f(s(

f(

f(

R

↳Dependency Pair Analysis

F(s(X),X) -> F(X, a(X))

F(X, c(X)) -> F(s(X),X)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

**F(s( X), X) -> F(X, a(X))**

f(s(X),X) -> f(X, a(X))

f(X, c(X)) -> f(s(X),X)

f(X,X) -> c(X)

The following dependency pair can be strictly oriented:

F(s(X),X) -> F(X, a(X))

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(a(x)_{1})= 0 _{ }^{ }_{ }^{ }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(s(X),X) -> f(X, a(X))

f(X, c(X)) -> f(s(X),X)

f(X,X) -> c(X)

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

**F( X, c(X)) -> F(s(X), X)**

f(s(X),X) -> f(X, a(X))

f(X, c(X)) -> f(s(X),X)

f(X,X) -> c(X)

The following dependency pair can be strictly oriented:

F(X, c(X)) -> F(s(X),X)

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

f(s(X),X) -> f(X, a(X))

f(X, c(X)) -> f(s(X),X)

f(X,X) -> c(X)

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes