f(0,

f(s(

R

↳Dependency Pair Analysis

F(s(x),y) -> F(f(x,y),y)

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

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

f(0,y) -> 0

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

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.

F(s(x),y) -> F(f(x,y),y)

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Forward Instantiation Transformation

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

f(0,y) -> 0

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

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),y) -> F(x,y)

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 3

↳Forward Instantiation Transformation

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

f(0,y) -> 0

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

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 4

↳Polynomial Ordering

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

f(0,y) -> 0

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

The following dependency pair can be strictly oriented:

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

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}, x_{2})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 5

↳Dependency Graph

f(0,y) -> 0

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes