f(1) -> f(g(1))

f(f(

g(0) -> g(f(0))

g(g(

R

↳Dependency Pair Analysis

F(1) -> F(g(1))

F(1) -> G(1)

G(0) -> G(f(0))

G(0) -> F(0)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

**F(1) -> F(g(1))**

f(1) -> f(g(1))

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

g(0) -> g(f(0))

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

The following dependency pair can be strictly oriented:

F(1) -> F(g(1))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

g(0) -> g(f(0))

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

f(1) -> f(g(1))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(g(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(1)= 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 3

↳Dependency Graph

→DP Problem 2

↳Polo

f(1) -> f(g(1))

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

g(0) -> g(f(0))

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

**G(0) -> G(f(0))**

f(1) -> f(g(1))

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

g(0) -> g(f(0))

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

The following dependency pair can be strictly oriented:

G(0) -> G(f(0))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

g(0) -> g(f(0))

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

f(1) -> f(g(1))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 1 _{ }^{ }_{ }^{ }POL(g(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(G(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }_{ }^{ }POL(f(x)_{1})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

f(1) -> f(g(1))

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

g(0) -> g(f(0))

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes