f(g(a)) -> f(s(g(b)))

f(f(

g(

R

↳Dependency Pair Analysis

F(g(a)) -> F(s(g(b)))

F(g(a)) -> G(b)

G(x) -> F(g(x))

G(x) -> G(x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

**G( x) -> G(x)**

f(g(a)) -> f(s(g(b)))

f(f(x)) -> b

g(x) -> f(g(x))

The following dependency pair can be strictly oriented:

G(x) -> F(g(x))

Additionally, the following usable rules using the Ce-refinement can be oriented:

g(x) -> f(g(x))

f(g(a)) -> f(s(g(b)))

f(f(x)) -> b

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(g(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(G(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(b)= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(a)= 1 _{ }^{ }_{ }^{ }POL(f(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(F(x)_{1})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Dependency Graph

**G( x) -> G(x)**

f(g(a)) -> f(s(g(b)))

f(f(x)) -> b

g(x) -> f(g(x))

Using the Dependency Graph the DP problem was split into 1 DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳DGraph

...

→DP Problem 3

↳Remaining Obligation(s)

The following remains to be proven:

**G( x) -> G(x)**

f(g(a)) -> f(s(g(b)))

f(f(x)) -> b

g(x) -> f(g(x))

Duration:

0:00 minutes