f(0) -> cons(0)

f(s(0)) -> f(p(s(0)))

p(s(0)) -> 0

R

↳Dependency Pair Analysis

F(s(0)) -> F(p(s(0)))

F(s(0)) -> P(s(0))

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

**F(s(0)) -> F(p(s(0)))**

f(0) -> cons(0)

f(s(0)) -> f(p(s(0)))

p(s(0)) -> 0

The following dependency pair can be strictly oriented:

F(s(0)) -> F(p(s(0)))

The following rules can be oriented:

p(s(0)) -> 0

f(0) -> cons(0)

f(s(0)) -> f(p(s(0)))

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

{cons, f}

s > p > 0

resulting in one new DP problem.

Used Argument Filtering System:

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

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

p(x) -> p_{1}

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Dependency Graph

f(0) -> cons(0)

f(s(0)) -> f(p(s(0)))

p(s(0)) -> 0

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes