f(+(

+(

R

↳Dependency Pair Analysis

F(+(x, 0)) -> F(x)

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

**F(+( x, 0)) -> F(x)**

f(+(x, 0)) -> f(x)

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

The following dependency pair can be strictly oriented:

F(+(x, 0)) -> F(x)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Polo

f(+(x, 0)) -> f(x)

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

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

f(+(x, 0)) -> f(x)

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

The following dependency pairs can be strictly oriented:

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

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

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

f(+(x, 0)) -> f(x)

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes