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))

Termination of R to be shown. 



   R
     ↳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. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Remaining


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))





The following dependency pair can be strictly oriented:

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


The following rules can be oriented:

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))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{nil, 0}
terms > recip
terms > {dbl, sqr} > add > s

resulting in one new DP problem.
Used Argument Filtering System:
ADD(x1, x2) -> ADD(x1, x2)
s(x1) -> s(x1)
terms(x1) -> terms(x1)
cons(x1, x2) -> x1
recip(x1) -> recip(x1)
sqr(x1) -> sqr(x1)
add(x1, x2) -> add(x1, x2)
dbl(x1) -> dbl(x1)
first(x1, x2) -> first(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 6
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Remaining


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))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Remaining


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))





The following dependency pair can be strictly oriented:

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


The following rules can be oriented:

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))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{nil, 0}
terms > recip
terms > sqr > add > s
terms > sqr > dbl > s

resulting in one new DP problem.
Used Argument Filtering System:
DBL(x1) -> DBL(x1)
s(x1) -> s(x1)
terms(x1) -> terms(x1)
cons(x1, x2) -> x1
recip(x1) -> recip(x1)
sqr(x1) -> sqr(x1)
add(x1, x2) -> add(x1, x2)
dbl(x1) -> dbl(x1)
first(x1, x2) -> first(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 7
             ↳Dependency Graph
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Remaining


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))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Argument Filtering and Ordering
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Remaining


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))





The following dependency pair can be strictly oriented:

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


The following rules can be oriented:

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))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{nil, 0}
{dbl, sqr} > add > s

resulting in one new DP problem.
Used Argument Filtering System:
FIRST(x1, x2) -> FIRST(x1, x2)
s(x1) -> s(x1)
cons(x1, x2) -> x2
terms(x1) -> terms
sqr(x1) -> sqr(x1)
add(x1, x2) -> add(x1, x2)
dbl(x1) -> dbl(x1)
first(x1, x2) -> first(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
           →DP Problem 8
             ↳Dependency Graph
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Remaining


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))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Argument Filtering and Ordering
       →DP Problem 5
         ↳Remaining


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))





The following dependency pair can be strictly oriented:

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


The following rules can be oriented:

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))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{nil, first}
terms > recip
terms > {dbl, sqr} > add > s

resulting in one new DP problem.
Used Argument Filtering System:
SQR(x1) -> SQR(x1)
s(x1) -> s(x1)
terms(x1) -> terms(x1)
cons(x1, x2) -> x1
recip(x1) -> recip(x1)
sqr(x1) -> sqr(x1)
add(x1, x2) -> add(x1, x2)
dbl(x1) -> dbl(x1)
first(x1, x2) -> first(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
           →DP Problem 9
             ↳Dependency Graph
       →DP Problem 5
         ↳Remaining


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))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Remaining Obligation(s)




The following remains to be proven:
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))




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