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

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

Furthermore, R contains four SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

one new Dependency Pair is created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

one new Dependency Pair is created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 6`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

DBL(s(X)) -> DBL(X)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 8`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

DBL(s(s(X''))) -> DBL(s(X''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 8`
`             ↳FwdInst`
`             ...`
`               →DP Problem 9`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 8`
`             ↳FwdInst`
`             ...`
`               →DP Problem 10`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

HALF(s(s(X))) -> HALF(X)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 11`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

HALF(s(s(s(s(X''))))) -> HALF(s(s(X'')))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 11`
`             ↳FwdInst`
`             ...`
`               →DP Problem 12`
`                 ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS 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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 11`
`             ↳FwdInst`
`             ...`
`               →DP Problem 13`
`                 ↳Dependency Graph`
`       →DP Problem 4`
`         ↳FwdInst`

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Forward Instantiation Transformation`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

SQR(s(X)) -> SQR(X)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`
`           →DP Problem 14`
`             ↳Forward Instantiation Transformation`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

SQR(s(s(X''))) -> SQR(s(X''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`
`           →DP Problem 14`
`             ↳FwdInst`
`             ...`
`               →DP Problem 15`
`                 ↳Polynomial Ordering`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳FwdInst`
`           →DP Problem 14`
`             ↳FwdInst`
`             ...`
`               →DP Problem 16`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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