f(f(

f(s(

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

R

↳Dependency Pair Analysis

F(s(x)) -> F(x)

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳FwdInst

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

f(f(x)) -> f(x)

f(s(x)) -> f(x)

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

innermost

The following dependency pair can be strictly oriented:

F(s(x)) -> F(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(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(F(x)_{1})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳FwdInst

f(f(x)) -> f(x)

f(s(x)) -> f(x)

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

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Forward Instantiation Transformation

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

f(f(x)) -> f(x)

f(s(x)) -> f(x)

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

innermost

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

As a result of transforming the rule

no new Dependency Pairs are created.

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

The transformation is resulting in no new DP problems.

Duration:

0:00 minutes