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))

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 rules 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.

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Remaining Obligation(s)

The following remains to be proven:

**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))

Duration:

0:00 minutes