Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

GE(s(x), s(y)) → GE(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
DIV(x, y) → GE(y, s(0))
IFY(true, x, y) → IF(ge(x, y), x, y)
IFY(true, x, y) → GE(x, y)
IF(true, x, y) → DIV(minus(x, y), y)
IF(true, x, y) → MINUS(x, y)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

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

Recursive path order with status [RPO].
Quasi-Precedence:

[MINUS₁, s₁]

Status:

s₁: [1]
MINUS₁: multiset

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

GE(s(x), s(y)) → GE(x, y)

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

Recursive path order with status [RPO].
Quasi-Precedence:

[GE₁, s₁]

Status:

s₁: [1]
GE₁: multiset

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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