f(g(

g(s(

g(0) -> 0

R

↳Dependency Pair Analysis

F(g(x), s(0),y) -> F(g(s(0)),y, g(x))

F(g(x), s(0),y) -> G(s(0))

G(s(x)) -> G(x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Rw

**G(s( x)) -> G(x)**

f(g(x), s(0),y) -> f(g(s(0)),y, g(x))

g(s(x)) -> s(g(x))

g(0) -> 0

innermost

The following dependency pair can be strictly oriented:

G(s(x)) -> G(x)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(G(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Rw

f(g(x), s(0),y) -> f(g(s(0)),y, g(x))

g(s(x)) -> s(g(x))

g(0) -> 0

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Rewriting Transformation

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

f(g(x), s(0),y) -> f(g(s(0)),y, g(x))

g(s(x)) -> s(g(x))

g(0) -> 0

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:

F(g(x), s(0),y) -> F(g(s(0)),y, g(x))

F(g(x), s(0),y) -> F(s(g(0)),y, g(x))

The transformation is resulting in no new DP problems.

Duration:

0:00 minutes