a(b(

R

↳Dependency Pair Analysis

A(b(x)) -> A(x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

**A(b( x)) -> A(x)**

a(b(x)) -> b(b(a(x)))

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:

A(b(x)) -> A(x)

A(b(b(x''))) -> A(b(x''))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Forward Instantiation Transformation

**A(b(b( x''))) -> A(b(x''))**

a(b(x)) -> b(b(a(x)))

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:

A(b(b(x''))) -> A(b(x''))

A(b(b(b(x'''')))) -> A(b(b(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

**A(b(b(b( x'''')))) -> A(b(b(x'''')))**

a(b(x)) -> b(b(a(x)))

innermost

The following dependency pair can be strictly oriented:

A(b(b(b(x'''')))) -> A(b(b(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(b(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(A(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

a(b(x)) -> b(b(a(x)))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes