plus(0,

plus(s(

min(

min(s(

min(min(

quot(0, s(

quot(s(

R

↳Dependency Pair Analysis

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

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

MIN(min(X,Y), Z) -> MIN(X, plus(Y, Z))

MIN(min(X,Y), Z) -> PLUS(Y, Z)

QUOT(s(X), s(Y)) -> QUOT(min(X,Y), s(Y))

QUOT(s(X), s(Y)) -> MIN(X,Y)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

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

plus(0,Y) ->Y

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

min(X, 0) ->X

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

min(min(X,Y), Z) -> min(X, plus(Y, Z))

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

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

innermost

The following dependency pair can be strictly oriented:

PLUS(s(X),Y) -> PLUS(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(PLUS(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 4

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

plus(0,Y) ->Y

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

min(X, 0) ->X

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

min(min(X,Y), Z) -> min(X, plus(Y, Z))

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

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

→DP Problem 3

↳Polo

**MIN(min( X, Y), Z) -> MIN(X, plus(Y, Z))**

plus(0,Y) ->Y

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

min(X, 0) ->X

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

min(min(X,Y), Z) -> min(X, plus(Y, Z))

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

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

innermost

The following dependency pair can be strictly oriented:

MIN(s(X), s(Y)) -> MIN(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(plus(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(Z)= 0 _{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(MIN(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(min(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 2

↳Polo

→DP Problem 5

↳Narrowing Transformation

→DP Problem 3

↳Polo

**MIN(min( X, Y), Z) -> MIN(X, plus(Y, Z))**

plus(0,Y) ->Y

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

min(X, 0) ->X

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

min(min(X,Y), Z) -> min(X, plus(Y, Z))

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

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

innermost

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

As a result of transforming the rule

one new Dependency Pair is created:

MIN(min(X,Y), Z) -> MIN(X, plus(Y, Z))

MIN(min(X, s(X'')), Z) -> MIN(X, s(plus(X'', Z)))

The transformation is resulting in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polynomial Ordering

**QUOT(s( X), s(Y)) -> QUOT(min(X, Y), s(Y))**

plus(0,Y) ->Y

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

min(X, 0) ->X

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

min(min(X,Y), Z) -> min(X, plus(Y, Z))

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

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

innermost

The following dependency pair can be strictly oriented:

QUOT(s(X), s(Y)) -> QUOT(min(X,Y), s(Y))

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

min(X, 0) ->X

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

min(min(X,Y), Z) -> min(X, plus(Y, Z))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(Z)= 0 _{ }^{ }_{ }^{ }POL(plus(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(QUOT(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 1 _{ }^{ }_{ }^{ }POL(min(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 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 6

↳Dependency Graph

plus(0,Y) ->Y

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

min(X, 0) ->X

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

min(min(X,Y), Z) -> min(X, plus(Y, Z))

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

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes