a(a(f(

f(a(

f(b(

R

↳Dependency Pair Analysis

A(a(f(x,y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))

A(a(f(x,y))) -> A(b(a(b(a(x)))))

A(a(f(x,y))) -> A(b(a(x)))

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

A(a(f(x,y))) -> A(b(a(b(a(y)))))

A(a(f(x,y))) -> A(b(a(y)))

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

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

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

F(b(x), b(y)) -> F(x,y)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

**F(b( x), b(y)) -> F(x, y)**

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

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

f(b(x), b(y)) -> b(f(x,y))

innermost

The following dependency pairs can be strictly oriented:

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

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

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

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

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

f(b(x), b(y)) -> b(f(x,y))

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Instantiation Transformation

**F(b( x), b(y)) -> F(x, y)**

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

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

f(b(x), b(y)) -> b(f(x,y))

innermost

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

As a result of transforming the rule

no new Dependency Pairs are created.

A(a(f(x,y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Inst

...

→DP Problem 3

↳Polynomial Ordering

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

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

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

f(b(x), b(y)) -> b(f(x,y))

innermost

The following dependency pair can be strictly oriented:

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

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Inst

...

→DP Problem 4

↳Polynomial Ordering

**F(b( x), b(y)) -> F(x, y)**

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

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

f(b(x), b(y)) -> b(f(x,y))

innermost

The following dependency pair can be strictly oriented:

F(b(x), b(y)) -> F(x,y)

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(b(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(F(x)_{1}, x_{2})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Inst

...

→DP Problem 5

↳Dependency Graph

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

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

f(b(x), b(y)) -> b(f(x,y))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes