terms(

sqr(0) -> 0

sqr(s(

dbl(0) -> 0

dbl(s(

add(0,

add(s(

first(0,

first(s(

R

↳Dependency Pair Analysis

TERMS(N) -> SQR(N)

TERMS(N) -> TERMS(s(N))

SQR(s(X)) -> ADD(sqr(X), dbl(X))

SQR(s(X)) -> SQR(X)

SQR(s(X)) -> DBL(X)

DBL(s(X)) -> DBL(X)

ADD(s(X),Y) -> ADD(X,Y)

FIRST(s(X), cons(Y,Z)) -> FIRST(X,Z)

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳AFS

→DP Problem 5

↳Remaining

**ADD(s( X), Y) -> ADD(X, Y)**

terms(N) -> cons(recip(sqr(N)), terms(s(N)))

sqr(0) -> 0

sqr(s(X)) -> s(add(sqr(X), dbl(X)))

dbl(0) -> 0

dbl(s(X)) -> s(s(dbl(X)))

add(0,X) ->X

add(s(X),Y) -> s(add(X,Y))

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

innermost

The following dependency pair can be strictly oriented:

ADD(s(X),Y) -> ADD(X,Y)

There are no usable rules for innermost 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:

ADD(x,_{1}x) -> ADD(_{2}x,_{1}x)_{2}

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 6

↳Dependency Graph

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳AFS

→DP Problem 5

↳Remaining

terms(N) -> cons(recip(sqr(N)), terms(s(N)))

sqr(0) -> 0

sqr(s(X)) -> s(add(sqr(X), dbl(X)))

dbl(0) -> 0

dbl(s(X)) -> s(s(dbl(X)))

add(0,X) ->X

add(s(X),Y) -> s(add(X,Y))

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

innermost

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

→DP Problem 4

↳AFS

→DP Problem 5

↳Remaining

**DBL(s( X)) -> DBL(X)**

terms(N) -> cons(recip(sqr(N)), terms(s(N)))

sqr(0) -> 0

sqr(s(X)) -> s(add(sqr(X), dbl(X)))

dbl(0) -> 0

dbl(s(X)) -> s(s(dbl(X)))

add(0,X) ->X

add(s(X),Y) -> s(add(X,Y))

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

innermost

The following dependency pair can be strictly oriented:

DBL(s(X)) -> DBL(X)

There are no usable rules for innermost 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:

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 7

↳Dependency Graph

→DP Problem 3

↳AFS

→DP Problem 4

↳AFS

→DP Problem 5

↳Remaining

terms(N) -> cons(recip(sqr(N)), terms(s(N)))

sqr(0) -> 0

sqr(s(X)) -> s(add(sqr(X), dbl(X)))

dbl(0) -> 0

dbl(s(X)) -> s(s(dbl(X)))

add(0,X) ->X

add(s(X),Y) -> s(add(X,Y))

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

innermost

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

→DP Problem 4

↳AFS

→DP Problem 5

↳Remaining

**FIRST(s( X), cons(Y, Z)) -> FIRST(X, Z)**

terms(N) -> cons(recip(sqr(N)), terms(s(N)))

sqr(0) -> 0

sqr(s(X)) -> s(add(sqr(X), dbl(X)))

dbl(0) -> 0

dbl(s(X)) -> s(s(dbl(X)))

add(0,X) ->X

add(s(X),Y) -> s(add(X,Y))

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

innermost

The following dependency pair can be strictly oriented:

FIRST(s(X), cons(Y,Z)) -> FIRST(X,Z)

There are no usable rules for innermost 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:

FIRST(x,_{1}x) -> FIRST(_{2}x,_{1}x)_{2}

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

cons(x,_{1}x) -> cons(_{2}x,_{1}x)_{2}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 8

↳Dependency Graph

→DP Problem 4

↳AFS

→DP Problem 5

↳Remaining

terms(N) -> cons(recip(sqr(N)), terms(s(N)))

sqr(0) -> 0

sqr(s(X)) -> s(add(sqr(X), dbl(X)))

dbl(0) -> 0

dbl(s(X)) -> s(s(dbl(X)))

add(0,X) ->X

add(s(X),Y) -> s(add(X,Y))

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Argument Filtering and Ordering

→DP Problem 5

↳Remaining

**SQR(s( X)) -> SQR(X)**

terms(N) -> cons(recip(sqr(N)), terms(s(N)))

sqr(0) -> 0

sqr(s(X)) -> s(add(sqr(X), dbl(X)))

dbl(0) -> 0

dbl(s(X)) -> s(s(dbl(X)))

add(0,X) ->X

add(s(X),Y) -> s(add(X,Y))

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

innermost

The following dependency pair can be strictly oriented:

SQR(s(X)) -> SQR(X)

There are no usable rules for innermost 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 4

↳AFS

→DP Problem 9

↳Dependency Graph

→DP Problem 5

↳Remaining

terms(N) -> cons(recip(sqr(N)), terms(s(N)))

sqr(0) -> 0

sqr(s(X)) -> s(add(sqr(X), dbl(X)))

dbl(0) -> 0

dbl(s(X)) -> s(s(dbl(X)))

add(0,X) ->X

add(s(X),Y) -> s(add(X,Y))

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳AFS

→DP Problem 5

↳Remaining Obligation(s)

The following remains to be proven:

**TERMS( N) -> TERMS(s(N))**

terms(N) -> cons(recip(sqr(N)), terms(s(N)))

sqr(0) -> 0

sqr(s(X)) -> s(add(sqr(X), dbl(X)))

dbl(0) -> 0

dbl(s(X)) -> s(s(dbl(X)))

add(0,X) ->X

add(s(X),Y) -> s(add(X,Y))

first(0,X) -> nil

first(s(X), cons(Y,Z)) -> cons(Y, first(X,Z))

innermost

Duration:

0:00 minutes