rev(

r1(empty,

r1(cons(

R

↳Dependency Pair Analysis

REV(ls) -> R1(ls, empty)

R1(cons(x,k),a) -> R1(k, cons(x,a))

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

**R1(cons( x, k), a) -> R1(k, cons(x, a))**

rev(ls) -> r1(ls, empty)

r1(empty,a) ->a

r1(cons(x,k),a) -> r1(k, cons(x,a))

The following dependency pair can be strictly oriented:

R1(cons(x,k),a) -> R1(k, cons(x,a))

There are no usable rules using the Ce-refinement that need to be oriented.

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

R1 > cons

resulting in one new DP problem.

Used Argument Filtering System:

R1(x,_{1}x) -> R1(_{2}x,_{1}x)_{2}

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Dependency Graph

rev(ls) -> r1(ls, empty)

r1(empty,a) ->a

r1(cons(x,k),a) -> r1(k, cons(x,a))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes