f(a, f(f(a,

R

↳Dependency Pair Analysis

F(a, f(f(a,x), a)) -> F(f(a, f(a,x)), a)

F(a, f(f(a,x), a)) -> F(a, f(a,x))

Furthermore,

R

↳DPs

→DP Problem 1

↳Narrowing Transformation

**F(a, f(f(a, x), a)) -> F(a, f(a, x))**

f(a, f(f(a,x), a)) -> f(f(a, f(a,x)), a)

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

As a result of transforming the rule

one new Dependency Pair is created:

F(a, f(f(a,x), a)) -> F(a, f(a,x))

F(a, f(f(a, f(f(a,x''), a)), a)) -> F(a, f(f(a, f(a,x'')), a))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Narrowing Transformation

**F(a, f(f(a, f(f(a, x''), a)), a)) -> F(a, f(f(a, f(a, x'')), a))**

f(a, f(f(a,x), a)) -> f(f(a, f(a,x)), a)

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

As a result of transforming the rule

one new Dependency Pair is created:

F(a, f(f(a, f(f(a,x''), a)), a)) -> F(a, f(f(a, f(a,x'')), a))

F(a, f(f(a, f(f(a, f(f(a,x'), a)), a)), a)) -> F(a, f(f(a, f(f(a, f(a,x')), a)), a))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 3

↳Polynomial Ordering

**F(a, f(f(a, f(f(a, f(f(a, x'), a)), a)), a)) -> F(a, f(f(a, f(f(a, f(a, x')), a)), a))**

f(a, f(f(a,x), a)) -> f(f(a, f(a,x)), a)

The following dependency pair can be strictly oriented:

F(a, f(f(a, f(f(a, f(f(a,x'), a)), a)), a)) -> F(a, f(f(a, f(f(a, f(a,x')), a)), a))

Additionally, the following usable rule w.r.t. to the implicit AFS can be oriented:

f(a, f(f(a,x), a)) -> f(f(a, f(a,x)), a)

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(a)= 0 _{ }^{ }_{ }^{ }POL(f(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(F(x)_{1}, x_{2})= 1 + x _{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 4

↳Dependency Graph

f(a, f(f(a,x), a)) -> f(f(a, f(a,x)), a)

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes