a(d(

a(c(

b(c(

b(d(

R

↳Dependency Pair Analysis

A(d(x)) -> B(a(x))

A(d(x)) -> A(x)

B(c(x)) -> A(b(x))

B(c(x)) -> B(x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Narrowing Transformation

**B(c( x)) -> B(x)**

a(d(x)) -> d(c(b(a(x))))

a(c(x)) ->x

b(c(x)) -> c(d(a(b(x))))

b(d(x)) ->x

innermost

On this DP problem, a Narrowing SCC transformation can be performed.

As a result of transforming the rule

two new Dependency Pairs are created:

A(d(x)) -> B(a(x))

A(d(d(x''))) -> B(d(c(b(a(x'')))))

A(d(c(x''))) -> B(x'')

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Narrowing Transformation

**A(d(c( x''))) -> B(x'')**

a(d(x)) -> d(c(b(a(x))))

a(c(x)) ->x

b(c(x)) -> c(d(a(b(x))))

b(d(x)) ->x

innermost

On this DP problem, a Narrowing SCC transformation can be performed.

As a result of transforming the rule

two new Dependency Pairs are created:

B(c(x)) -> A(b(x))

B(c(c(x''))) -> A(c(d(a(b(x'')))))

B(c(d(x''))) -> A(x'')

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 3

↳Forward Instantiation Transformation

**A(d( x)) -> A(x)**

a(d(x)) -> d(c(b(a(x))))

a(c(x)) ->x

b(c(x)) -> c(d(a(b(x))))

b(d(x)) ->x

innermost

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

As a result of transforming the rule

two new Dependency Pairs are created:

A(d(x)) -> A(x)

A(d(d(x''))) -> A(d(x''))

A(d(d(c(x'''')))) -> A(d(c(x'''')))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 4

↳Forward Instantiation Transformation

**A(d(d(c( x'''')))) -> A(d(c(x'''')))**

a(d(x)) -> d(c(b(a(x))))

a(c(x)) ->x

b(c(x)) -> c(d(a(b(x))))

b(d(x)) ->x

innermost

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

As a result of transforming the rule

two new Dependency Pairs are created:

B(c(x)) -> B(x)

B(c(c(x''))) -> B(c(x''))

B(c(c(d(x'''')))) -> B(c(d(x'''')))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 5

↳Forward Instantiation Transformation

**B(c(c(d( x'''')))) -> B(c(d(x'''')))**

a(d(x)) -> d(c(b(a(x))))

a(c(x)) ->x

b(c(x)) -> c(d(a(b(x))))

b(d(x)) ->x

innermost

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

As a result of transforming the rule

three new Dependency Pairs are created:

A(d(c(x''))) -> B(x'')

A(d(c(c(d(x''''))))) -> B(c(d(x'''')))

A(d(c(c(c(x''''))))) -> B(c(c(x'''')))

A(d(c(c(c(d(x'''''')))))) -> B(c(c(d(x''''''))))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 6

↳Forward Instantiation Transformation

**A(d(c(c(c(d( x'''''')))))) -> B(c(c(d(x''''''))))**

a(d(x)) -> d(c(b(a(x))))

a(c(x)) ->x

b(c(x)) -> c(d(a(b(x))))

b(d(x)) ->x

innermost

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

As a result of transforming the rule

five new Dependency Pairs are created:

B(c(d(x''))) -> A(x'')

B(c(d(d(d(x''''))))) -> A(d(d(x'''')))

B(c(d(d(d(c(x'''''')))))) -> A(d(d(c(x''''''))))

B(c(d(d(c(c(d(x''''''))))))) -> A(d(c(c(d(x'''''')))))

B(c(d(d(c(c(c(x''''''))))))) -> A(d(c(c(c(x'''''')))))

B(c(d(d(c(c(c(d(x'''''''')))))))) -> A(d(c(c(c(d(x''''''''))))))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 7

↳Polynomial Ordering

**B(c(d(d(c(c(c(d( x'''''''')))))))) -> A(d(c(c(c(d(x''''''''))))))**

a(d(x)) -> d(c(b(a(x))))

a(c(x)) ->x

b(c(x)) -> c(d(a(b(x))))

b(d(x)) ->x

innermost

The following dependency pairs can be strictly oriented:

B(c(d(d(c(c(c(d(x'''''''')))))))) -> A(d(c(c(c(d(x''''''''))))))

A(d(c(c(c(x''''))))) -> B(c(c(x'''')))

B(c(d(d(c(c(c(x''''''))))))) -> A(d(c(c(c(x'''''')))))

B(c(d(d(c(c(d(x''''''))))))) -> A(d(c(c(d(x'''''')))))

B(c(d(d(d(c(x'''''')))))) -> A(d(d(c(x''''''))))

A(d(c(c(d(x''''))))) -> B(c(d(x'''')))

B(c(d(d(d(x''''))))) -> A(d(d(x'''')))

B(c(c(d(x'''')))) -> B(c(d(x'''')))

B(c(c(x''))) -> B(c(x''))

A(d(c(c(c(d(x'''''')))))) -> B(c(c(d(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(c(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(B(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(d(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(A(x)_{1})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 8

↳Dependency Graph

**A(d(d(c( x'''')))) -> A(d(c(x'''')))**

a(d(x)) -> d(c(b(a(x))))

a(c(x)) ->x

b(c(x)) -> c(d(a(b(x))))

b(d(x)) ->x

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 9

↳Polynomial Ordering

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

a(d(x)) -> d(c(b(a(x))))

a(c(x)) ->x

b(c(x)) -> c(d(a(b(x))))

b(d(x)) ->x

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 10

↳Dependency Graph

a(d(x)) -> d(c(b(a(x))))

a(c(x)) ->x

b(c(x)) -> c(d(a(b(x))))

b(d(x)) ->x

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:01 minutes