O(0) -> 0

+(0,

+(

+(O(

+(O(

+(I(

+(I(

*(0,

*(

*(O(

*(I(

-(

-(0,

-(O(

-(O(

-(I(

-(I(

R

↳Dependency Pair Analysis

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

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

+'(O(x), I(y)) -> +'(x,y)

+'(I(x), O(y)) -> +'(x,y)

+'(I(x), I(y)) -> O'(+(+(x,y), I(0)))

+'(I(x), I(y)) -> +'(+(x,y), I(0))

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

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

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

*'(I(x),y) -> +'(O(*(x,y)),y)

*'(I(x),y) -> O'(*(x,y))

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

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

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

-'(O(x), I(y)) -> -'(-(x,y), I(1))

-'(O(x), I(y)) -> -'(x,y)

-'(I(x), O(y)) -> -'(x,y)

-'(I(x), I(y)) -> O'(-(x,y))

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Narrowing Transformation

→DP Problem 2

↳Nar

→DP Problem 3

↳Polo

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

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

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

As a result of transforming the rule

six new Dependency Pairs are created:

+'(I(x), I(y)) -> +'(+(x,y), I(0))

+'(I(0), I(y')) -> +'(y', I(0))

+'(I(x''), I(0)) -> +'(x'', I(0))

+'(I(O(x'')), I(O(y''))) -> +'(O(+(x'',y'')), I(0))

+'(I(O(x'')), I(I(y''))) -> +'(I(+(x'',y'')), I(0))

+'(I(I(x'')), I(O(y''))) -> +'(I(+(x'',y'')), I(0))

+'(I(I(x'')), I(I(y''))) -> +'(O(+(+(x'',y''), I(0))), I(0))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 4

↳Polynomial Ordering

→DP Problem 2

↳Nar

→DP Problem 3

↳Polo

**+'(I(I( x'')), I(I(y''))) -> +'(O(+(+(x'', y''), I(0))), I(0))**

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

The following dependency pairs can be strictly oriented:

+'(I(I(x'')), I(O(y''))) -> +'(I(+(x'',y'')), I(0))

+'(I(O(x'')), I(O(y''))) -> +'(O(+(x'',y'')), I(0))

+'(I(x), O(y)) -> +'(x,y)

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(I(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(O(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(+'(x)_{1}, x_{2})= x _{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 4

↳Polo

...

→DP Problem 5

↳Polynomial Ordering

→DP Problem 2

↳Nar

→DP Problem 3

↳Polo

**+'(I(I( x'')), I(I(y''))) -> +'(O(+(+(x'', y''), I(0))), I(0))**

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

The following dependency pairs can be strictly oriented:

+'(I(I(x'')), I(I(y''))) -> +'(O(+(+(x'',y''), I(0))), I(0))

+'(I(O(x'')), I(I(y''))) -> +'(I(+(x'',y'')), I(0))

+'(O(x), I(y)) -> +'(x,y)

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(I(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(O(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(+'(x)_{1}, x_{2})= x _{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 4

↳Polo

...

→DP Problem 6

↳Instantiation Transformation

→DP Problem 2

↳Nar

→DP Problem 3

↳Polo

**+'(I( x''), I(0)) -> +'(x'', I(0))**

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

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

As a result of transforming the rule

one new Dependency Pair is created:

+'(I(0), I(y')) -> +'(y', I(0))

+'(I(0), I(0)) -> +'(0, I(0))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 4

↳Polo

...

→DP Problem 7

↳Polynomial Ordering

→DP Problem 2

↳Nar

→DP Problem 3

↳Polo

**+'(I( x''), I(0)) -> +'(x'', I(0))**

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

The following dependency pair can be strictly oriented:

+'(I(x''), I(0)) -> +'(x'', I(0))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(I(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(+'(x)_{1}, x_{2})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 4

↳Polo

...

→DP Problem 8

↳Dependency Graph

→DP Problem 2

↳Nar

→DP Problem 3

↳Polo

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Narrowing Transformation

→DP Problem 3

↳Polo

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

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

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

As a result of transforming the rule

six new Dependency Pairs are created:

-'(O(x), I(y)) -> -'(-(x,y), I(1))

-'(O(x''), I(0)) -> -'(x'', I(1))

-'(O(0), I(y')) -> -'(0, I(1))

-'(O(O(x'')), I(O(y''))) -> -'(O(-(x'',y'')), I(1))

-'(O(O(x'')), I(I(y''))) -> -'(I(-(-(x'',y''), I(1))), I(1))

-'(O(I(x'')), I(O(y''))) -> -'(I(-(x'',y'')), I(1))

-'(O(I(x'')), I(I(y''))) -> -'(O(-(x'',y'')), I(1))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

→DP Problem 9

↳Polynomial Ordering

→DP Problem 3

↳Polo

**-'(O(I( x'')), I(I(y''))) -> -'(O(-(x'', y'')), I(1))**

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

The following dependency pair can be strictly oriented:

-'(O(x''), I(0)) -> -'(x'', I(1))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(I(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 1 _{ }^{ }_{ }^{ }POL(-'(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }_{ }^{ }POL(O(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(-(x)_{1}, x_{2})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

→DP Problem 9

↳Polo

...

→DP Problem 10

↳Polynomial Ordering

→DP Problem 3

↳Polo

**-'(O(I( x'')), I(I(y''))) -> -'(O(-(x'', y'')), I(1))**

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

The following dependency pairs can be strictly oriented:

-'(O(I(x'')), I(O(y''))) -> -'(I(-(x'',y'')), I(1))

-'(O(O(x'')), I(O(y''))) -> -'(O(-(x'',y'')), I(1))

-'(I(x), O(y)) -> -'(x,y)

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(I(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(-'(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }_{ }^{ }POL(O(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(-(x)_{1}, x_{2})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

→DP Problem 9

↳Polo

...

→DP Problem 11

↳Polynomial Ordering

→DP Problem 3

↳Polo

**-'(O(I( x'')), I(I(y''))) -> -'(O(-(x'', y'')), I(1))**

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

The following dependency pairs can be strictly oriented:

-'(O(I(x'')), I(I(y''))) -> -'(O(-(x'',y'')), I(1))

-'(O(O(x'')), I(I(y''))) -> -'(I(-(-(x'',y''), I(1))), I(1))

-'(O(x), I(y)) -> -'(x,y)

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(I(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(-'(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }_{ }^{ }POL(O(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(-(x)_{1}, x_{2})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

→DP Problem 9

↳Polo

...

→DP Problem 12

↳Dependency Graph

→DP Problem 3

↳Polo

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

→DP Problem 3

↳Polynomial Ordering

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

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(I(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(*'(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(O(x)_{1})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

→DP Problem 3

↳Polo

→DP Problem 13

↳Polynomial Ordering

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

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(*'(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(O(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

→DP Problem 3

↳Polo

→DP Problem 13

↳Polo

...

→DP Problem 14

↳Dependency Graph

O(0) -> 0

+(0,x) ->x

+(x, 0) ->x

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

+(O(x), I(y)) -> I(+(x,y))

+(I(x), O(y)) -> I(+(x,y))

+(I(x), I(y)) -> O(+(+(x,y), I(0)))

*(0,x) -> 0

*(x, 0) -> 0

*(O(x),y) -> O(*(x,y))

*(I(x),y) -> +(O(*(x,y)),y)

-(x, 0) ->x

-(0,x) -> 0

-(O(x), O(y)) -> O(-(x,y))

-(O(x), I(y)) -> I(-(-(x,y), I(1)))

-(I(x), O(y)) -> I(-(x,y))

-(I(x), I(y)) -> O(-(x,y))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes