f(f(f(a, b), c),

f(

R

↳Dependency Pair Analysis

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

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

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Modular Removal of Rules

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

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

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

We have the following set of usable rules:

To remove rules and DPs from this DP problem we used the following monotonic and C

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

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

Polynomial interpretation:

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

We have the following set D of usable symbols: {c, b, a, F, f}

The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

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

No Rules can be deleted.

The result of this processor delivers one new DP problem.

R

↳DPs

→DP Problem 1

↳MRR

→DP Problem 2

↳Negative Polynomial Order

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

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

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

The following Dependency Pair can be strictly oriented using the given order.

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

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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

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

Used ordering:

Polynomial Order with Interpretation:

**POL( **F(x_{1}, x_{2})** )** = x_{1}

**POL( **f(x_{1}, x_{2})** )** = x_{1}

**POL( **a** )** = 1

**POL( **b** )** = 0

This results in one new DP problem.

R

↳DPs

→DP Problem 1

↳MRR

→DP Problem 2

↳Neg POLO

...

→DP Problem 3

↳Remaining Obligation(s)

The following remains to be proven:

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

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

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

Duration:

0:41 minutes