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

↳Forward Instantiation Transformation

→DP Problem 2

↳Nar

**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

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳Forward Instantiation Transformation

→DP Problem 2

↳Nar

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

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳FwdInst

...

→DP Problem 4

↳Polynomial Ordering

→DP Problem 2

↳Nar

**PLUS(s(s(s( x''''))), y'''') -> PLUS(s(s(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(s(s(x''''))),y'''') -> PLUS(s(s(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})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳FwdInst

...

→DP Problem 5

↳Dependency Graph

→DP Problem 2

↳Nar

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

↳FwdInst

→DP Problem 2

↳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 one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳Forward Instantiation Transformation

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

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 Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

three new Dependency Pairs are created:

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

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

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

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

The transformation is resulting in two new DP problems:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳FwdInst

...

→DP Problem 7

↳Forward Instantiation Transformation

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

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 Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳FwdInst

...

→DP Problem 9

↳Forward Instantiation Transformation

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

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 Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳FwdInst

...

→DP Problem 11

↳Polynomial Ordering

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

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

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳FwdInst

...

→DP Problem 14

↳Dependency Graph

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

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

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳FwdInst

...

→DP Problem 8

↳Narrowing Transformation

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

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

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳FwdInst

...

→DP Problem 10

↳Forward Instantiation Transformation

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

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 Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

two new Dependency Pairs are created:

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

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳FwdInst

...

→DP Problem 12

↳Forward Instantiation Transformation

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

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 Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

two new Dependency Pairs are created:

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

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳FwdInst

...

→DP Problem 13

↳Polynomial Ordering

**QUOT(s(s(s( x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), 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 pairs can be strictly oriented:

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

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

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

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

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(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

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳FwdInst

...

→DP Problem 15

↳Dependency Graph

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

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.

Duration:

0:00 minutes