p(s(

fac(0) -> s(0)

fac(s(

R

↳Dependency Pair Analysis

FAC(s(x)) -> FAC(p(s(x)))

FAC(s(x)) -> P(s(x))

Furthermore,

R

↳DPs

→DP Problem 1

↳Rewriting Transformation

**FAC(s( x)) -> FAC(p(s(x)))**

p(s(x)) ->x

fac(0) -> s(0)

fac(s(x)) -> times(s(x), fac(p(s(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:

FAC(s(x)) -> FAC(p(s(x)))

FAC(s(x)) -> FAC(x)

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Rw

→DP Problem 2

↳Polynomial Ordering

**FAC(s( x)) -> FAC(x)**

p(s(x)) ->x

fac(0) -> s(0)

fac(s(x)) -> times(s(x), fac(p(s(x))))

innermost

The following dependency pair can be strictly oriented:

FAC(s(x)) -> FAC(x)

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(FAC(x)_{1})= x _{1}_{ }^{ }_{ }^{ }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

p(s(x)) ->x

fac(0) -> s(0)

fac(s(x)) -> times(s(x), fac(p(s(x))))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes