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

↳Argument Filtering and Ordering

→DP Problem 2

↳Remaining

**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)))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

first(0,X) -> nil

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

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

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

0 > nil

resulting in one new DP problem.

Used Argument Filtering System:

FIRST(x,_{1}x) -> FIRST(_{2}x,_{1}x)_{2}

s(x) -> s(_{1}x)_{1}

cons(x,_{1}x) ->_{2}x_{2}

first(x,_{1}x) -> first(_{2}x,_{1}x)_{2}

from(x) -> from_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Remaining

first(0,X) -> nil

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

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Remaining Obligation(s)

The following remains to be proven:

**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)))

Duration:

0:00 minutes