-(0,

-(

-(

p(0) -> 0

p(s(

R

↳Dependency Pair Analysis

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Rewriting Transformation

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

-(0,y) -> 0

-(x, 0) ->x

-(x, s(y)) -> if(greater(x, s(y)), s(-(x, p(s(y)))), 0)

p(0) -> 0

p(s(x)) ->x

innermost

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Rw

→DP Problem 2

↳Polynomial Ordering

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

-(0,y) -> 0

-(x, 0) ->x

-(x, s(y)) -> if(greater(x, s(y)), s(-(x, p(s(y)))), 0)

p(0) -> 0

p(s(x)) ->x

innermost

The following dependency pair can be strictly oriented:

-'(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 _{2}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Rw

→DP Problem 2

↳Polo

...

→DP Problem 3

↳Dependency Graph

-(0,y) -> 0

-(x, 0) ->x

-(x, s(y)) -> if(greater(x, s(y)), s(-(x, p(s(y)))), 0)

p(0) -> 0

p(s(x)) ->x

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes