i(0) -> 0

i(i(

i(+(

+(0,

+(

+(i(

+(

+(

+(+(

+(+(

R

↳Dependency Pair Analysis

I(+(x,y)) -> +'(i(x), i(y))

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

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

i(0) -> 0

i(i(x)) ->x

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

+(0,y) ->y

+(x, 0) ->x

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

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

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

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

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

The following dependency pairs can be strictly oriented:

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

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

Additionally, the following rules can be oriented:

i(0) -> 0

i(i(x)) ->x

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

+(0,y) ->y

+(x, 0) ->x

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

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

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(i(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }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 3

↳Dependency Graph

→DP Problem 2

↳Polo

i(0) -> 0

i(i(x)) ->x

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

+(0,y) ->y

+(x, 0) ->x

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

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

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

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

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

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

i(0) -> 0

i(i(x)) ->x

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

+(0,y) ->y

+(x, 0) ->x

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

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

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

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

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

The following dependency pairs can be strictly oriented:

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

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

Additionally, the following rules can be oriented:

i(0) -> 0

i(i(x)) ->x

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

+(0,y) ->y

+(x, 0) ->x

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

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

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

i(0) -> 0

i(i(x)) ->x

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

+(0,y) ->y

+(x, 0) ->x

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

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

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

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

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes