f(f(

f(f(

g(c(

g(d(

g(c(h(0))) -> g(d(1))

g(c(1)) -> g(d(h(0)))

g(h(

R

↳Dependency Pair Analysis

F(f(x)) -> F(c(f(x)))

F(f(x)) -> F(d(f(x)))

G(c(h(0))) -> G(d(1))

G(c(1)) -> G(d(h(0)))

G(h(x)) -> G(x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

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

f(f(x)) -> f(c(f(x)))

f(f(x)) -> f(d(f(x)))

g(c(x)) ->x

g(d(x)) ->x

g(c(h(0))) -> g(d(1))

g(c(1)) -> g(d(h(0)))

g(h(x)) -> g(x)

The following dependency pair can be strictly oriented:

G(h(x)) -> G(x)

Additionally, the following rules can be oriented:

f(f(x)) -> f(c(f(x)))

f(f(x)) -> f(d(f(x)))

g(c(x)) ->x

g(d(x)) ->x

g(c(h(0))) -> g(d(1))

g(c(1)) -> g(d(h(0)))

g(h(x)) -> g(x)

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(c(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(g(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(G(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(1)= 1 _{ }^{ }_{ }^{ }POL(d(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(h(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(f(x)_{1})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Dependency Graph

f(f(x)) -> f(c(f(x)))

f(f(x)) -> f(d(f(x)))

g(c(x)) ->x

g(d(x)) ->x

g(c(h(0))) -> g(d(1))

g(c(1)) -> g(d(h(0)))

g(h(x)) -> g(x)

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes