fac(0) -> 1

fac(s(

fac(0) -> s(0)

floop(0,

floop(s(

*(

*(

+(

+(

1 -> s(0)

R

↳Dependency Pair Analysis

FAC(0) -> 1'

FAC(s(x)) -> *'(s(x), fac(x))

FAC(s(x)) -> FAC(x)

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

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

→DP Problem 4

↳Polo

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

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 5

↳Forward Instantiation Transformation

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

→DP Problem 4

↳Polo

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

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 5

↳FwdInst

...

→DP Problem 6

↳Polynomial Ordering

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

→DP Problem 4

↳Polo

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

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 5

↳FwdInst

...

→DP Problem 7

↳Dependency Graph

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

→DP Problem 4

↳Polo

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Forward Instantiation Transformation

→DP Problem 3

↳FwdInst

→DP Problem 4

↳Polo

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

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 8

↳Forward Instantiation Transformation

→DP Problem 3

↳FwdInst

→DP Problem 4

↳Polo

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

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 8

↳FwdInst

...

→DP Problem 9

↳Polynomial Ordering

→DP Problem 3

↳FwdInst

→DP Problem 4

↳Polo

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

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

innermost

The following dependency pair can be strictly oriented:

*'(x'''', s(s(s(y'''')))) -> *'(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(*'(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 2

↳FwdInst

→DP Problem 8

↳FwdInst

...

→DP Problem 10

↳Dependency Graph

→DP Problem 3

↳FwdInst

→DP Problem 4

↳Polo

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 3

↳Forward Instantiation Transformation

→DP Problem 4

↳Polo

**FAC(s( x)) -> FAC(x)**

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

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:

FAC(s(x)) -> FAC(x)

FAC(s(s(x''))) -> FAC(s(x''))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

→DP Problem 11

↳Forward Instantiation Transformation

→DP Problem 4

↳Polo

**FAC(s(s( x''))) -> FAC(s(x''))**

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

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:

FAC(s(s(x''))) -> FAC(s(x''))

FAC(s(s(s(x'''')))) -> FAC(s(s(x'''')))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

→DP Problem 11

↳FwdInst

...

→DP Problem 12

↳Polynomial Ordering

→DP Problem 4

↳Polo

**FAC(s(s(s( x'''')))) -> FAC(s(s(x'''')))**

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

innermost

The following dependency pair can be strictly oriented:

FAC(s(s(s(x'''')))) -> FAC(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(FAC(x)_{1})= 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

↳FwdInst

→DP Problem 3

↳FwdInst

→DP Problem 11

↳FwdInst

...

→DP Problem 13

↳Dependency Graph

→DP Problem 4

↳Polo

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

→DP Problem 4

↳Polynomial Ordering

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

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

innermost

The following dependency pair can be strictly oriented:

FLOOP(s(x),y) -> FLOOP(x, *(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(0)= 0 _{ }^{ }_{ }^{ }POL(*(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(FLOOP(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

→DP Problem 4

↳Polo

→DP Problem 14

↳Dependency Graph

fac(0) -> 1

fac(s(x)) -> *(s(x), fac(x))

fac(0) -> s(0)

floop(0,y) ->y

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

*(x, 0) -> 0

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

+(x, 0) ->x

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

1 -> s(0)

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes