f(

f(s(

g(0,

R

↳Dependency Pair Analysis

F(s(x), s(y)) -> F(x,y)

G(0,x) -> G(f(x,x),x)

G(0,x) -> F(x,x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

**F(s( x), s(y)) -> F(x, y)**

f(x, 0) -> s(0)

f(s(x), s(y)) -> s(f(x,y))

g(0,x) -> g(f(x,x),x)

The following dependency pair can be strictly oriented:

F(s(x), s(y)) -> F(x,y)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Polo

f(x, 0) -> s(0)

f(s(x), s(y)) -> s(f(x,y))

g(0,x) -> g(f(x,x),x)

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

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

f(x, 0) -> s(0)

f(s(x), s(y)) -> s(f(x,y))

g(0,x) -> g(f(x,x),x)

The following dependency pair can be strictly oriented:

G(0,x) -> G(f(x,x),x)

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

f(x, 0) -> s(0)

f(s(x), s(y)) -> s(f(x,y))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 1 _{ }^{ }_{ }^{ }POL(G(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(f(x)_{1}, x_{2})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

f(x, 0) -> s(0)

f(s(x), s(y)) -> s(f(x,y))

g(0,x) -> g(f(x,x),x)

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes