Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

TERMS(N) -> SQR(N)
TERMS(N) -> TERMS(s(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)) -> FIRST(X, Z)

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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 6

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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 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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Inst

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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 9

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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 9

             ↳FwdInst

...

               →DP Problem 10

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳FwdInst

...

               →DP Problem 11

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Inst

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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)) -> FIRST(X, Z)

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 12

             ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

Strategy:

innermost

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

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

one new Dependency Pair is created:

FIRST(s(s(s(X''''))), cons(Y, cons(Y''0, cons(Y'''', Z'''')))) -> FIRST(s(s(X'''')), cons(Y''0, cons(Y'''', Z'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 12

             ↳FwdInst

...

               →DP Problem 13

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Inst

Dependency Pair:

FIRST(s(s(s(X''''))), cons(Y, cons(Y''0, cons(Y'''', Z'''')))) -> FIRST(s(s(X'''')), cons(Y''0, cons(Y'''', Z'''')))

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

FIRST(s(s(s(X''''))), cons(Y, cons(Y''0, cons(Y'''', Z'''')))) -> FIRST(s(s(X'''')), cons(Y''0, cons(Y'''', Z'''')))

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(cons(x₁, x₂)) = 0

POL(FIRST(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 12

             ↳FwdInst

...

               →DP Problem 14

                 ↳Dependency Graph

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Inst

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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 4

         ↳FwdInst

           →DP Problem 15

             ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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 4

         ↳FwdInst

           →DP Problem 15

             ↳FwdInst

...

               →DP Problem 16

                 ↳Polynomial Ordering

       →DP Problem 5

         ↳Inst

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 15

             ↳FwdInst

...

               →DP Problem 17

                 ↳Dependency Graph

       →DP Problem 5

         ↳Inst

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

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

         ↳FwdInst

       →DP Problem 5

         ↳Instantiation Transformation

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

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 5

         ↳Inst

           →DP Problem 18

             ↳Instantiation Transformation

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

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 5

         ↳Inst

           →DP Problem 18

             ↳Inst

...

               →DP Problem 19

                 ↳Instantiation Transformation

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

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 5

         ↳Inst

           →DP Problem 18

             ↳Inst

...

               →DP Problem 20

                 ↳Instantiation Transformation

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

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 5

         ↳Inst

           →DP Problem 18

             ↳Inst

...

               →DP Problem 21

                 ↳Instantiation Transformation

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

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 5

         ↳Inst

           →DP Problem 18

             ↳Inst

...

               →DP Problem 22

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
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, first(X, Z))

Strategy:

innermost

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

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

POL(cons(x₁, x₂))	= 0
POL(FIRST(x₁, x₂))	= x₁
POL(s(x₁))	= 1 + x₁