Termination w.r.t. Q proof of /tmp/tmp7Cprui/Ex49_GM04

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 QDPOrderProof (⇔)

    ↳11 QDP

    ↳12 QDPOrderProof (⇔)

    ↳13 QDP

    ↳14 QDPOrderProof (⇔)

    ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 PisEmptyProof (⇔)

    ↳19 TRUE

  ↳20 QDP

    ↳21 QDPOrderProof (⇔)

    ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 QDPOrderProof (⇔)

    ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 PisEmptyProof (⇔)

    ↳30 TRUE

  ↳31 QDP

    ↳32 QDPOrderProof (⇔)

    ↳33 QDP

    ↳34 QDPOrderProof (⇔)

    ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 PisEmptyProof (⇔)

    ↳41 TRUE

  ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 PisEmptyProof (⇔)

    ↳48 TRUE

  ↳49 QDP

    ↳50 QDPOrderProof (⇔)

    ↳51 QDP

    ↳52 QDPOrderProof (⇔)

    ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 PisEmptyProof (⇔)

    ↳59 TRUE

  ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 QDPOrderProof (⇔)

    ↳64 QDP

    ↳65 QDPOrderProof (⇔)

    ↳66 QDP

    ↳67 DependencyGraphProof (⇔)

    ↳68 TRUE

(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, 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) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(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(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)

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) 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)

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)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) 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.

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(X1, active(X2), 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) = x2
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(11) 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)

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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(X1, mark(X2), 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) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(active(X1), X2, 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) = x1
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

IF(mark(X1), 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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(mark(X1), X2, 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) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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(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.

(18) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(19) TRUE

(20) 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.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DIV(X1, active(X2)) → DIV(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2) = x2
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(22) 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)

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.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DIV(X1, mark(X2)) → DIV(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

DIV(mark(X1), X2) → DIV(X1, X2)
DIV(active(X1), 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.

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DIV(active(X1), X2) → DIV(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2) = x1
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(26) Obligation:

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

DIV(mark(X1), 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.

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DIV(mark(X1), X2) → DIV(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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(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.

(29) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(30) TRUE

(31) 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.

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

GEQ(X1, active(X2)) → GEQ(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GEQ(x1, x2) = x2
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(33) 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)

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.

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

GEQ(X1, mark(X2)) → GEQ(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GEQ(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

GEQ(mark(X1), X2) → GEQ(X1, X2)
GEQ(active(X1), 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.

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

GEQ(active(X1), X2) → GEQ(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GEQ(x1, x2) = x1
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(37) Obligation:

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

GEQ(mark(X1), 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.

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

GEQ(mark(X1), X2) → GEQ(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GEQ(x1, x2) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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(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.

(40) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(41) TRUE

(42) 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.

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
active(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

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

S(active(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.

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

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(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.

(47) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(48) TRUE

(49) 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.

(50) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(X1, active(X2)) → MINUS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = x2
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(51) 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)

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.

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(X1, mark(X2)) → MINUS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(53) Obligation:

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

MINUS(mark(X1), X2) → MINUS(X1, X2)
MINUS(active(X1), 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.

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(active(X1), X2) → MINUS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = x1
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

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

MINUS(mark(X1), 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.

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(mark(X1), X2) → MINUS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

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(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.

(58) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(59) TRUE

(60) 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.

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [LPO].
Precedence:

div₁ > geq > true
div₁ > geq > false
div₁ > s₁ > minus > 0
div₁ > if₃

Status:

trivial

The following usable rules [FROCOS05] were oriented:

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

(62) 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)))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(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.

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(geq(s(X), s(Y))) → MARK(geq(X, Y))

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

Lexicographic path order with status [LPO].
Precedence:

div₁ > s₁ > minus > 0 > true
div₁ > s₁ > false
div₁ > if₂

Status:

trivial

The following usable rules [FROCOS05] were oriented:

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

(64) 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))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(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.

(65) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))

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

Lexicographic path order with status [LPO].
Precedence:

geq > true
geq > false
div₁ > s₁
div₁ > if₂
div₁ > 0

Status:

trivial

The following usable rules [FROCOS05] were oriented:

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

(66) Obligation:

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

MARK(minus(X1, X2)) → ACTIVE(minus(X1, X2))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(geq(X1, X2)) → ACTIVE(geq(X1, X2))
MARK(div(X1, X2)) → ACTIVE(div(mark(X1), X2))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(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.

(67) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 5 less nodes.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPOrderProof (EQUIVALENT transformation)

(9) Obligation:

(10) QDPOrderProof (EQUIVALENT transformation)

(11) Obligation:

(12) QDPOrderProof (EQUIVALENT transformation)

(13) Obligation:

(14) QDPOrderProof (EQUIVALENT transformation)

(15) Obligation:

(16) QDPOrderProof (EQUIVALENT transformation)

(17) Obligation:

(18) PisEmptyProof (EQUIVALENT transformation)

(19) TRUE

(20) Obligation:

(21) QDPOrderProof (EQUIVALENT transformation)

(22) Obligation:

(23) QDPOrderProof (EQUIVALENT transformation)

(24) Obligation:

(25) QDPOrderProof (EQUIVALENT transformation)

(26) Obligation:

(27) QDPOrderProof (EQUIVALENT transformation)

(28) Obligation:

(29) PisEmptyProof (EQUIVALENT transformation)

(30) TRUE

(31) Obligation:

(32) QDPOrderProof (EQUIVALENT transformation)

(33) Obligation:

(34) QDPOrderProof (EQUIVALENT transformation)

(35) Obligation:

(36) QDPOrderProof (EQUIVALENT transformation)

(37) Obligation:

(38) QDPOrderProof (EQUIVALENT transformation)

(39) Obligation:

(40) PisEmptyProof (EQUIVALENT transformation)

(41) TRUE

(42) Obligation:

(43) QDPOrderProof (EQUIVALENT transformation)

(44) Obligation:

(45) QDPOrderProof (EQUIVALENT transformation)

(46) Obligation:

(47) PisEmptyProof (EQUIVALENT transformation)

(48) TRUE

(49) Obligation:

(50) QDPOrderProof (EQUIVALENT transformation)

(51) Obligation:

(52) QDPOrderProof (EQUIVALENT transformation)

(53) Obligation:

(54) QDPOrderProof (EQUIVALENT transformation)

(55) Obligation:

(56) QDPOrderProof (EQUIVALENT transformation)

(57) Obligation:

(58) PisEmptyProof (EQUIVALENT transformation)

(59) TRUE

(60) Obligation:

(61) QDPOrderProof (EQUIVALENT transformation)

(62) Obligation:

(63) QDPOrderProof (EQUIVALENT transformation)

(64) Obligation:

(65) QDPOrderProof (EQUIVALENT transformation)

(66) Obligation:

(67) DependencyGraphProof (EQUIVALENT transformation)

(68) TRUE