quot(0, s(

quot(s(

quot(

plus(0,

plus(s(

R

↳Dependency Pair Analysis

QUOT(s(x), s(y),z) -> QUOT(x,y,z)

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))

QUOT(x, 0, s(z)) -> PLUS(z, s(0))

PLUS(s(x),y) -> PLUS(x,y)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

**PLUS(s( x), y) -> PLUS(x, y)**

quot(0, s(y), s(z)) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

The following dependency pair can be strictly oriented:

PLUS(s(x),y) -> PLUS(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(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 3

↳Dependency Graph

→DP Problem 2

↳Polo

quot(0, s(y), s(z)) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

**QUOT( x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))**

quot(0, s(y), s(z)) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

The following dependency pair can be strictly oriented:

QUOT(s(x), s(y),z) -> QUOT(x,y,z)

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(plus(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(QUOT(x)_{1}, x_{2}, x_{3})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }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 4

↳Polynomial Ordering

**QUOT( x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))**

quot(0, s(y), s(z)) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

The following dependency pair can be strictly oriented:

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))

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

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(plus(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(QUOT(x)_{1}, x_{2}, x_{3})= 1 + x _{2}_{ }^{ }_{ }^{ }POL(0)= 1 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Polo

...

→DP Problem 5

↳Dependency Graph

quot(0, s(y), s(z)) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes