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

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

  ↳8 QDP

  ↳9 QDP

  ↳10 QDP

  ↳11 QDP

  ↳12 QDP

(0) Obligation:

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

ACTIVE(minus(0, Y)) → MARK(0)
ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
ACTIVE(minus(s(X), s(Y))) → 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(geq(s(X), s(Y))) → 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(div(s(X), s(Y))) → IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
ACTIVE(div(s(X), s(Y))) → GEQ(X, Y)
ACTIVE(div(s(X), s(Y))) → S(div(minus(X, Y), s(Y)))
ACTIVE(div(s(X), s(Y))) → DIV(minus(X, Y), s(Y))
ACTIVE(div(s(X), s(Y))) → MINUS(X, Y)
ACTIVE(if(true, X, Y)) → MARK(X)
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
MARK(0) → ACTIVE(0)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
MARK(true) → ACTIVE(true)
MARK(false) → ACTIVE(false)
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
MARK(div(X1, X2)) → DIV(mark(X1), X2)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(if(X1, X2, X3)) → IF(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MINUS(mark(X1), X2) → MINUS(X1, X2)
MINUS(X1, mark(X2)) → MINUS(X1, X2)
MINUS(active(X1), X2) → MINUS(X1, X2)
MINUS(X1, active(X2)) → MINUS(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
GEQ(mark(X1), X2) → GEQ(X1, X2)
GEQ(X1, mark(X2)) → GEQ(X1, X2)
GEQ(active(X1), X2) → GEQ(X1, X2)
GEQ(X1, active(X2)) → GEQ(X1, X2)
DIV(mark(X1), X2) → DIV(X1, X2)
DIV(X1, mark(X2)) → DIV(X1, X2)
DIV(active(X1), X2) → DIV(X1, X2)
DIV(X1, active(X2)) → DIV(X1, X2)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, active(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)
mark(minus(X1, X2)) → active(minus(X1, X2))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(geq(X1, X2)) → active(geq(X1, X2))
mark(true) → active(true)
mark(false) → active(false)
mark(div(X1, X2)) → active(div(mark(X1), X2))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
geq(mark(X1), X2) → geq(X1, X2)
geq(X1, mark(X2)) → geq(X1, X2)
geq(active(X1), X2) → geq(X1, X2)
geq(X1, active(X2)) → geq(X1, X2)
div(mark(X1), X2) → div(X1, X2)
div(X1, mark(X2)) → div(X1, X2)
div(active(X1), X2) → div(X1, X2)
div(X1, active(X2)) → div(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 17 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3) = x3
mark(x1) = mark(x1)
active(x1) = active(x1)
minus(x1, x2) = minus
0 = 0
s(x1) = s
geq(x1, x2) = geq
true = true
false = false
div(x1, x2) = div
if(x1, x2, x3) = if(x2, x3)

Lexicographic path order with status [LPO].
Quasi-Precedence:

[mark₁, active₁]
minus > 0
geq > true
geq > false
div > 0
div > if₂
div > s

Status:

trivial

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

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

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

(8) Obligation:

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

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

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

(9) Obligation:

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

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

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

(10) Obligation:

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

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

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

(11) Obligation:

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

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

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

(12) Obligation:

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

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

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