f(a, f(

R

↳Dependency Pair Analysis

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Narrowing Transformation

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

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

innermost

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Narrowing Transformation

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

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

innermost

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 3

↳Narrowing Transformation

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

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

innermost

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 4

↳Narrowing Transformation

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

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

innermost

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

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

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

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

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

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

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

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

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

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

innermost

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

As a result of transforming the rule

no new Dependency Pairs are created.

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

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

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

innermost

The following dependency pairs can be strictly oriented:

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

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

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

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↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 8

↳Dependency Graph

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

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes