Termination Proof using AProVE

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.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

GE(s(s(x'')), s(s(y''))) -> GE(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 Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

The following dependency pair can be strictly oriented:

GE(s(s(s(x''''))), s(s(s(y'''')))) -> GE(s(s(x'''')), s(s(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)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(s(s(x'')), s(s(y''))) -> MINUS(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 Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

The following dependency pair can be strictly oriented:

MINUS(s(s(s(x''''))), s(s(s(y'''')))) -> MINUS(s(s(x'''')), s(s(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)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →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:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳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:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 11

                 ↳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:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 12

                 ↳Narrowing 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, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 13

                 ↳Instantiation Transformation

Dependency Pairs:

IFY(true, s(s(x')), s(s(y'))) -> IF(ge(x', y'), s(s(x')), s(s(y')))
IFY(true, s(x'''), s(0)) -> IF(true, s(x'''), s(0))
DIV(x, s(x'')) -> IFY(true, 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, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 14

                 ↳Rewriting Transformation

Dependency Pairs:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 15

                 ↳Rewriting Transformation

Dependency Pairs:

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 16

                 ↳Rewriting Transformation

Dependency Pairs:

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

one new Dependency Pair is created:

IF(true, s(x'''''), s(0)) -> DIV(x''''', s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 17

                 ↳Rewriting Transformation

Dependency Pairs:

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 18

                 ↳Polynomial Ordering

Dependency Pairs:

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

The following dependency pairs can be strictly oriented:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 19

                 ↳Dependency Graph

Dependency Pairs:

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

Using the Dependency Graph resulted in no new DP problems.

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

POL(MINUS(x₁, x₂))	= 1 + x₁
POL(s(x₁))	= 1 + x₁

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