+(0,

+(s(

-(0,

-(

-(s(

R

↳Dependency Pair Analysis

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

→DP Problem 2

↳FwdInst

**+'(s( x), y) -> +'(x, y)**

+(0,y) ->y

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

-(0,y) -> 0

-(x, 0) ->x

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

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳Forward Instantiation Transformation

→DP Problem 2

↳FwdInst

**+'(s(s( x'')), y'') -> +'(s(x''), y'')**

+(0,y) ->y

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

-(0,y) -> 0

-(x, 0) ->x

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

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳FwdInst

...

→DP Problem 4

↳Polynomial Ordering

→DP Problem 2

↳FwdInst

**+'(s(s(s( x''''))), y'''') -> +'(s(s(x'''')), y'''')**

+(0,y) ->y

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

-(0,y) -> 0

-(x, 0) ->x

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

innermost

The following dependency pair can be strictly oriented:

+'(s(s(s(x''''))),y'''') -> +'(s(s(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(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(+'(x)_{1}, x_{2})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳FwdInst

...

→DP Problem 5

↳Dependency Graph

→DP Problem 2

↳FwdInst

+(0,y) ->y

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

-(0,y) -> 0

-(x, 0) ->x

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

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Forward Instantiation Transformation

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

+(0,y) ->y

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

-(0,y) -> 0

-(x, 0) ->x

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

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 6

↳Forward Instantiation Transformation

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

+(0,y) ->y

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

-(0,y) -> 0

-(x, 0) ->x

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

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 6

↳FwdInst

...

→DP Problem 7

↳Polynomial Ordering

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

+(0,y) ->y

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

-(0,y) -> 0

-(x, 0) ->x

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

innermost

The following dependency pair can be strictly oriented:

-'(s(s(s(x''''))), s(s(s(y'''')))) -> -'(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(-'(x)_{1}, x_{2})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 6

↳FwdInst

...

→DP Problem 8

↳Dependency Graph

+(0,y) ->y

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

-(0,y) -> 0

-(x, 0) ->x

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

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes