f(a, f(a,

R

↳Dependency Pair Analysis

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

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

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

Furthermore,

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↳DPs

→DP Problem 1

↳Polynomial Ordering

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

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

The following dependency pair can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Narrowing Transformation

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

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

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

The transformation is resulting in one new DP problem:

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Nar

...

→DP Problem 3

↳Remaining Obligation(s)

The following remains to be proven:

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

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

Duration:

0:00 minutes