quot(0, s(

quot(s(

quot(

R

↳Dependency Pair Analysis

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Forward Instantiation Transformation

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

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

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

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

↳FwdInst

...

→DP Problem 3

↳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, s(z), s(z)))

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 4

↳Forward Instantiation Transformation

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

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

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

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

The transformation is resulting in two new DP problems:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 5

↳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, s(z), s(z)))

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 7

↳Polynomial Ordering

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

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

↳FwdInst

...

→DP Problem 9

↳Dependency Graph

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

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 6

↳Forward Instantiation Transformation

**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, s(z), s(z)))

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

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 8

↳Polynomial Ordering

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

innermost

The following dependency pairs can be strictly oriented:

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

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

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

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})= 1 + 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

↳FwdInst

...

→DP Problem 10

↳Dependency Graph

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

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes