pred(s(

minus(

minus(

quot(0, s(

quot(s(

R

↳Dependency Pair Analysis

MINUS(x, s(y)) -> PRED(minus(x,y))

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

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

pred(s(x)) ->x

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

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

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

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(MINUS(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:

MINUS(x,_{1}x) -> MINUS(_{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

pred(s(x)) ->x

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

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

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and Ordering

**QUOT(s( x), s(y)) -> QUOT(minus(x, y), s(y))**

pred(s(x)) ->x

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

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

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

The following dependency pair can be strictly oriented:

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

The following usable rules w.r.t. to the AFS can be oriented:

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

pred(s(x)) ->x

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 4

↳Dependency Graph

pred(s(x)) ->x

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

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

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes