Termination w.r.t. Q proof of /tmp/tmpQf6bKs/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 PisEmptyProof (⇔)

        ↳9 TRUE

    ↳10 QDP

        ↳11 QDPOrderProof (⇔)

        ↳12 QDP

        ↳13 PisEmptyProof (⇔)

        ↳14 TRUE

    ↳15 QDP

        ↳16 QDPOrderProof (⇔)

        ↳17 QDP

        ↳18 QDPOrderProof (⇔)

        ↳19 QDP

        ↳20 PisEmptyProof (⇔)

        ↳21 TRUE

    ↳22 QDP

        ↳23 QDPOrderProof (⇔)

        ↳24 QDP

        ↳25 QDPOrderProof (⇔)

        ↳26 QDP

        ↳27 PisEmptyProof (⇔)

        ↳28 TRUE

    ↳29 QDP

        ↳30 QDPOrderProof (⇔)

        ↳31 QDP

        ↳32 QDPOrderProof (⇔)

        ↳33 QDP

        ↳34 PisEmptyProof (⇔)

        ↳35 TRUE

    ↳36 QDP

        ↳37 QDPOrderProof (⇔)

        ↳38 QDP

        ↳39 QDPOrderProof (⇔)

        ↳40 QDP

        ↳41 PisEmptyProof (⇔)

        ↳42 TRUE

    ↳43 QDP

        ↳44 QDPOrderProof (⇔)

        ↳45 QDP

        ↳46 QDPOrderProof (⇔)

        ↳47 QDP

        ↳48 PisEmptyProof (⇔)

        ↳49 TRUE

    ↳50 QDP

        ↳51 QDPOrderProof (⇔)

        ↳52 QDP

        ↳53 QDPOrderProof (⇔)

        ↳54 QDP

        ↳55 PisEmptyProof (⇔)

        ↳56 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)
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.

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 17 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

GEQ(ok(X1), ok(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) = GEQ(x2)
ok(x1) = ok(x1)

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

trivial

Status:

GEQ₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) 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)
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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(ok(X1), ok(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) = MINUS(x2)
ok(x1) = ok(x1)

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

trivial

Status:

MINUS₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(12) 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)
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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(ok(X1), ok(X2), ok(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) = IF(x1, x2, x3)
ok(x1) = ok(x1)
mark(x1) = x1

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

[IF₃, ok₁]

Status:

IF₃: [3,2,1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(17) 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)
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.

(18) 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) = IF(x1)
mark(x1) = mark(x1)

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

trivial

Status:

IF₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(19) 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)
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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

DIV(ok(X1), ok(X2)) → DIV(X1, X2)
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)
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.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DIV(ok(X1), ok(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) = DIV(x1, x2)
ok(x1) = ok(x1)
mark(x1) = x1

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

trivial

Status:

DIV₂: [2,1]
ok₁: [1]

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)

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.

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

[DIV₂, mark₁]

Status:

DIV₂: [2,1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(26) 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)
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.

(27) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) Obligation:

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))

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(ok(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
ok(x1) = ok(x1)
mark(x1) = x1

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

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

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

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))

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.

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
mark(x1) = mark(x1)

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

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(33) 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)
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.

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) Obligation:

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

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

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

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

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

trivial

Status:

PROPER₁: [1]
minus₂: [2,1]
geq₂: [1,2]
div₂: [1,2]
if₃: [2,1,3]

The following usable rules [FROCOS05] were oriented: none

(38) Obligation:

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

PROPER(s(X)) → PROPER(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.

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(s(X)) → PROPER(X)

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

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

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(40) 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)
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.

(41) PisEmptyProof (EQUIVALENT transformation)

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

(42) TRUE

(43) Obligation:

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

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

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

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
div(x1, x2) = div(x1, x2)
s(x1) = x1
if(x1, x2, x3) = if(x1, x2, x3)

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

trivial

Status:

div₂: [2,1]
if₃: [3,2,1]

The following usable rules [FROCOS05] were oriented: none

(45) Obligation:

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

ACTIVE(s(X)) → 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.

(46) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(s(X)) → ACTIVE(X)

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

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

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(47) 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)
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.

(48) PisEmptyProof (EQUIVALENT transformation)

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

(49) TRUE

(50) Obligation:

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

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

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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

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

TOP₁ > 0
[geq₁, true, div₁] > s₁ > [mark₁, minus₁] > 0
[geq₁, true, div₁] > false > [mark₁, minus₁] > 0
[geq₁, true, div₁] > if₃ > [mark₁, minus₁] > 0

Status:

TOP₁: [1]
mark₁: [1]
minus₁: [1]
0: []
s₁: [1]
geq₁: [1]
true: []
false: []
div₁: [1]
if₃: [1,3,2]

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

(52) Obligation:

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)
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.

(53) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(ok(X)) → TOP(active(X))

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

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

TOP₁ > [active, mark]
minus₁ > ok > [active, mark]
minus₁ > 0 > true > [active, mark]
false > [active, mark]

Status:

TOP₁: [1]
ok: []
active: []
minus₁: [1]
0: []
mark: []
true: []
false: []

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)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(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))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))

(54) 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)
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.

(55) PisEmptyProof (EQUIVALENT transformation)

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

(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) PisEmptyProof (EQUIVALENT transformation)

(9) TRUE

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) PisEmptyProof (EQUIVALENT transformation)

(14) TRUE

(15) Obligation:

(16) QDPOrderProof (EQUIVALENT transformation)

(17) Obligation:

(18) QDPOrderProof (EQUIVALENT transformation)

(19) Obligation:

(20) PisEmptyProof (EQUIVALENT transformation)

(21) TRUE

(22) Obligation:

(23) QDPOrderProof (EQUIVALENT transformation)

(24) Obligation:

(25) QDPOrderProof (EQUIVALENT transformation)

(26) Obligation:

(27) PisEmptyProof (EQUIVALENT transformation)

(28) TRUE

(29) Obligation:

(30) QDPOrderProof (EQUIVALENT transformation)

(31) Obligation:

(32) QDPOrderProof (EQUIVALENT transformation)

(33) Obligation:

(34) PisEmptyProof (EQUIVALENT transformation)

(35) TRUE

(36) Obligation:

(37) QDPOrderProof (EQUIVALENT transformation)

(38) Obligation:

(39) QDPOrderProof (EQUIVALENT transformation)

(40) Obligation:

(41) PisEmptyProof (EQUIVALENT transformation)

(42) TRUE

(43) Obligation:

(44) QDPOrderProof (EQUIVALENT transformation)

(45) Obligation:

(46) QDPOrderProof (EQUIVALENT transformation)

(47) Obligation:

(48) PisEmptyProof (EQUIVALENT transformation)

(49) TRUE

(50) Obligation:

(51) QDPOrderProof (EQUIVALENT transformation)

(52) Obligation:

(53) QDPOrderProof (EQUIVALENT transformation)

(54) Obligation:

(55) PisEmptyProof (EQUIVALENT transformation)

(56) TRUE