₃(x1, x2, x3) = x1
div₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(div₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
if₃(x1, x2, x3) = x1
div₂(x1, x2) = div₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

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

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

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: ACTIVE₁(x1) = x1
if₃(x1, x2, x3) = if₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ 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

            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

Used argument filtering: TOP₁(x1) = x1
ok₁(x1) = x1
active₁(x1) = x1
mark₁(x1) = mark₁(x1)
proper₁(x1) = x1
minus₂(x1, x2) = minus₁(x1)
0 = 0
s₁(x1) = s₁(x1)
geq₂(x1, x2) = x2
true = true
false = false
div₂(x1, x2) = div₂(x1, x2)
if₃(x1, x2, x3) = if₃(x1, x2, x3)
Used ordering: Precedence:

div₂ > 0 > mark₁
div₂ > 0 > true
div₂ > s₁ > minus₁ > mark₁
div₂ > s₁ > false
div₂ > if₃ > mark₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

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

TOP₁(ok₁(X)) -> TOP₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

Used argument filtering: TOP₁(x1) = x1
ok₁(x1) = ok₁(x1)
active₁(x1) = x1
minus₂(x1, x2) = minus
mark₁(x1) = mark
geq₂(x1, x2) = geq
div₂(x1, x2) = x2
s₁(x1) = x1
if₃(x1, x2, x3) = x3
Used ordering: Precedence:

ok₁ > mark
minus > mark
geq > mark

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

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.