first(0,

first(s(

from(

R

↳Dependency Pair Analysis

FIRST(s(X), cons(Y,Z)) -> FIRST(X,Z)

FROM(X) -> FROM(s(X))

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Inst

**FIRST(s( X), cons(Y, Z)) -> FIRST(X, Z)**

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

from(X) -> cons(X, from(s(X)))

innermost

The following dependency pair can be strictly oriented:

FIRST(s(X), cons(Y,Z)) -> FIRST(X,Z)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(cons(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(FIRST(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 3

↳Dependency Graph

→DP Problem 2

↳Inst

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

from(X) -> cons(X, from(s(X)))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Instantiation Transformation

**FROM( X) -> FROM(s(X))**

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

from(X) -> cons(X, from(s(X)))

innermost

On this DP problem, an Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

FROM(X) -> FROM(s(X))

FROM(s(X'')) -> FROM(s(s(X'')))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Inst

→DP Problem 4

↳Remaining Obligation(s)

The following remains to be proven:

**FROM(s( X'')) -> FROM(s(s(X'')))**

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

from(X) -> cons(X, from(s(X)))

innermost

Duration:

0:00 minutes