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 PisEmptyProof (⇔)
    ↳19 TRUE

(0) Obligation:

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)
A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__DIV(s(X), s(Y)) → A__GEQ(X, Y)
A__IF(true, X, Y) → MARK(X)
A__IF(false, X, Y) → MARK(Y)
MARK(minus(X1, X2)) → A__MINUS(X1, X2)
MARK(geq(X1, X2)) → A__GEQ(X1, X2)
MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__GEQ(x1, x2)  =  x2
s(x1)  =  s(x1)
a__minus(x1, x2)  =  a__minus
0  =  0
a__geq(x1, x2)  =  a__geq
true  =  true
false  =  false
a__div(x1, x2)  =  a__div(x1)
a__if(x1, x2, x3)  =  a__if(x2, x3)
div(x1, x2)  =  div(x1)
minus(x1, x2)  =  minus
mark(x1)  =  x1
geq(x1, x2)  =  geq
if(x1, x2, x3)  =  if(x2, x3)

Lexicographic Path Order [LPO].
Precedence:
[ageq, geq] > false > [true, aif2, if2]
[adiv1, div1] > s1 > [aminus, 0, minus] > [true, aif2, if2]


The following usable rules [FROCOS05] were oriented:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

(7) Obligation:

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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:

A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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.


A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__MINUS(x1, x2)  =  x2
s(x1)  =  s(x1)
a__minus(x1, x2)  =  a__minus
0  =  0
a__geq(x1, x2)  =  a__geq
true  =  true
false  =  false
a__div(x1, x2)  =  a__div(x1)
a__if(x1, x2, x3)  =  a__if(x2, x3)
div(x1, x2)  =  div(x1)
minus(x1, x2)  =  minus
mark(x1)  =  x1
geq(x1, x2)  =  geq
if(x1, x2, x3)  =  if(x2, x3)

Lexicographic Path Order [LPO].
Precedence:
[ageq, geq] > false > [true, aif2, if2]
[adiv1, div1] > s1 > [aminus, 0, minus] > [true, aif2, if2]


The following usable rules [FROCOS05] were oriented:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

(12) Obligation:

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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:

MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__IF(true, X, Y) → MARK(X)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(false, X, Y) → MARK(Y)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__IF(true, X, Y) → MARK(X)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(false, X, Y) → MARK(Y)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
div(x1, x2)  =  div(x1)
A__DIV(x1, x2)  =  A__DIV(x1)
mark(x1)  =  x1
s(x1)  =  s(x1)
A__IF(x1, x2, x3)  =  A__IF(x1, x2, x3)
a__geq(x1, x2)  =  a__geq(x1)
minus(x1, x2)  =  x1
0  =  0
true  =  true
if(x1, x2, x3)  =  if(x1, x2, x3)
false  =  false
a__minus(x1, x2)  =  x1
a__div(x1, x2)  =  a__div(x1)
a__if(x1, x2, x3)  =  a__if(x1, x2, x3)
geq(x1, x2)  =  geq(x1)

Lexicographic Path Order [LPO].
Precedence:
[div1, ADIV1, ageq1, adiv1, geq1] > [AIF3, if3, aif3] > [MARK1, true] > [s1, 0]
[div1, ADIV1, ageq1, adiv1, geq1] > false > [MARK1, true] > [s1, 0]


The following usable rules [FROCOS05] were oriented:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

(17) Obligation:

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE