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

↳Forward Instantiation Transformation

→DP Problem 2

↳Nar

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

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

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

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

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(x)) -> F(x)

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳Forward Instantiation Transformation

→DP Problem 2

↳Nar

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

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

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

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

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳FwdInst

...

→DP Problem 4

↳Polynomial Ordering

→DP Problem 2

↳Nar

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

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

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

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

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳FwdInst

...

→DP Problem 5

↳Dependency Graph

→DP Problem 2

↳Nar

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

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

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Narrowing 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)))

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳Narrowing Transformation

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

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

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

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

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

As a result of transforming the rule

no new Dependency Pairs are created.

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

The transformation is resulting in no new DP problems.

Duration:

0:00 minutes