-(

-(s(

+(0,

+(s(

*(

*(

f(s(

R

↳Dependency Pair Analysis

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

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

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

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

F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))

F(s(x)) -> -'(*(s(s(0)), s(x)), s(s(x)))

F(s(x)) -> *'(s(s(0)), s(x))

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 4

↳Remaining

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

The following dependency pair can be strictly oriented:

-'(s(x), s(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(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 5

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 4

↳Remaining

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

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

→DP Problem 4

↳Remaining

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

The following dependency pair can be strictly oriented:

+'(s(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(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 6

↳Dependency Graph

→DP Problem 3

↳Polo

→DP Problem 4

↳Remaining

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

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

→DP Problem 4

↳Remaining

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

The following dependency pair can be strictly oriented:

*'(x, s(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 _{2}_{ }^{ }_{ }^{ }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 3

↳Polo

→DP Problem 7

↳Dependency Graph

→DP Problem 4

↳Remaining

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 4

↳Remaining Obligation(s)

The following remains to be proven:

**F(s( x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))**

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Duration:

0:00 minutes