f(0) -> 1

f(s(

f(s(

g(

R

↳Dependency Pair Analysis

F(s(x)) -> G(f(x))

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

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

f(0) -> 1

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

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

g(x) -> +(x, s(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(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 2

↳Forward Instantiation Transformation

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

f(0) -> 1

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

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

g(x) -> +(x, s(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(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 2

↳FwdInst

...

→DP Problem 3

↳Polynomial Ordering

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

f(0) -> 1

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

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

g(x) -> +(x, s(x))

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 2

↳FwdInst

...

→DP Problem 4

↳Dependency Graph

f(0) -> 1

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

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

g(x) -> +(x, s(x))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes