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

↳Polynomial 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))

Additionally, the following rules can be oriented:

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

r1(empty,a) ->a

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(rev(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(cons(x)_{1}, x_{2})= 1 + x _{2}_{ }^{ }_{ }^{ }POL(r1(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(R1(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(empty)= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→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