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.

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 strictly oriented and are deleted.


GEQ2(ok1(X1), ok1(X2)) -> GEQ2(X1, X2)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
GEQ2(x1, x2)  =  GEQ1(x1)
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ 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 strictly oriented and are deleted.


MINUS2(ok1(X1), ok1(X2)) -> MINUS2(X1, X2)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS2(x1, x2)  =  MINUS1(x1)
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ 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 strictly oriented and are deleted.


IF3(ok1(X1), ok1(X2), ok1(X3)) -> IF3(X1, X2, X3)
The remaining pairs can at least by weakly be oriented.

IF3(mark1(X1), X2, X3) -> IF3(X1, X2, X3)
Used ordering: Combined order from the following AFS and order.
IF3(x1, x2, x3)  =  IF3(x1, x2, x3)
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
ok1 > IF3


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 strictly oriented and are deleted.


IF3(mark1(X1), X2, X3) -> IF3(X1, X2, X3)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
IF3(x1, x2, x3)  =  IF1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
[IF1, mark1]


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 strictly oriented and are deleted.


DIV2(ok1(X1), ok1(X2)) -> DIV2(X1, X2)
The remaining pairs can at least by weakly be oriented.

DIV2(mark1(X1), X2) -> DIV2(X1, X2)
Used ordering: Combined order from the following AFS and order.
DIV2(x1, x2)  =  x2
mark1(x1)  =  mark
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DIV2(mark1(X1), X2) -> DIV2(X1, X2)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DIV2(x1, x2)  =  DIV1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
[DIV1, mark1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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 strictly oriented and are deleted.


S1(mark1(X)) -> S1(X)
The remaining pairs can at least by weakly be oriented.

S1(ok1(X)) -> S1(X)
Used ordering: Combined order from the following AFS and order.
S1(x1)  =  S1(x1)
ok1(x1)  =  x1
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


S1(ok1(X)) -> S1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
S1(x1)  =  S1(x1)
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
ok1 > S1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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 strictly oriented 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(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 by weakly be oriented.

PROPER1(s1(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
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)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(s1(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > PROPER1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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 strictly oriented and are deleted.


ACTIVE1(div2(X1, X2)) -> ACTIVE1(X1)
The remaining pairs can at least by weakly be oriented.

ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
if3(x1, x2, x3)  =  x1
div2(x1, x2)  =  div1(x1)
s1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[ACTIVE1, div1]


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(if3(X1, X2, X3)) -> 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 strictly oriented and are deleted.


ACTIVE1(s1(X)) -> ACTIVE1(X)
The remaining pairs can at least by weakly be oriented.

ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X1)
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
if3(x1, x2, x3)  =  x1
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > ACTIVE1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X1)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  x1
if3(x1, x2, x3)  =  if1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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
            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


TOP1(mark1(X)) -> TOP1(proper1(X))
The remaining pairs can at least by weakly be oriented.

TOP1(ok1(X)) -> TOP1(active1(X))
Used ordering: Combined order from the following AFS and order.
TOP1(x1)  =  TOP1(x1)
ok1(x1)  =  x1
active1(x1)  =  x1
mark1(x1)  =  mark1(x1)
proper1(x1)  =  x1
if3(x1, x2, x3)  =  if3(x1, x2, x3)
geq2(x1, x2)  =  geq1(x1)
0  =  0
true  =  true
div2(x1, x2)  =  div2(x1, x2)
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  minus1(x1)
false  =  false

Lexicographic Path Order [19].
Precedence:
TOP1 > false
[geq1, 0, true, div2] > if3 > [mark1, minus1] > false
[geq1, 0, true, div2] > s1 > [mark1, minus1] > false


The following usable rules [14] were oriented:

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)
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
div2(mark1(X1), X2) -> mark1(div2(X1, X2))
div2(ok1(X1), ok1(X2)) -> ok1(div2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
s1(ok1(X)) -> ok1(s1(X))
geq2(ok1(X1), ok1(X2)) -> ok1(geq2(X1, X2))
minus2(ok1(X1), ok1(X2)) -> ok1(minus2(X1, X2))
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))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


TOP1(ok1(X)) -> TOP1(active1(X))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TOP1(x1)  =  TOP1(x1)
ok1(x1)  =  ok1(x1)
active1(x1)  =  x1
if3(x1, x2, x3)  =  if2(x2, x3)
geq2(x1, x2)  =  geq1(x2)
0  =  0
mark1(x1)  =  x1
true  =  true
div2(x1, x2)  =  div2(x1, x2)
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  x1
false  =  false

Lexicographic Path Order [19].
Precedence:
div2 > if2 > [ok1, 0, s1] > false
div2 > [geq1, true] > [ok1, 0, s1] > false


The following usable rules [14] were oriented:

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)
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
div2(mark1(X1), X2) -> mark1(div2(X1, X2))
div2(ok1(X1), ok1(X2)) -> ok1(div2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
s1(ok1(X)) -> ok1(s1(X))
minus2(ok1(X1), ok1(X2)) -> ok1(minus2(X1, X2))
geq2(ok1(X1), ok1(X2)) -> ok1(geq2(X1, X2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
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.