(0) Obligation:

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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), 0) → GE(x, 0)
GE(0, s(s(x))) → GE(0, s(x))
GE(s(x), s(y)) → GE(x, y)
MINUS(0, s(x)) → MINUS(0, x)
MINUS(s(x), 0) → MINUS(x, 0)
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(0, s(x)) → PLUS(0, x)
PLUS(s(x), y) → PLUS(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(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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 9 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(0, s(x)) → PLUS(0, x)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.


PLUS(0, s(x)) → PLUS(0, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[0, s1] > PLUS2

Status:
PLUS2: [1,2]
0: []
s1: [1]


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(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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:

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.


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

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

Status:
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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:

MINUS(s(x), 0) → MINUS(x, 0)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(s(x), 0) → MINUS(x, 0)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[MINUS2, s1] > 0

Status:
MINUS2: [2,1]
s1: [1]
0: []


The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(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:

MINUS(0, s(x)) → MINUS(0, x)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(0, s(x)) → MINUS(0, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[0, s1] > MINUS2

Status:
MINUS2: [1,2]
0: []
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) 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(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(28) 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)

Lexicographic path order with status [LPO].
Quasi-Precedence:
s1 > MINUS1

Status:
MINUS1: [1]
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(29) Obligation:

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE

(32) Obligation:

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

GE(0, s(s(x))) → GE(0, s(x))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


GE(0, s(s(x))) → GE(0, s(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GE(x1, x2)  =  GE(x2)
0  =  0
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[GE1, 0] > s1

Status:
GE1: [1]
0: []
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) Obligation:

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

GE(s(x), 0) → GE(x, 0)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


GE(s(x), 0) → GE(x, 0)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[GE2, s1] > 0

Status:
GE2: [2,1]
s1: [1]
0: []


The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(40) PisEmptyProof (EQUIVALENT transformation)

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

(41) TRUE

(42) 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(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(43) 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)

Lexicographic path order with status [LPO].
Quasi-Precedence:
s1 > GE1

Status:
GE1: [1]
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) 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(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(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(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), 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.