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

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

**+'( 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))

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.

Used Argument Filtering System:

+'(x,_{1}x) -> +'(_{2}x,_{1}x)_{2}

s(x) -> s(_{1}x)_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

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))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and Ordering

→DP Problem 3

↳AFS

**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))

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.

Used Argument Filtering System:

DOUBLE(x) -> DOUBLE(_{1}x)_{1}

s(x) -> s(_{1}x)_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳AFS

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))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳Argument Filtering and 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))

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.

Used Argument Filtering System:

SQR(x) -> SQR(_{1}x)_{1}

s(x) -> s(_{1}x)_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→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))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes