f(f(a, a),

f(

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↳Dependency Pair Analysis

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

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

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

F(x, f(y,z)) -> F(f(x,y),z)

F(x, f(y,z)) -> F(x,y)

Furthermore,

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↳DPs

→DP Problem 1

↳Polynomial Ordering

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

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

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

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules using the Ce-refinement can be oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

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

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

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

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules using the Ce-refinement can be oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

...

→DP Problem 3

↳Remaining Obligation(s)

The following remains to be proven:

**F( x, f(y, z)) -> F(x, y)**

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

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

Duration:

0:00 minutes