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

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

**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))

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(PLUS(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

Used Argument Filtering System:

PLUS(x,_{1}x) -> PLUS(_{2}x,_{1}x)_{2}

s(x) -> s(_{1}x)_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳AFS

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))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and 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))

innermost

The following dependency pair can be strictly oriented:

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

The following usable rules for innermost 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 _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

Used Argument Filtering System:

QUOT(x,_{1}x,_{2}x) ->_{3}x_{1}

s(x) -> s(_{1}x)_{1}

plus(x,_{1}x) -> plus(_{2}x,_{1}x)_{2}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 4

↳Narrowing Transformation

**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))

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:

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

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

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

The transformation is resulting in no new DP problems.

Duration:

0:00 minutes