f(s(

R

↳Dependency Pair Analysis

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Instantiation Transformation

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

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

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

As a result of transforming the rule

one new Dependency Pair is created:

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

F(s(x0), s(x'''), s(x''')) -> F(s(x'''),x0, s(x0))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳Forward Instantiation Transformation

**F(s( x0), s(x'''), s(x''')) -> F(s(x'''), x0, s(x0))**

f(s(x),y,y) -> f(y,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(x0), s(x'''), s(x''')) -> F(s(x'''),x0, s(x0))

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳FwdInst

...

→DP Problem 3

↳Polynomial Ordering

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

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

The following dependency pair can be strictly oriented:

F(s(s(x'''''')), s(x'''0), s(x'''0)) -> F(s(x'''0), s(x''''''), 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}, x_{2}, x_{3})= 1 + x _{1}+ x_{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳FwdInst

...

→DP Problem 4

↳Dependency Graph

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes