f(0) -> s(0)

f(s(0)) -> s(s(0))

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

f(+(

f(+(

R

↳Dependency Pair Analysis

F(s(0)) -> F(0)

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

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

f(0) -> s(0)

f(s(0)) -> s(s(0))

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

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

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

The following dependency pairs can be strictly oriented:

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

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

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

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Dependency Graph

f(0) -> s(0)

f(s(0)) -> s(s(0))

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

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

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes