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

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

**+'( 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

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

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

↳Polo

→DP Problem 2

↳Polynomial Ordering

→DP Problem 3

↳Polo

**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

The following dependency pair can be strictly oriented:

DOUBLE(s(x)) -> DOUBLE(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})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Polo

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

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polynomial Ordering

**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

The following dependency pair can be strictly oriented:

SQR(s(x)) -> SQR(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})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 6

↳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