f(0, 1,

f(

R

↳Dependency Pair Analysis

F(0, 1,x) -> F(s(x),x,x)

F(x,y, s(z)) -> F(0, 1,z)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

**F( x, y, s(z)) -> F(0, 1, z)**

f(0, 1,x) -> f(s(x),x,x)

f(x,y, s(z)) -> s(f(0, 1,z))

The following dependency pair can be strictly oriented:

F(x,y, s(z)) -> F(0, 1,z)

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(0)= 0 _{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(F(x)_{1}, x_{2}, x_{3})= x _{3}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Dependency Graph

**F(0, 1, x) -> F(s(x), x, x)**

f(0, 1,x) -> f(s(x),x,x)

f(x,y, s(z)) -> s(f(0, 1,z))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes