f(

f(

g(empty,

g(cons(

R

↳Dependency Pair Analysis

F(a, empty) -> G(a, empty)

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

G(cons(x,k),d) -> G(k, cons(x,d))

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

**G(cons( x, k), d) -> G(k, cons(x, d))**

f(a, empty) -> g(a, empty)

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

g(empty,d) ->d

g(cons(x,k),d) -> g(k, cons(x,d))

The following dependency pair can be strictly oriented:

G(cons(x,k),d) -> G(k, cons(x,d))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(G(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(cons(x)_{1}, x_{2})= 1 + x _{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Polo

f(a, empty) -> g(a, empty)

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

g(empty,d) ->d

g(cons(x,k),d) -> g(k, cons(x,d))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

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

f(a, empty) -> g(a, empty)

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

g(empty,d) ->d

g(cons(x,k),d) -> g(k, cons(x,d))

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(cons(x)_{1}, x_{2})= 1 + x _{2}_{ }^{ }_{ }^{ }POL(F(x)_{1}, x_{2})= x _{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

f(a, empty) -> g(a, empty)

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

g(empty,d) ->d

g(cons(x,k),d) -> g(k, cons(x,d))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes