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

↳Forward Instantiation Transformation

→DP Problem 2

↳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)))

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:

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

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

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

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:

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

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

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

innermost

The following dependency pair can be strictly oriented:

MINUS(x'''', s(s(s(y'''')))) -> MINUS(x'''', s(s(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(MINUS(x)_{1}, x_{2})= 1 + x _{2}_{ }^{ }_{ }^{ }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

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)))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Narrowing Transformation

**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)))

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

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

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

The transformation is resulting in two new DP problems:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 6

↳Forward Instantiation Transformation

→DP Problem 7

↳Nar

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

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

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), 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 8

↳Polynomial Ordering

→DP Problem 7

↳Nar

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

innermost

The following dependency pair can be strictly oriented:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), 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})= 1 + 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 11

↳Dependency Graph

→DP Problem 7

↳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)))

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 7

↳Narrowing Transformation

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

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

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

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

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

↳Nar

...

→DP Problem 9

↳Polynomial Ordering

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

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

pred(s(x)) ->x

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(QUOT(x)_{1}, x_{2})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(pred(x)_{1})= x _{1}_{ }^{ }_{ }^{ }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 7

↳Nar

...

→DP Problem 10

↳Polynomial Ordering

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

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

pred(s(x)) ->x

minus(x, 0) ->x

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

Duration:

0:00 minutes