+(

+(

double(

f(0, s(0),

g(

g(

R

↳Dependency Pair Analysis

+'(X, s(Y)) -> +'(X,Y)

DOUBLE(X) -> +'(X,X)

F(0, s(0),X) -> F(X, double(X),X)

F(0, s(0),X) -> DOUBLE(X)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Rw

**+'( X, s(Y)) -> +'(X, Y)**

+(X, 0) ->X

+(X, s(Y)) -> s(+(X,Y))

double(X) -> +(X,X)

f(0, s(0),X) -> f(X, double(X),X)

g(X,Y) ->X

g(X,Y) ->Y

innermost

The following dependency pair can be strictly oriented:

+'(X, s(Y)) -> +'(X,Y)

There are no usable rules for innermost 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(+'(x)_{1}, x_{2})= x _{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Rw

+(X, 0) ->X

+(X, s(Y)) -> s(+(X,Y))

double(X) -> +(X,X)

f(0, s(0),X) -> f(X, double(X),X)

g(X,Y) ->X

g(X,Y) ->Y

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Rewriting Transformation

**F(0, s(0), X) -> F(X, double(X), X)**

+(X, 0) ->X

+(X, s(Y)) -> s(+(X,Y))

double(X) -> +(X,X)

f(0, s(0),X) -> f(X, double(X),X)

g(X,Y) ->X

g(X,Y) ->Y

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(0, s(0),X) -> F(X, double(X),X)

F(0, s(0),X) -> F(X, +(X,X),X)

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Rw

→DP Problem 4

↳Narrowing Transformation

**F(0, s(0), X) -> F(X, +(X, X), X)**

+(X, 0) ->X

+(X, s(Y)) -> s(+(X,Y))

double(X) -> +(X,X)

f(0, s(0),X) -> f(X, double(X),X)

g(X,Y) ->X

g(X,Y) ->Y

innermost

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

As a result of transforming the rule

two new Dependency Pairs are created:

F(0, s(0),X) -> F(X, +(X,X),X)

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

F(0, s(0), s(Y')) -> F(s(Y'), s(+(s(Y'),Y')), s(Y'))

The transformation is resulting in no new DP problems.

Duration:

0:00 minutes