.(1,

.(

.(i(

.(

.(i(

.(

.(.(

i(1) -> 1

i(i(

i(.(

R

↳Dependency Pair Analysis

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

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

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

.(1,x) ->x

.(x, 1) ->x

.(i(x),x) -> 1

.(x, i(x)) -> 1

.(i(y), .(y,z)) ->z

.(y, .(i(y),z)) ->z

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

i(1) -> 1

i(i(x)) ->x

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

The following dependency pairs can be strictly oriented:

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

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Polo

.(1,x) ->x

.(x, 1) ->x

.(i(x),x) -> 1

.(x, i(x)) -> 1

.(i(y), .(y,z)) ->z

.(y, .(i(y),z)) ->z

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

i(1) -> 1

i(i(x)) ->x

i(.(x,y)) -> .(i(y), i(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(x)**

.(1,x) ->x

.(x, 1) ->x

.(i(x),x) -> 1

.(x, i(x)) -> 1

.(i(y), .(y,z)) ->z

.(y, .(i(y),z)) ->z

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

i(1) -> 1

i(i(x)) ->x

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

The following dependency pairs can be strictly oriented:

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

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

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(I(x)_{1})= x _{1}_{ }^{ }_{ }^{ }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

.(1,x) ->x

.(x, 1) ->x

.(i(x),x) -> 1

.(x, i(x)) -> 1

.(i(y), .(y,z)) ->z

.(y, .(i(y),z)) ->z

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

i(1) -> 1

i(i(x)) ->x

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes