Term Rewriting System R:
[x, y]
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))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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)

Furthermore, R contains three SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

GE(s(x), s(y)) -> GE(x, y)

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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

GE(s(x), s(y)) -> GE(x, y)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(GE(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

MINUS(s(x), s(y)) -> MINUS(x, y)

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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(s(x), s(y)) -> MINUS(x, y)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(MINUS(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Nar`

Dependency Pair:

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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Narrowing Transformation`

Dependency Pairs:

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

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))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

DIV(x, y) -> IFY(ge(y, s(0)), x, y)
two new Dependency Pairs are created:

DIV(x, 0) -> IFY(false, x, 0)
DIV(x, s(x'')) -> IFY(ge(x'', 0), x, s(x''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rewriting Transformation`

Dependency Pairs:

IFY(true, x, y) -> IF(ge(x, y), x, y)
DIV(x, s(x'')) -> IFY(ge(x'', 0), x, s(x''))
IF(true, x, y) -> DIV(minus(x, y), y)

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))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

DIV(x, s(x'')) -> IFY(ge(x'', 0), x, s(x''))
one new Dependency Pair is created:

DIV(x, s(x'')) -> IFY(true, x, s(x''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 7`
`                 ↳Narrowing Transformation`

Dependency Pairs:

DIV(x, s(x'')) -> IFY(true, x, s(x''))
IF(true, x, y) -> DIV(minus(x, y), y)
IFY(true, x, y) -> IF(ge(x, y), x, y)

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))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IFY(true, x, y) -> IF(ge(x, y), x, y)
three new Dependency Pairs are created:

IFY(true, x'', 0) -> IF(true, x'', 0)
IFY(true, 0, s(x'')) -> IF(false, 0, s(x''))
IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 8`
`                 ↳Instantiation Transformation`

Dependency Pairs:

IF(true, x, y) -> DIV(minus(x, y), y)
IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))
DIV(x, s(x'')) -> IFY(true, x, s(x''))

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))

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF(true, x, y) -> DIV(minus(x, y), y)
one new Dependency Pair is created:

IF(true, s(x'''''), s(y'''')) -> DIV(minus(s(x'''''), s(y'''')), s(y''''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 9`
`                 ↳Rewriting Transformation`

Dependency Pairs:

DIV(x, s(x'')) -> IFY(true, x, s(x''))
IF(true, s(x'''''), s(y'''')) -> DIV(minus(s(x'''''), s(y'''')), s(y''''))
IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))

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))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF(true, s(x'''''), s(y'''')) -> DIV(minus(s(x'''''), s(y'''')), s(y''''))
one new Dependency Pair is created:

IF(true, s(x'''''), s(y'''')) -> DIV(minus(x''''', y''''), s(y''''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 10`
`                 ↳Polynomial Ordering`

Dependency Pairs:

IF(true, s(x'''''), s(y'''')) -> DIV(minus(x''''', y''''), s(y''''))
IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))
DIV(x, s(x'')) -> IFY(true, x, s(x''))

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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF(true, s(x'''''), s(y'''')) -> DIV(minus(x''''', y''''), s(y''''))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(IFY(x1, x2, x3)) =  x2 POL(0) =  0 POL(false) =  0 POL(DIV(x1, x2)) =  x1 POL(minus(x1, x2)) =  x1 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(ge(x1, x2)) =  0 POL(IF(x1, x2, x3)) =  x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 11`
`                 ↳Dependency Graph`

Dependency Pairs:

IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))
DIV(x, s(x'')) -> IFY(true, x, s(x''))

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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes