(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)  =  x2
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
minus(x1, x2)  =  minus(x1)
0  =  0
mark(x1)  =  x1
s(x1)  =  x1
geq(x1, x2)  =  geq(x2)
true  =  true
false  =  false
div(x1, x2)  =  div(x2)
if(x1, x2, x3)  =  if(x1, x2, x3)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
proper1 > div1 > active1 > if3 > ok1
proper1 > div1 > minus1 > ok1
proper1 > div1 > 0 > ok1
proper1 > div1 > geq1 > true > ok1
proper1 > div1 > geq1 > false > ok1
top > active1 > if3 > ok1

Status:
ok1: [1]
active1: [1]
minus1: [1]
0: []
geq1: [1]
true: []
false: []
div1: [1]
if3: [2,3,1]
proper1: [1]
top: []

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)
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))

(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)  =  x2
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
minus(x1, x2)  =  minus(x1)
0  =  0
mark(x1)  =  x1
s(x1)  =  x1
geq(x1, x2)  =  geq(x2)
true  =  true
false  =  false
div(x1, x2)  =  div(x2)
if(x1, x2, x3)  =  if(x1, x2, x3)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
proper1 > div1 > active1 > if3 > ok1
proper1 > div1 > minus1 > ok1
proper1 > div1 > 0 > ok1
proper1 > div1 > geq1 > true > ok1
proper1 > div1 > geq1 > false > ok1
top > active1 > if3 > ok1

Status:
ok1: [1]
active1: [1]
minus1: [1]
0: []
geq1: [1]
true: []
false: []
div1: [1]
if3: [2,3,1]
proper1: [1]
top: []

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)
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))

(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(x3)
ok(x1)  =  ok(x1)
mark(x1)  =  mark
active(x1)  =  active(x1)
minus(x1, x2)  =  x2
0  =  0
s(x1)  =  x1
geq(x1, x2)  =  x2
true  =  true
false  =  false
div(x1, x2)  =  x1
if(x1, x2, x3)  =  x3
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
IF1 > mark
0 > ok1 > mark
true > ok1 > mark
false > ok1 > mark
top > active1 > mark
top > proper1 > ok1 > mark

Status:
IF1: [1]
ok1: [1]
mark: []
active1: [1]
0: []
true: []
false: []
proper1: [1]
top: []

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)
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))

(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, x2, x3)
mark(x1)  =  mark(x1)
active(x1)  =  x1
minus(x1, x2)  =  minus(x1)
0  =  0
s(x1)  =  s(x1)
geq(x1, x2)  =  x2
true  =  true
false  =  false
div(x1, x2)  =  div(x1, x2)
if(x1, x2, x3)  =  if(x1, x2, x3)
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top(x1)

Lexicographic path order with status [LPO].
Precedence:
IF3 > mark1
div2 > s1 > minus1 > mark1
div2 > s1 > 0 > true > mark1
div2 > s1 > false > mark1
div2 > s1 > if3 > mark1
top1 > mark1

Status:
IF3: [3,1,2]
mark1: [1]
minus1: [1]
0: []
s1: [1]
true: []
false: []
div2: [1,2]
if3: [3,1,2]
top1: [1]

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)
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))

(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)  =  x2
ok(x1)  =  ok(x1)
mark(x1)  =  mark
active(x1)  =  x1
minus(x1, x2)  =  minus(x1)
0  =  0
s(x1)  =  s(x1)
geq(x1, x2)  =  x2
true  =  true
false  =  false
div(x1, x2)  =  x1
if(x1, x2, x3)  =  if(x1, x2, x3)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
proper1 > minus1 > ok1 > mark
proper1 > 0 > ok1 > mark
proper1 > s1 > ok1 > mark
proper1 > true > ok1 > mark
proper1 > false > ok1 > mark
proper1 > if3 > ok1 > mark
top > mark

Status:
ok1: [1]
mark: []
minus1: [1]
0: []
s1: [1]
true: []
false: []
if3: [3,2,1]
proper1: [1]
top: []

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)
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))

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

Lexicographic path order with status [LPO].
Precedence:
DIV2 > mark1
active1 > 0 > mark1
active1 > s1 > if3 > mark1
active1 > true > mark1
active1 > false > mark1
active1 > div2 > if3 > mark1
top > mark1

Status:
DIV2: [1,2]
mark1: [1]
active1: [1]
0: []
s1: [1]
true: []
false: []
div2: [1,2]
if3: [3,1,2]
top: []

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)
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))

(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)  =  S(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
minus(x1, x2)  =  x2
0  =  0
s(x1)  =  x1
geq(x1, x2)  =  geq(x2)
true  =  true
false  =  false
div(x1, x2)  =  x2
if(x1, x2, x3)  =  if(x2, x3)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
S1 > 0
proper1 > geq1 > true > ok1 > 0
proper1 > geq1 > false > ok1 > 0
proper1 > if2 > ok1 > 0
top > active1 > if2 > ok1 > 0

Status:
S1: [1]
ok1: [1]
active1: [1]
0: []
geq1: [1]
true: []
false: []
if2: [2,1]
proper1: [1]
top: []

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)
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))

(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)
active(x1)  =  x1
minus(x1, x2)  =  minus(x1)
0  =  0
s(x1)  =  s(x1)
geq(x1, x2)  =  geq(x2)
true  =  true
false  =  false
div(x1, x2)  =  div(x1, x2)
if(x1, x2, x3)  =  if(x1, x2, x3)
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top(x1)

Lexicographic path order with status [LPO].
Precedence:
div2 > s1 > minus1 > mark1
div2 > s1 > 0 > mark1
div2 > s1 > false > mark1
div2 > s1 > if3 > mark1
div2 > geq1 > true > mark1
div2 > geq1 > false > mark1
top1 > mark1

Status:
mark1: [1]
minus1: [1]
0: []
s1: [1]
geq1: [1]
true: []
false: []
div2: [1,2]
if3: [3,1,2]
top1: [1]

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)
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))

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

Lexicographic path order with status [LPO].
Precedence:
active1 > geq2 > 0
active1 > mark > s1 > minus2 > 0
active1 > mark > div2 > minus2 > 0
active1 > mark > if3 > 0
active1 > mark > top > 0
active1 > true > 0
active1 > false > 0

Status:
minus2: [1,2]
s1: [1]
geq2: [1,2]
div2: [1,2]
if3: [3,1,2]
active1: [1]
0: []
mark: []
true: []
false: []
top: []

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)
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))

(38) 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.

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE

(41) 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.

(42) 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)
active(x1)  =  active(x1)
minus(x1, x2)  =  x2
0  =  0
mark(x1)  =  x1
geq(x1, x2)  =  x2
true  =  true
false  =  false
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
top > active1 > div2 > 0
top > active1 > if3 > 0
top > active1 > true > 0
top > active1 > false > 0

Status:
div2: [1,2]
if3: [2,3,1]
active1: [1]
0: []
true: []
false: []
top: []

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)
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))

(43) 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.

(44) 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)
active(x1)  =  x1
minus(x1, x2)  =  x1
0  =  0
mark(x1)  =  x1
geq(x1, x2)  =  geq(x2)
true  =  true
false  =  false
div(x1, x2)  =  div(x1, x2)
if(x1, x2, x3)  =  if(x2, x3)
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
geq1 > true > 0
geq1 > false > 0
div2 > s1 > false > 0
div2 > if2 > 0
top > 0

Status:
s1: [1]
0: []
geq1: [1]
true: []
false: []
div2: [1,2]
if2: [2,1]
top: []

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)
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))

(45) 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.

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) 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.

(49) 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)  =  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(x2)
true  =  true
false  =  false
div(x1, x2)  =  div(x1, x2)
if(x1, x2, x3)  =  if(x1, x2, x3)
top(x1)  =  top(x1)

Lexicographic path order with status [LPO].
Precedence:
div2 > s1 > minus1 > mark1 > 0
div2 > s1 > false > mark1 > 0
div2 > s1 > if3 > mark1 > 0
div2 > geq1 > true > mark1 > 0
div2 > geq1 > false > mark1 > 0
top1 > 0

Status:
mark1: [1]
minus1: [1]
0: []
s1: [1]
geq1: [1]
true: []
false: []
div2: [1,2]
if3: [1,2,3]
top1: [1]

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)
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))

(50) 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.

(51) 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)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
minus(x1, x2)  =  x2
0  =  0
mark(x1)  =  mark
s(x1)  =  x1
geq(x1, x2)  =  x2
true  =  true
false  =  false
div(x1, x2)  =  x1
if(x1, x2, x3)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
true > ok1 > active1 > mark > 0
false > ok1 > active1 > mark > 0
top > proper1 > ok1 > active1 > mark > 0

Status:
ok1: [1]
active1: [1]
0: []
mark: []
true: []
false: []
proper1: [1]
top: []

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)
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))

(52) 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.

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE