f(

f(

g(0) -> s(0)

g(s(

sel(0, cons(

sel(s(

activate(n

activate(

R

↳Dependency Pair Analysis

F(X) -> G(X)

G(s(X)) -> G(X)

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

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

ACTIVATE(n_{f}(X)) -> F(X)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Remaining

**G(s( X)) -> G(X)**

f(X) -> cons(X, n_{f}(g(X)))

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

g(0) -> s(0)

g(s(X)) -> s(s(g(X)))

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

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

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

activate(X) ->X

innermost

The following dependency pair can be strictly oriented:

G(s(X)) -> G(X)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(G(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Remaining

f(X) -> cons(X, n_{f}(g(X)))

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

g(0) -> s(0)

g(s(X)) -> s(s(g(X)))

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

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

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

activate(X) ->X

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Remaining Obligation(s)

The following remains to be proven:

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

f(X) -> cons(X, n_{f}(g(X)))

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

g(0) -> s(0)

g(s(X)) -> s(s(g(X)))

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

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

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

activate(X) ->X

innermost

Duration:

0:00 minutes