minus(

minus(s(

p(s(

div(0, s(

div(s(

R

↳Dependency Pair Analysis

MINUS(s(X), s(Y)) -> P(minus(X,Y))

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

DIV(s(X), s(Y)) -> DIV(minus(X,Y), s(Y))

DIV(s(X), s(Y)) -> MINUS(X,Y)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

**MINUS(s( X), s(Y)) -> MINUS(X, Y)**

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

div(0, s(Y)) -> 0

div(s(X), s(Y)) -> s(div(minus(X,Y), s(Y)))

The following dependency pair can be strictly oriented:

MINUS(s(X), s(Y)) -> MINUS(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(MINUS(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Polo

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

div(0, s(Y)) -> 0

div(s(X), s(Y)) -> s(div(minus(X,Y), s(Y)))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

**DIV(s( X), s(Y)) -> DIV(minus(X, Y), s(Y))**

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

div(0, s(Y)) -> 0

div(s(X), s(Y)) -> s(div(minus(X,Y), s(Y)))

The following dependency pair can be strictly oriented:

DIV(s(X), s(Y)) -> DIV(minus(X,Y), s(Y))

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

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 1 _{ }^{ }_{ }^{ }POL(DIV(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(minus(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(p(x)_{1})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

div(0, s(Y)) -> 0

div(s(X), s(Y)) -> s(div(minus(X,Y), s(Y)))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes