sqr(0) -> 0

sqr(s(

sqr(s(

double(0) -> 0

double(s(

+(

+(

R

↳Dependency Pair Analysis

SQR(s(x)) -> +'(sqr(x), s(double(x)))

SQR(s(x)) -> SQR(x)

SQR(s(x)) -> DOUBLE(x)

SQR(s(x)) -> +'(sqr(x), double(x))

DOUBLE(s(x)) -> DOUBLE(x)

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

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

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

+'(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 4

↳Forward Instantiation Transformation

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

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

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

+'(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 4

↳FwdInst

...

→DP Problem 5

↳Polynomial Ordering

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

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

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

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 4

↳FwdInst

...

→DP Problem 6

↳Dependency Graph

→DP Problem 2

↳FwdInst

→DP Problem 3

↳FwdInst

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

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

**DOUBLE(s( x)) -> DOUBLE(x)**

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

DOUBLE(s(x)) -> DOUBLE(x)

DOUBLE(s(s(x''))) -> DOUBLE(s(x''))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 7

↳Forward Instantiation Transformation

→DP Problem 3

↳FwdInst

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

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

DOUBLE(s(s(x''))) -> DOUBLE(s(x''))

DOUBLE(s(s(s(x'''')))) -> DOUBLE(s(s(x'''')))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

→DP Problem 7

↳FwdInst

...

→DP Problem 8

↳Polynomial Ordering

→DP Problem 3

↳FwdInst

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

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

innermost

The following dependency pair can be strictly oriented:

DOUBLE(s(s(s(x'''')))) -> DOUBLE(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(DOUBLE(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 7

↳FwdInst

...

→DP Problem 9

↳Dependency Graph

→DP Problem 3

↳FwdInst

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

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

**SQR(s( x)) -> SQR(x)**

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

SQR(s(x)) -> SQR(x)

SQR(s(s(x''))) -> SQR(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 10

↳Forward Instantiation Transformation

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

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

SQR(s(s(x''))) -> SQR(s(x''))

SQR(s(s(s(x'''')))) -> SQR(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 10

↳FwdInst

...

→DP Problem 11

↳Polynomial Ordering

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

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

innermost

The following dependency pair can be strictly oriented:

SQR(s(s(s(x'''')))) -> SQR(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(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(SQR(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 10

↳FwdInst

...

→DP Problem 12

↳Dependency Graph

sqr(0) -> 0

sqr(s(x)) -> +(sqr(x), s(double(x)))

sqr(s(x)) -> s(+(sqr(x), double(x)))

double(0) -> 0

double(s(x)) -> s(s(double(x)))

+(x, 0) ->x

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

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes