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.



   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
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →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




The following dependency pair can be strictly oriented:

ADD(s(X), Y) -> ADD(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))=  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
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.


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




The following dependency pair can be strictly oriented:

DBL(s(X)) -> DBL(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))=  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
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.


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




The following dependency pair can be strictly oriented:

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



   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
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.


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




The following dependency pair can be strictly oriented:

SQR(s(X)) -> SQR(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))=  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
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.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →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:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Inst
           →DP Problem 10
Remaining Obligation(s)




The following remains to be proven:
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



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