f(s(

f(0) -> 0

p(s(

R

↳Dependency Pair Analysis

F(s(x)) -> F(p(s(x)))

F(s(x)) -> P(s(x))

Furthermore,

R

↳DPs

→DP Problem 1

↳Rewriting Transformation

**F(s( x)) -> F(p(s(x)))**

f(s(x)) -> s(s(f(p(s(x)))))

f(0) -> 0

p(s(x)) ->x

innermost

On this DP problem, a Rewriting SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

F(s(x)) -> F(p(s(x)))

F(s(x)) -> F(x)

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Rw

→DP Problem 2

↳Forward Instantiation Transformation

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

f(s(x)) -> s(s(f(p(s(x)))))

f(0) -> 0

p(s(x)) ->x

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

F(s(x)) -> F(x)

F(s(s(x''))) -> F(s(x''))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Rw

→DP Problem 2

↳FwdInst

...

→DP Problem 3

↳Polynomial Ordering

**F(s(s( x''))) -> F(s(x''))**

f(s(x)) -> s(s(f(p(s(x)))))

f(0) -> 0

p(s(x)) ->x

innermost

The following dependency pair can be strictly oriented:

F(s(s(x''))) -> F(s(x''))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Rw

→DP Problem 2

↳FwdInst

...

→DP Problem 4

↳Dependency Graph

f(s(x)) -> s(s(f(p(s(x)))))

f(0) -> 0

p(s(x)) ->x

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes