Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


GEQ2(ok1(X1), ok1(X2)) -> GEQ2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(GEQ2(x1, x2)) = x2   
POL(ok1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MINUS2(ok1(X1), ok1(X2)) -> MINUS2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MINUS2(x1, x2)) = x2   
POL(ok1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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

IF3(mark1(X1), X2, X3) -> IF3(X1, X2, X3)
Used ordering: Polynomial interpretation [21]:

POL(IF3(x1, x2, x3)) = x2 + x3   
POL(mark1(x1)) = 0   
POL(ok1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


IF3(mark1(X1), X2, X3) -> IF3(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(IF3(x1, x2, x3)) = x1   
POL(mark1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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

DIV2(mark1(X1), X2) -> DIV2(X1, X2)
Used ordering: Polynomial interpretation [21]:

POL(DIV2(x1, x2)) = x2   
POL(mark1(x1)) = 0   
POL(ok1(x1)) = 1 + x1   

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 oriented strictly and are deleted.


DIV2(mark1(X1), X2) -> DIV2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DIV2(x1, x2)) = x1   
POL(mark1(x1)) = 1 + x1   

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 oriented strictly and are deleted.


S1(ok1(X)) -> S1(X)
S1(mark1(X)) -> S1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(S1(x1)) = x1   
POL(mark1(x1)) = 1 + x1   
POL(ok1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(div2(X1, X2)) -> PROPER1(X2)
PROPER1(minus2(X1, X2)) -> PROPER1(X2)
PROPER1(div2(X1, X2)) -> PROPER1(X1)
PROPER1(geq2(X1, X2)) -> PROPER1(X1)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(geq2(X1, X2)) -> PROPER1(X2)
PROPER1(minus2(X1, X2)) -> PROPER1(X1)
PROPER1(if3(X1, X2, X3)) -> PROPER1(X3)
PROPER1(if3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(if3(X1, X2, X3)) -> PROPER1(X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = x1   
POL(div2(x1, x2)) = 1 + x1 + x2   
POL(geq2(x1, x2)) = 1 + x1 + x2   
POL(if3(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(minus2(x1, x2)) = 1 + x1 + x2   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X1)
ACTIVE1(div2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = x1   
POL(div2(x1, x2)) = 1 + x1   
POL(if3(x1, x2, x3)) = 1 + x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

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

The TRS R consists of the following rules:

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

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