pred(s(

minus(

minus(

quot(0, s(

quot(s(

log(s(0)) -> 0

log(s(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)

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

→DP Problem 3

↳Nar

**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)))

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

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)))

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(pred(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(log(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(MINUS(x)_{1}, x_{2})= 1 + 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}

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

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳AFS

→DP Problem 3

↳Nar

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)))

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and Ordering

→DP Problem 3

↳Nar

**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)))

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

minus(x, 0) ->x

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

pred(s(x)) ->x

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

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

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(QUOT(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(pred(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(log(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}

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Nar

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)))

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳Narrowing Transformation

**LOG(s(s( x))) -> LOG(s(quot(x, s(s(0)))))**

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)))

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

As a result of transforming the rule

two new Dependency Pairs are created:

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

LOG(s(s(0))) -> LOG(s(0))

LOG(s(s(s(x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳Nar

→DP Problem 6

↳Narrowing Transformation

**LOG(s(s(s( x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))**

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)))

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

As a result of transforming the rule

one new Dependency Pair is created:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))

LOG(s(s(s(x''')))) -> LOG(s(s(quot(pred(minus(x''', 0)), s(s(0))))))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳Nar

→DP Problem 6

↳Nar

...

→DP Problem 7

↳Argument Filtering and Ordering

**LOG(s(s(s( x''')))) -> LOG(s(s(quot(pred(minus(x''', 0)), s(s(0))))))**

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)))

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The following dependency pair can be strictly oriented:

LOG(s(s(s(x''')))) -> LOG(s(s(quot(pred(minus(x''', 0)), s(s(0))))))

The following rules can be oriented:

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

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

pred(s(x)) ->x

minus(x, 0) ->x

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

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(pred(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(log(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(LOG(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳Nar

→DP Problem 6

↳Nar

...

→DP Problem 8

↳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)))

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes