Termination Proof using AProVE

Term Rewriting System R:
[N, X, Y, Z, X1, X2]
terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳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)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_terms}(X)) -> TERMS(X)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳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:

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

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 6

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(ADD(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.
Used Argument Filtering System:

ADD(x₁, x₂) -> ADD(x₁, x₂)
s(x₁) -> s(x₁)

     ↳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)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳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)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(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:

     ↳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)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(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:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 9

                 ↳Argument Filtering and 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)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(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 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.
Used Argument Filtering System:

DBL(x₁) -> DBL(x₁)
s(x₁) -> s(x₁)

     ↳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)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(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:

     ↳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:

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

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(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:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 12

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(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 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.
Used Argument Filtering System:

SQR(x₁) -> SQR(x₁)
s(x₁) -> s(x₁)

     ↳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)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Forward Instantiation Transformation

Dependency Pairs:

ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

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

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

one new Dependency Pair is created:

FIRST(s(X), cons(Y, n_{_first}(X1'', X2''))) -> ACTIVATE(n_{_first}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳Forward Instantiation Transformation

Dependency Pairs:

FIRST(s(X), cons(Y, n_{_first}(X1'', X2''))) -> ACTIVATE(n_{_first}(X1'', X2''))
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

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

ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)

one new Dependency Pair is created:

ACTIVATE(n_{_first}(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 15

                 ↳Forward Instantiation Transformation

Dependency Pairs:

ACTIVATE(n_{_first}(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))
FIRST(s(X), cons(Y, n_{_first}(X1'', X2''))) -> ACTIVATE(n_{_first}(X1'', X2''))

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

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

FIRST(s(X), cons(Y, n_{_first}(X1'', X2''))) -> ACTIVATE(n_{_first}(X1'', X2''))

one new Dependency Pair is created:

FIRST(s(X), cons(Y, n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 16

                 ↳Forward Instantiation Transformation

Dependency Pairs:

FIRST(s(X), cons(Y, n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))
ACTIVATE(n_{_first}(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

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

ACTIVATE(n_{_first}(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))

one new Dependency Pair is created:

ACTIVATE(n_{_first}(s(X'''), cons(Y''', n_{_first}(s(X''''''), cons(Y'''''', n_{_first}(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', n_{_first}(s(X''''''), cons(Y'''''', n_{_first}(X1'''''''', X2'''''''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 17

                 ↳Argument Filtering and Ordering

Dependency Pairs:

ACTIVATE(n_{_first}(s(X'''), cons(Y''', n_{_first}(s(X''''''), cons(Y'''''', n_{_first}(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', n_{_first}(s(X''''''), cons(Y'''''', n_{_first}(X1'''''''', X2'''''''')))))
FIRST(s(X), cons(Y, n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

FIRST(s(X), cons(Y, n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(cons(x₁, x₂)) = x₁ + x₂

POL(FIRST(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

POL(ACTIVATE(x₁)) = x₁

POL(n__first(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.
Used Argument Filtering System:

FIRST(x₁, x₂) -> FIRST(x₁, x₂)
ACTIVATE(x₁) -> ACTIVATE(x₁)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
n_{_first}(x₁, x₂) -> n_{_first}(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 18

                 ↳Dependency Graph

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(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
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(s(x₁))	= 1 + x₁
POL(ADD(x₁, x₂))	= 1 + x₁ + x₂

POL(cons(x₁, x₂))	= x₁ + x₂
POL(FIRST(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁
POL(ACTIVATE(x₁))	= x₁
POL(n__first(x₁, x₂))	= x₁ + x₂