from(

from(

sel(0, cons(

sel(s(

activate(n

activate(

R

↳Dependency Pair Analysis

SEL(s(X), cons(Y,Z)) -> SEL(X, activate(Z))

SEL(s(X), cons(Y,Z)) -> ACTIVATE(Z)

ACTIVATE(n_{from}(X)) -> FROM(X)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

**SEL(s( X), cons(Y, Z)) -> SEL(X, activate(Z))**

from(X) -> cons(X, n_{from}(s(X)))

from(X) -> n_{from}(X)

sel(0, cons(X,Y)) ->X

sel(s(X), cons(Y,Z)) -> sel(X, activate(Z))

activate(n_{from}(X)) -> from(X)

activate(X) ->X

The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y,Z)) -> SEL(X, activate(Z))

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

activate(n_{from}(X)) -> from(X)

activate(X) ->X

from(X) -> cons(X, n_{from}(s(X)))

from(X) -> n_{from}(X)

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(n__from(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(from(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(activate(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(cons(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(SEL(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Dependency Graph

from(X) -> cons(X, n_{from}(s(X)))

from(X) -> n_{from}(X)

sel(0, cons(X,Y)) ->X

sel(s(X), cons(Y,Z)) -> sel(X, activate(Z))

activate(n_{from}(X)) -> from(X)

activate(X) ->X

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes