+(

+(+(

*(

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↳Dependency Pair Analysis

+'(x, +(y,z)) -> +'(+(x,y),z)

+'(x, +(y,z)) -> +'(x,y)

+'(+(x, *(y,z)), *(y,u)) -> +'(x, *(y, +(z,u)))

+'(+(x, *(y,z)), *(y,u)) -> *'(y, +(z,u))

+'(+(x, *(y,z)), *(y,u)) -> +'(z,u)

*'(x, +(y,z)) -> +'(*(x,y), *(x,z))

*'(x, +(y,z)) -> *'(x,y)

*'(x, +(y,z)) -> *'(x,z)

Furthermore,

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↳DPs

→DP Problem 1

↳Polynomial Ordering

***'( x, +(y, z)) -> *'(x, z)**

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

+(+(x, *(y,z)), *(y,u)) -> +(x, *(y, +(z,u)))

*(x, +(y,z)) -> +(*(x,y), *(x,z))

The following dependency pairs can be strictly oriented:

*'(x, +(y,z)) -> *'(x,z)

*'(x, +(y,z)) -> *'(x,y)

+'(+(x, *(y,z)), *(y,u)) -> +'(z,u)

*'(x, +(y,z)) -> +'(*(x,y), *(x,z))

+'(x, +(y,z)) -> +'(x,y)

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

*(x, +(y,z)) -> +(*(x,y), *(x,z))

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

+(+(x, *(y,z)), *(y,u)) -> +(x, *(y, +(z,u)))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(*'(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(*(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(+'(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }

resulting in one new DP problem.

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Dependency Graph

**+'(+( x, *(y, z)), *(y, u)) -> *'(y, +(z, u))**

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

+(+(x, *(y,z)), *(y,u)) -> +(x, *(y, +(z,u)))

*(x, +(y,z)) -> +(*(x,y), *(x,z))

Using the Dependency Graph the DP problem was split into 1 DP problems.

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳DGraph

...

→DP Problem 3

↳Remaining Obligation(s)

The following remains to be proven:

**+'( x, +(y, z)) -> +'(+(x, y), z)**

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

+(+(x, *(y,z)), *(y,u)) -> +(x, *(y, +(z,u)))

*(x, +(y,z)) -> +(*(x,y), *(x,z))

Duration:

0:00 minutes