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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →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 that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

GE(x₁, x₂) -> GE(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →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

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and 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 that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →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.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 10

                 ↳Remaining Obligation(s)

The following remains to be proven:
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

Innermost Termination of R could not be shown.
Duration:
0:00 minutes