Termination w.r.t. Q proof of /tmp/tmpT8dOQN/Ex49_GM04

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

S(ok(X)) → S(X)
PROPER(geq(X1, X2)) → GEQ(proper(X1), proper(X2))
TOP(mark(X)) → PROPER(X)
ACTIVE(minus(s(X), s(Y))) → MINUS(X, Y)
ACTIVE(div(s(X), s(Y))) → IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
PROPER(if(X1, X2, X3)) → PROPER(X1)
ACTIVE(div(s(X), s(Y))) → S(div(minus(X, Y), s(Y)))
ACTIVE(div(X1, X2)) → DIV(active(X1), X2)
TOP(ok(X)) → ACTIVE(X)
PROPER(geq(X1, X2)) → PROPER(X1)
PROPER(div(X1, X2)) → PROPER(X1)
PROPER(div(X1, X2)) → DIV(proper(X1), proper(X2))
PROPER(minus(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(if(X1, X2, X3)) → IF(proper(X1), proper(X2), proper(X3))
ACTIVE(div(s(X), s(Y))) → DIV(minus(X, Y), s(Y))
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
DIV(ok(X1), ok(X2)) → DIV(X1, X2)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
TOP(ok(X)) → TOP(active(X))
ACTIVE(if(X1, X2, X3)) → IF(active(X1), X2, X3)
S(mark(X)) → S(X)
PROPER(minus(X1, X2)) → PROPER(X2)
ACTIVE(div(s(X), s(Y))) → GEQ(X, Y)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(geq(X1, X2)) → PROPER(X2)
GEQ(ok(X1), ok(X2)) → GEQ(X1, X2)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
PROPER(s(X)) → S(proper(X))
DIV(mark(X1), X2) → DIV(X1, X2)
ACTIVE(div(s(X), s(Y))) → MINUS(X, Y)
ACTIVE(div(X1, X2)) → ACTIVE(X1)
MINUS(ok(X1), ok(X2)) → MINUS(X1, X2)
PROPER(minus(X1, X2)) → MINUS(proper(X1), proper(X2))
ACTIVE(s(X)) → ACTIVE(X)
PROPER(div(X1, X2)) → PROPER(X2)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(geq(s(X), s(Y))) → GEQ(X, Y)
ACTIVE(s(X)) → S(active(X))

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

S(ok(X)) → S(X)
PROPER(geq(X1, X2)) → GEQ(proper(X1), proper(X2))
TOP(mark(X)) → PROPER(X)
ACTIVE(minus(s(X), s(Y))) → MINUS(X, Y)
ACTIVE(div(s(X), s(Y))) → IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
PROPER(if(X1, X2, X3)) → PROPER(X1)
ACTIVE(div(s(X), s(Y))) → S(div(minus(X, Y), s(Y)))
ACTIVE(div(X1, X2)) → DIV(active(X1), X2)
TOP(ok(X)) → ACTIVE(X)
PROPER(geq(X1, X2)) → PROPER(X1)
PROPER(div(X1, X2)) → PROPER(X1)
PROPER(div(X1, X2)) → DIV(proper(X1), proper(X2))
PROPER(minus(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(if(X1, X2, X3)) → IF(proper(X1), proper(X2), proper(X3))
ACTIVE(div(s(X), s(Y))) → DIV(minus(X, Y), s(Y))
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
DIV(ok(X1), ok(X2)) → DIV(X1, X2)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
TOP(ok(X)) → TOP(active(X))
ACTIVE(if(X1, X2, X3)) → IF(active(X1), X2, X3)
S(mark(X)) → S(X)
PROPER(minus(X1, X2)) → PROPER(X2)
ACTIVE(div(s(X), s(Y))) → GEQ(X, Y)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(geq(X1, X2)) → PROPER(X2)
GEQ(ok(X1), ok(X2)) → GEQ(X1, X2)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
PROPER(s(X)) → S(proper(X))
DIV(mark(X1), X2) → DIV(X1, X2)
ACTIVE(div(s(X), s(Y))) → MINUS(X, Y)
ACTIVE(div(X1, X2)) → ACTIVE(X1)
MINUS(ok(X1), ok(X2)) → MINUS(X1, X2)
PROPER(minus(X1, X2)) → MINUS(proper(X1), proper(X2))
ACTIVE(s(X)) → ACTIVE(X)
PROPER(div(X1, X2)) → PROPER(X2)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(geq(s(X), s(Y))) → GEQ(X, Y)
ACTIVE(s(X)) → S(active(X))

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 8 SCCs with 17 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

GEQ(ok(X1), ok(X2)) → GEQ(X1, X2)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

GEQ(ok(X1), ok(X2)) → GEQ(X1, X2)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

GEQ(ok(X1), ok(X2)) → GEQ(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS(ok(X1), ok(X2)) → MINUS(X1, X2)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS(ok(X1), ok(X2)) → MINUS(X1, X2)

From the DPs we obtained the following set of size-change graphs:

MINUS(ok(X1), ok(X2)) → MINUS(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)

From the DPs we obtained the following set of size-change graphs:

IF(mark(X1), X2, X3) → IF(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 > 2, 3 > 3

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DIV(mark(X1), X2) → DIV(X1, X2)
DIV(ok(X1), ok(X2)) → DIV(X1, X2)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DIV(mark(X1), X2) → DIV(X1, X2)
DIV(ok(X1), ok(X2)) → DIV(X1, X2)

From the DPs we obtained the following set of size-change graphs:

DIV(mark(X1), X2) → DIV(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
DIV(ok(X1), ok(X2)) → DIV(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

From the DPs we obtained the following set of size-change graphs:

S(ok(X)) → S(X)
The graph contains the following edges 1 > 1
S(mark(X)) → S(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

PROPER(div(X1, X2)) → PROPER(X1)
PROPER(minus(X1, X2)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(geq(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(minus(X1, X2)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(div(X1, X2)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(geq(X1, X2)) → PROPER(X1)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

PROPER(minus(X1, X2)) → PROPER(X2)
PROPER(div(X1, X2)) → PROPER(X1)
PROPER(minus(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(geq(X1, X2)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(div(X1, X2)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(geq(X1, X2)) → PROPER(X1)

From the DPs we obtained the following set of size-change graphs:

PROPER(div(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(minus(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(if(X1, X2, X3)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(geq(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(s(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(minus(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(if(X1, X2, X3)) → PROPER(X3)
The graph contains the following edges 1 > 1
PROPER(div(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(if(X1, X2, X3)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(geq(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(div(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(div(X1, X2)) → ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)

From the DPs we obtained the following set of size-change graphs:

ACTIVE(s(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
ACTIVE(div(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = x₁
POL(active(x₁)) = 2·x₁
POL(div(x₁, x₂)) = x₁ + 2·x₂
POL(false) = 0
POL(geq(x₁, x₂)) = x₁ + x₂
POL(if(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(mark(x₁)) = x₁
POL(minus(x₁, x₂)) = x₁ + x₂
POL(ok(x₁)) = 2·x₁
POL(proper(x₁)) = x₁
POL(s(x₁)) = x₁
POL(true) = 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ QDPOrderProof

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
div(mark(X1), X2) → mark(div(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

TOP(mark(X)) → TOP(proper(X))

The remaining pairs can at least be oriented weakly.

TOP(ok(X)) → TOP(active(X))

Used ordering: Combined order from the following AFS and order.
TOP(x1) = TOP(x1)
mark(x1) = mark(x1)
proper(x1) = x1
ok(x1) = x1
active(x1) = x1
minus(x1, x2) = minus(x1)
s(x1) = s(x1)
geq(x1, x2) = geq(x2)
0 = 0
true = true
false = false
div(x1, x2) = div(x1, x2)
if(x1, x2, x3) = if(x1, x2, x3)

Recursive path order with status [2].
Quasi-Precedence:

[div₂, if₃] > [s₁, geq₁, true, false] > [mark₁, minus₁] > 0

Status:

if₃: [3,1,2]
div₂: [1,2]
minus₁: multiset
true: multiset
geq₁: multiset
mark₁: multiset
false: multiset
TOP₁: [1]
s₁: [1]
0: multiset

The following usable rules [17] were oriented:

active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(minus(0, Y)) → mark(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
s(ok(X)) → ok(s(X))
proper(true) → ok(true)
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(div(X1, X2)) → div(active(X1), X2)
active(s(X)) → s(active(X))
s(mark(X)) → mark(s(X))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
div(mark(X1), X2) → mark(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
div(mark(X1), X2) → mark(div(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))

POL(0) = 0
POL(TOP(x₁)) = 2·x₁
POL(active(x₁)) = 2·x₁
POL(div(x₁, x₂)) = x₁ + 2·x₂
POL(false) = 0
POL(geq(x₁, x₂)) = x₁ + x₂
POL(if(x₁, x₂, x₃)) = x₁ + x₂ + 2·x₃
POL(mark(x₁)) = x₁
POL(minus(x₁, x₂)) = x₁ + x₂
POL(ok(x₁)) = 2·x₁
POL(s(x₁)) = x₁
POL(true) = 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
div(mark(X1), X2) → mark(div(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

TOP(ok(X)) → TOP(active(X))

Strictly oriented rules of the TRS R:

active(div(0, s(Y))) → mark(0)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = 2·x₁
POL(active(x₁)) = 2·x₁
POL(div(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(false) = 0
POL(geq(x₁, x₂)) = x₁ + x₂
POL(if(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃
POL(mark(x₁)) = x₁
POL(minus(x₁, x₂)) = x₁ + x₂
POL(ok(x₁)) = 1 + 2·x₁
POL(s(x₁)) = x₁
POL(true) = 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ PisEmptyProof

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

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(s(X)) → s(active(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
div(mark(X1), X2) → mark(div(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(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.