Term Rewriting System R:
[N, X, Y]
terms(N) -> cons(recip(sqr(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)) -> cons(Y)

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

TERMS(N) -> SQR(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)

Furthermore, R contains three SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(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)) -> cons(Y)

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(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)) -> cons(Y)

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(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)) -> cons(Y)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(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)) -> cons(Y)

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(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)) -> cons(Y)

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(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)) -> cons(Y)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(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)) -> cons(Y)

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(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)) -> cons(Y)

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(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)) -> cons(Y)

Termination of R could not be shown.
Duration:
0:12 minutes