Term Rewriting System R:
[N, X, Y, Z]
terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

TERMS(N) -> SQR(N)
TERMS(N) -> TERMS(s(N))
SQR(s(X)) -> SQR(X)
SQR(s(X)) -> DBL(X)
DBL(s(X)) -> DBL(X)
FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)
HALF(s(s(X))) -> HALF(X)

Furthermore, R contains six SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`
`       →DP Problem 6`
`         ↳Remaining`

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

The following dependency pair can be strictly oriented:

The following rules can be oriented:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
terms > recip
terms > sqr > add > s > half
terms > sqr > dbl > s > half
first > nil

resulting in one new DP problem.
Used Argument Filtering System:
s(x1) -> s(x1)
terms(x1) -> terms(x1)
cons(x1, x2) -> x1
recip(x1) -> recip(x1)
sqr(x1) -> sqr(x1)
dbl(x1) -> dbl(x1)
first(x1, x2) -> first(x1, x2)
half(x1) -> half(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 7`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`
`       →DP Problem 6`
`         ↳Remaining`

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

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`
`         ↳AFS`
`       →DP Problem 6`
`         ↳Remaining`

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
terms > {dbl, sqr} > add > {half, s}
terms > recip
first > nil

resulting in one new DP problem.
Used Argument Filtering System:
DBL(x1) -> DBL(x1)
s(x1) -> s(x1)
terms(x1) -> terms(x1)
cons(x1, x2) -> x1
recip(x1) -> recip(x1)
sqr(x1) -> sqr(x1)
dbl(x1) -> dbl(x1)
first(x1, x2) -> first(x1, x2)
half(x1) -> half(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`           →DP Problem 8`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`
`       →DP Problem 6`
`         ↳Remaining`

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

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`
`         ↳AFS`
`       →DP Problem 6`
`         ↳Remaining`

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{nil, 0}
sqr > add > {half, s}
sqr > dbl > {half, s}

resulting in one new DP problem.
Used Argument Filtering System:
FIRST(x1, x2) -> FIRST(x1, x2)
s(x1) -> s(x1)
cons(x1, x2) -> x2
terms(x1) -> terms
sqr(x1) -> sqr(x1)
dbl(x1) -> dbl(x1)
first(x1, x2) -> first(x1, x2)
half(x1) -> half(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`           →DP Problem 9`
`             ↳Dependency Graph`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`
`       →DP Problem 6`
`         ↳Remaining`

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

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`
`         ↳AFS`
`       →DP Problem 6`
`         ↳Remaining`

Dependency Pair:

HALF(s(s(X))) -> HALF(X)

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

The following dependency pair can be strictly oriented:

HALF(s(s(X))) -> HALF(X)

The following rules can be oriented:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{nil, first}
terms > {dbl, sqr} > add > {half, s}
terms > recip

resulting in one new DP problem.
Used Argument Filtering System:
HALF(x1) -> HALF(x1)
s(x1) -> s(x1)
terms(x1) -> terms(x1)
cons(x1, x2) -> x1
recip(x1) -> recip(x1)
sqr(x1) -> sqr(x1)
dbl(x1) -> dbl(x1)
first(x1, x2) -> first(x1, x2)
half(x1) -> half(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`           →DP Problem 10`
`             ↳Dependency Graph`
`       →DP Problem 5`
`         ↳AFS`
`       →DP Problem 6`
`         ↳Remaining`

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

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`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 6`
`         ↳Remaining`

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
half > s
terms > {dbl, sqr} > add > s
terms > recip
0 > nil

resulting in one new DP problem.
Used Argument Filtering System:
SQR(x1) -> SQR(x1)
s(x1) -> s(x1)
terms(x1) -> terms(x1)
cons(x1, x2) -> x1
recip(x1) -> recip(x1)
sqr(x1) -> sqr(x1)
dbl(x1) -> dbl(x1)
first(x1, x2) -> first(x1, x2)
half(x1) -> half(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`
`           →DP Problem 11`
`             ↳Dependency Graph`
`       →DP Problem 6`
`         ↳Remaining`

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

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`
`         ↳AFS`
`       →DP Problem 6`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))