f(g(

g(s(

g(0) -> 0

R

↳Dependency Pair Analysis

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

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

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

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

g(0) -> 0

innermost

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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Forward Instantiation Transformation

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

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

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

g(0) -> 0

innermost

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:

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

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

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

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

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

g(0) -> 0

innermost

The following dependency pair can be strictly oriented:

G(s(s(s(x'''')))) -> G(s(s(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(G(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(s(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(g(x), s(0),y) -> f(y,y, g(x))

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

g(0) -> 0

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes