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 DependencyGraphProof (⇔)
    ↳19 TRUE

(0) Obligation:

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

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

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:

MINUS_ACTIVE(s(x), s(y)) → MINUS_ACTIVE(x, y)
MARK(s(x)) → MARK(x)
MARK(minus(x, y)) → MINUS_ACTIVE(x, y)
MARK(ge(x, y)) → GE_ACTIVE(x, y)
GE_ACTIVE(s(x), s(y)) → GE_ACTIVE(x, y)
MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
MARK(div(x, y)) → MARK(x)
MARK(if(x, y, z)) → IF_ACTIVE(mark(x), y, z)
MARK(if(x, y, z)) → MARK(x)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
DIV_ACTIVE(s(x), s(y)) → GE_ACTIVE(x, y)
IF_ACTIVE(true, x, y) → MARK(x)
IF_ACTIVE(false, x, y) → MARK(y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

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:

GE_ACTIVE(s(x), s(y)) → GE_ACTIVE(x, y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

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.


GE_ACTIVE(s(x), s(y)) → GE_ACTIVE(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GE_ACTIVE(x1, x2)  =  x2
s(x1)  =  s(x1)
minus_active(x1, x2)  =  minus_active(x1)
0  =  0
mark(x1)  =  x1
ge_active(x1, x2)  =  x1
true  =  true
minus(x1, x2)  =  minus(x1)
false  =  false
ge(x1, x2)  =  x1
div(x1, x2)  =  div(x1, x2)
div_active(x1, x2)  =  div_active(x1, x2)
if(x1, x2, x3)  =  if(x1, x2, x3)
if_active(x1, x2, x3)  =  if_active(x1, x2, x3)

Lexicographic Path Order [LPO].
Precedence:
[div2, divactive2, if3, ifactive3] > s1 > [minusactive1, minus1] > true
[div2, divactive2, if3, ifactive3] > [0, false] > true


The following usable rules [FROCOS05] were oriented:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

(7) Obligation:

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

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

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_ACTIVE(s(x), s(y)) → MINUS_ACTIVE(x, y)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

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_ACTIVE(s(x), s(y)) → MINUS_ACTIVE(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS_ACTIVE(x1, x2)  =  x2
s(x1)  =  s(x1)
minus_active(x1, x2)  =  minus_active(x1)
0  =  0
mark(x1)  =  x1
ge_active(x1, x2)  =  x1
true  =  true
minus(x1, x2)  =  minus(x1)
false  =  false
ge(x1, x2)  =  x1
div(x1, x2)  =  div(x1)
div_active(x1, x2)  =  div_active(x1)
if(x1, x2, x3)  =  if(x1, x2, x3)
if_active(x1, x2, x3)  =  if_active(x1, x2, x3)

Lexicographic Path Order [LPO].
Precedence:
[div1, divactive1] > s1 > [minusactive1, minus1] > true
[div1, divactive1] > [0, false] > true
[div1, divactive1] > [if3, ifactive3] > true


The following usable rules [FROCOS05] were oriented:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

(12) Obligation:

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

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

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(x, y)) → DIV_ACTIVE(mark(x), y)
DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
IF_ACTIVE(true, x, y) → MARK(x)
MARK(s(x)) → MARK(x)
MARK(div(x, y)) → MARK(x)
MARK(if(x, y, z)) → IF_ACTIVE(mark(x), y, z)
IF_ACTIVE(false, x, y) → MARK(y)
MARK(if(x, y, z)) → MARK(x)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

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.


DIV_ACTIVE(s(x), s(y)) → IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
IF_ACTIVE(true, x, y) → MARK(x)
MARK(s(x)) → MARK(x)
MARK(div(x, y)) → MARK(x)
IF_ACTIVE(false, x, y) → MARK(y)
MARK(if(x, y, z)) → MARK(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
div(x1, x2)  =  div(x1, x2)
DIV_ACTIVE(x1, x2)  =  DIV_ACTIVE(x1, x2)
mark(x1)  =  x1
s(x1)  =  s(x1)
IF_ACTIVE(x1, x2, x3)  =  IF_ACTIVE(x1, x2, x3)
ge_active(x1, x2)  =  ge_active(x1)
minus(x1, x2)  =  x1
0  =  0
true  =  true
if(x1, x2, x3)  =  if(x1, x2, x3)
false  =  false
minus_active(x1, x2)  =  x1
ge(x1, x2)  =  ge(x1)
div_active(x1, x2)  =  div_active(x1, x2)
if_active(x1, x2, x3)  =  if_active(x1, x2, x3)

Lexicographic Path Order [LPO].
Precedence:
[div2, DIVACTIVE2, IFACTIVE3, 0, if3, divactive2, ifactive3] > s1 > [geactive1, ge1] > true
[div2, DIVACTIVE2, IFACTIVE3, 0, if3, divactive2, ifactive3] > s1 > [geactive1, ge1] > false


The following usable rules [FROCOS05] were oriented:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

(17) Obligation:

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

MARK(div(x, y)) → DIV_ACTIVE(mark(x), y)
MARK(if(x, y, z)) → IF_ACTIVE(mark(x), y, z)

The TRS R consists of the following rules:

minus_active(0, y) → 0
mark(0) → 0
minus_active(s(x), s(y)) → minus_active(x, y)
mark(s(x)) → s(mark(x))
ge_active(x, 0) → true
mark(minus(x, y)) → minus_active(x, y)
ge_active(0, s(y)) → false
mark(ge(x, y)) → ge_active(x, y)
ge_active(s(x), s(y)) → ge_active(x, y)
mark(div(x, y)) → div_active(mark(x), y)
div_active(0, s(y)) → 0
mark(if(x, y, z)) → if_active(mark(x), y, z)
div_active(s(x), s(y)) → if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
if_active(true, x, y) → mark(x)
minus_active(x, y) → minus(x, y)
if_active(false, x, y) → mark(y)
ge_active(x, y) → ge(x, y)
if_active(x, y, z) → if(x, y, z)
div_active(x, y) → div(x, y)

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

(18) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(19) TRUE