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

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

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

↳Polynomial Ordering

**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

The following dependency pairs can be strictly oriented:

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

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

A(d(c(x''))) -> B(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(c(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(B(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(d(x)_{1})= 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 4

↳Polynomial Ordering

**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

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 5

↳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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes