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
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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)) -> ADD(sqr(X), dbl(X))
SQR(s(X)) -> SQR(X)
SQR(s(X)) -> DBL(X)
DBL(s(X)) -> DBL(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
HALF(s(s(X))) -> HALF(X)
ACTIVATE(nterms(X)) -> TERMS(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)

Furthermore, R contains five SCCs.

R
DPs
→DP Problem 1
Polynomial Ordering
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 4
Polo
→DP Problem 5
Polo

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(s(x1)) =  1 + x1 POL(ADD(x1, x2)) =  x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 6
Dependency Graph
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 4
Polo
→DP Problem 5
Polo

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polynomial Ordering
→DP Problem 3
Polo
→DP Problem 4
Polo
→DP Problem 5
Polo

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

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

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(s(x1)) =  1 + x1 POL(DBL(x1)) =  x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 7
Dependency Graph
→DP Problem 3
Polo
→DP Problem 4
Polo
→DP Problem 5
Polo

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polynomial Ordering
→DP Problem 4
Polo
→DP Problem 5
Polo

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

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

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(HALF(x1)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 8
Dependency Graph
→DP Problem 4
Polo
→DP Problem 5
Polo

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 4
Polynomial Ordering
→DP Problem 5
Polo

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

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

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(s(x1)) =  1 + x1 POL(SQR(x1)) =  x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 4
Polo
→DP Problem 9
Dependency Graph
→DP Problem 5
Polo

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
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
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 4
Polo
→DP Problem 5
Polynomial Ordering

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
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(cons(x1, x2)) =  x2 POL(FIRST(x1, x2)) =  x2 POL(s(x1)) =  0 POL(ACTIVATE(x1)) =  x1 POL(n__first(x1, x2)) =  1 + x2

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 4
Polo
→DP Problem 5
Polo
→DP Problem 10
Dependency Graph

Dependency Pair:

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
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes