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

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

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

As we are in the innermost case, we can delete all 10 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1
2	>	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

s(x₁) -> s(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

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

As we are in the innermost case, we can delete all 10 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1
2	>	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

s(x₁) -> s(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

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

As we are in the innermost case, we can delete all 5 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳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:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
ge(x, 0) -> true
ge(s(x), s(y)) -> ge(x, y)
ge(0, s(x)) -> false

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
ge(x, 0) -> true
ge(s(x), s(y)) -> ge(x, y)
ge(0, s(x)) -> false

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Rewriting Transformation

Dependency Pairs:

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

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
ge(x, 0) -> true
ge(s(x), s(y)) -> ge(x, y)
ge(0, s(x)) -> false

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

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳Instantiation Transformation

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

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
ge(x, 0) -> true
ge(s(x), s(y)) -> ge(x, y)
ge(0, s(x)) -> false

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

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳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:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
ge(x, 0) -> true
ge(s(x), s(y)) -> ge(x, y)
ge(0, s(x)) -> false

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

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 11

                 ↳Negative Polynomial Order

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:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
ge(x, 0) -> true
ge(s(x), s(y)) -> ge(x, y)
ge(0, s(x)) -> false

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

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

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
ge(x, 0) -> true
ge(s(x), s(y)) -> ge(x, y)
ge(0, s(x)) -> false

Used ordering:
Polynomial Order with Interpretation:

POL( IF(x₁, ..., x₃) ) = x₂

POL( s(x₁) ) = x₁ + 1

POL( DIV(x₁, x₂) ) = x₁

POL( minus(x₁, x₂) ) = x₁

POL( IFY(x₁, ..., x₃) ) = x₂

POL( ge(x₁, x₂) ) = 0

POL( true ) = 0

POL( false ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 12

                 ↳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:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
ge(x, 0) -> true
ge(s(x), s(y)) -> ge(x, y)
ge(0, s(x)) -> false

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:07 minutes