:(

:(

:(:(

:(e,

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:(

:(i(

:(i(

i(:(

i(i(

i(e) -> e

R

↳Dependency Pair Analysis

:'(:(x,y),z) -> :'(x, :(z, i(y)))

:'(:(x,y),z) -> :'(z, i(y))

:'(:(x,y),z) -> I(y)

:'(e,x) -> I(x)

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

:'(x, :(y, :(i(x),z))) -> :'(i(z),y)

:'(x, :(y, :(i(x),z))) -> I(z)

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

:'(i(x), :(y, :(x,z))) -> :'(i(z),y)

:'(i(x), :(y, :(x,z))) -> I(z)

I(:(x,y)) -> :'(y,x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Modular Removal of Rules

**:'(i( x), :(y, :(x, z))) -> I(z)**

:(x,x) -> e

:(x, e) ->x

:(:(x,y),z) -> :(x, :(z, i(y)))

:(e,x) -> i(x)

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

:(x, :(y, :(i(x),z))) -> :(i(z),y)

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

:(i(x), :(y, :(x,z))) -> :(i(z),y)

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

i(i(x)) ->x

i(e) -> e

innermost

We have the following set of usable rules:

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

:(i(x), :(y, :(x,z))) -> :(i(z),y)

:(x,x) -> e

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

i(e) -> e

:(e,x) -> i(x)

:(x, :(y, :(i(x),z))) -> :(i(z),y)

i(i(x)) ->x

:(:(x,y),z) -> :(x, :(z, i(y)))

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

:(x, e) ->x

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

Polynomial interpretation:

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

We have the following set D of usable symbols: {:, I, i, e, :'}

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

:'(i(x), :(y, :(x,z))) -> I(z)

:'(i(x), :(y, :(x,z))) -> :'(i(z),y)

:'(x, :(y, :(i(x),z))) -> I(z)

I(:(x,y)) -> :'(y,x)

:'(:(x,y),z) -> I(y)

:'(:(x,y),z) -> :'(z, i(y))

:'(x, :(y, :(i(x),z))) -> :'(i(z),y)

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

:(i(x), :(y, :(x,z))) -> :(i(z),y)

:(x,x) -> e

:(e,x) -> i(x)

:(x, :(y, :(i(x),z))) -> :(i(z),y)

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

:(x, e) ->x

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

The result of this processor delivers one new DP problem.

R

↳DPs

→DP Problem 1

↳MRR

→DP Problem 2

↳Modular Removal of Rules

**:'(e, x) -> I(x)**

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

i(e) -> e

i(i(x)) ->x

:(:(x,y),z) -> :(x, :(z, i(y)))

innermost

We have the following set of usable rules:

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

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

i(e) -> e

i(i(x)) ->x

:(:(x,y),z) -> :(x, :(z, i(y)))

Polynomial interpretation:

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

We have the following set D of usable symbols: {:, I, i, e, :'}

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

:'(e,x) -> I(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

↳MRR

...

→DP Problem 3

↳Dependency Graph

**:'(:( x, y), z) -> :'(x, :(z, i(y)))**

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

i(e) -> e

i(i(x)) ->x

:(:(x,y),z) -> :(x, :(z, i(y)))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes