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))

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)

Furthermore, R contains five SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`
`       →DP Problem 5`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• 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))

• 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))

• 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))

• 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))

• 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)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`
`       →DP Problem 5`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• 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))

• 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))

• 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))

• 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))

• 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)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`
`       →DP Problem 5`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• 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))

• 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))

• 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))

• 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))

• 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)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`
`       →DP Problem 5`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• 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))

• 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))

• 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))

• 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))

• 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)))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`
`       →DP Problem 5`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• 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))

• 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))

• 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))

• 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))

• Dependency Pair:

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

Rules:

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