Term Rewriting System R:
[N, X, Y, Z, X1, X2]
terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

TERMS(N) -> SQR(N)
SQR(s(X)) -> SQR(X)
SQR(s(X)) -> DBL(X)
DBL(s(X)) -> DBL(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nterms(X)) -> TERMS(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)

Furthermore, R contains four 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)`

The following remains to be proven:
• Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pairs:

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

`   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)`

The following remains to be proven:
• Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pairs:

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

`   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)`

The following remains to be proven:
• Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pairs:

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

`   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)`

The following remains to be proven:
• Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

• Dependency Pairs:

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Termination of R could not be shown.
Duration:
0:00 minutes