-(

-(0, s(

-(s(

lt(

lt(0, s(

lt(s(

if(true,

if(false,

div(

div(0,

div(s(

R

↳Dependency Pair Analysis

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

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

DIV(s(x), s(y)) -> IF(lt(x,y), 0, s(div(-(x,y), s(y))))

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

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

-(x, 0) ->x

-(0, s(y)) -> 0

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

lt(x, 0) -> false

lt(0, s(y)) -> true

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

if(true,x,y) ->x

if(false,x,y) ->y

div(x, 0) -> 0

div(0,y) -> 0

div(s(x), s(y)) -> if(lt(x,y), 0, s(div(-(x,y), s(y))))

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

-(x, 0) ->x

-(0, s(y)) -> 0

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

lt(x, 0) -> false

lt(0, s(y)) -> true

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

if(true,x,y) ->x

if(false,x,y) ->y

div(x, 0) -> 0

div(0,y) -> 0

div(s(x), s(y)) -> if(lt(x,y), 0, s(div(-(x,y), s(y))))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

→DP Problem 3

↳Polo

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

-(x, 0) ->x

-(0, s(y)) -> 0

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

lt(x, 0) -> false

lt(0, s(y)) -> true

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

if(true,x,y) ->x

if(false,x,y) ->y

div(x, 0) -> 0

div(0,y) -> 0

div(s(x), s(y)) -> if(lt(x,y), 0, s(div(-(x,y), s(y))))

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Polo

-(x, 0) ->x

-(0, s(y)) -> 0

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

lt(x, 0) -> false

lt(0, s(y)) -> true

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

if(true,x,y) ->x

if(false,x,y) ->y

div(x, 0) -> 0

div(0,y) -> 0

div(s(x), s(y)) -> if(lt(x,y), 0, s(div(-(x,y), s(y))))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polynomial Ordering

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

-(x, 0) ->x

-(0, s(y)) -> 0

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

lt(x, 0) -> false

lt(0, s(y)) -> true

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

if(true,x,y) ->x

if(false,x,y) ->y

div(x, 0) -> 0

div(0,y) -> 0

div(s(x), s(y)) -> if(lt(x,y), 0, s(div(-(x,y), s(y))))

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

-(x, 0) ->x

-(0, s(y)) -> 0

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(DIV(x)_{1}, x_{2})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(-(x)_{1}, x_{2})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 6

↳Dependency Graph

-(x, 0) ->x

-(0, s(y)) -> 0

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

lt(x, 0) -> false

lt(0, s(y)) -> true

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

if(true,x,y) ->x

if(false,x,y) ->y

div(x, 0) -> 0

div(0,y) -> 0

div(s(x), s(y)) -> if(lt(x,y), 0, s(div(-(x,y), s(y))))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes