f(s(0), g(

g(s(

R

↳Dependency Pair Analysis

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳FwdInst

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

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

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

The following dependency pair can be strictly oriented:

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

↳FwdInst

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

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Forward Instantiation Transformation

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

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

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

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳FwdInst

→DP Problem 4

↳Remaining Obligation(s)

The following remains to be proven:

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

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

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

Duration:

0:00 minutes