from(

from(

after(0,

after(s(

activate(n

activate(

R

↳Dependency Pair Analysis

AFTER(s(N), cons(X,XS)) -> AFTER(N, activate(XS))

AFTER(s(N), cons(X,XS)) -> ACTIVATE(XS)

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

**AFTER(s( N), cons(X, XS)) -> AFTER(N, activate(XS))**

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

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

after(0,XS) ->XS

after(s(N), cons(X,XS)) -> after(N, activate(XS))

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

activate(X) ->X

innermost

The following dependency pair can be strictly oriented:

AFTER(s(N), cons(X,XS)) -> AFTER(N, activate(XS))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

after(0,XS) ->XS

after(s(N), cons(X,XS)) -> after(N, activate(XS))

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

activate(X) ->X

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes