Term Rewriting System R:
[N, X, Y, X1, X2, Z]
aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

ATERMS(N) -> ASQR(mark(N))
ATERMS(N) -> MARK(N)
AADD(0, X) -> MARK(X)
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
MARK(terms(X)) -> ATERMS(mark(X))
MARK(terms(X)) -> MARK(X)
MARK(sqr(X)) -> ASQR(mark(X))
MARK(sqr(X)) -> MARK(X)
MARK(add(X1, X2)) -> AADD(mark(X1), mark(X2))
MARK(add(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X2)
MARK(dbl(X)) -> ADBL(mark(X))
MARK(dbl(X)) -> MARK(X)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(recip(X)) -> MARK(X)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

MARK(recip(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
MARK(dbl(X)) -> MARK(X)
MARK(add(X1, X2)) -> MARK(X2)
MARK(add(X1, X2)) -> MARK(X1)
AADD(0, X) -> MARK(X)
MARK(add(X1, X2)) -> AADD(mark(X1), mark(X2))
MARK(sqr(X)) -> MARK(X)
MARK(terms(X)) -> MARK(X)
MARK(terms(X)) -> ATERMS(mark(X))
ATERMS(N) -> MARK(N)


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




The following dependency pair can be strictly oriented:

MARK(dbl(X)) -> MARK(X)


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(sqr(x1))=  x1  
  POL(a__terms(x1))=  x1  
  POL(a__dbl(x1))=  1 + x1  
  POL(terms(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__add(x1, x2))=  x1 + x2  
  POL(a__first(x1, x2))=  x1 + x2  
  POL(add(x1, x2))=  x1 + x2  
  POL(A__ADD(x1, x2))=  x2  
  POL(A__FIRST(x1, x2))=  x2  
  POL(0)=  0  
  POL(first(x1, x2))=  x1 + x2  
  POL(a__sqr(x1))=  x1  
  POL(cons(x1, x2))=  x1  
  POL(dbl(x1))=  1 + x1  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(recip(x1))=  x1  
  POL(A__TERMS(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polynomial Ordering


Dependency Pairs:

MARK(recip(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
MARK(add(X1, X2)) -> MARK(X2)
MARK(add(X1, X2)) -> MARK(X1)
AADD(0, X) -> MARK(X)
MARK(add(X1, X2)) -> AADD(mark(X1), mark(X2))
MARK(sqr(X)) -> MARK(X)
MARK(terms(X)) -> MARK(X)
MARK(terms(X)) -> ATERMS(mark(X))
ATERMS(N) -> MARK(N)


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MARK(add(X1, X2)) -> MARK(X2)
MARK(add(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> AADD(mark(X1), mark(X2))


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(sqr(x1))=  x1  
  POL(a__terms(x1))=  x1  
  POL(a__dbl(x1))=  0  
  POL(terms(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__add(x1, x2))=  1 + x1 + x2  
  POL(a__first(x1, x2))=  x1 + x2  
  POL(add(x1, x2))=  1 + x1 + x2  
  POL(A__ADD(x1, x2))=  x2  
  POL(A__FIRST(x1, x2))=  x2  
  POL(0)=  0  
  POL(first(x1, x2))=  x1 + x2  
  POL(a__sqr(x1))=  x1  
  POL(cons(x1, x2))=  x1  
  POL(dbl(x1))=  0  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(recip(x1))=  x1  
  POL(A__TERMS(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 3
                 ↳Dependency Graph


Dependency Pairs:

MARK(recip(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
AADD(0, X) -> MARK(X)
MARK(sqr(X)) -> MARK(X)
MARK(terms(X)) -> MARK(X)
MARK(terms(X)) -> ATERMS(mark(X))
ATERMS(N) -> MARK(N)


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 4
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
MARK(sqr(X)) -> MARK(X)
MARK(terms(X)) -> MARK(X)
ATERMS(N) -> MARK(N)
MARK(terms(X)) -> ATERMS(mark(X))
MARK(recip(X)) -> MARK(X)


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(sqr(x1))=  x1  
  POL(a__terms(x1))=  x1  
  POL(a__dbl(x1))=  0  
  POL(terms(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__add(x1, x2))=  x2  
  POL(a__first(x1, x2))=  1 + x1 + x2  
  POL(add(x1, x2))=  x2  
  POL(A__FIRST(x1, x2))=  x2  
  POL(0)=  0  
  POL(first(x1, x2))=  1 + x1 + x2  
  POL(a__sqr(x1))=  x1  
  POL(cons(x1, x2))=  x1  
  POL(dbl(x1))=  0  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(recip(x1))=  x1  
  POL(A__TERMS(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
MARK(sqr(X)) -> MARK(X)
MARK(terms(X)) -> MARK(X)
ATERMS(N) -> MARK(N)
MARK(terms(X)) -> ATERMS(mark(X))
MARK(recip(X)) -> MARK(X)


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 6
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(recip(X)) -> MARK(X)
MARK(sqr(X)) -> MARK(X)
MARK(terms(X)) -> MARK(X)
ATERMS(N) -> MARK(N)
MARK(terms(X)) -> ATERMS(mark(X))
MARK(cons(X1, X2)) -> MARK(X1)


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MARK(terms(X)) -> MARK(X)
MARK(terms(X)) -> ATERMS(mark(X))


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(sqr(x1))=  x1  
  POL(a__terms(x1))=  1 + x1  
  POL(a__dbl(x1))=  0  
  POL(terms(x1))=  1 + x1  
  POL(mark(x1))=  x1  
  POL(a__add(x1, x2))=  x2  
  POL(a__first(x1, x2))=  x2  
  POL(add(x1, x2))=  x2  
  POL(0)=  0  
  POL(first(x1, x2))=  x2  
  POL(cons(x1, x2))=  x1  
  POL(a__sqr(x1))=  x1  
  POL(dbl(x1))=  0  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(recip(x1))=  x1  
  POL(A__TERMS(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pairs:

MARK(recip(X)) -> MARK(X)
MARK(sqr(X)) -> MARK(X)
ATERMS(N) -> MARK(N)
MARK(cons(X1, X2)) -> MARK(X1)


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 8
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(sqr(X)) -> MARK(X)
MARK(recip(X)) -> MARK(X)


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




The following dependency pair can be strictly oriented:

MARK(cons(X1, X2)) -> MARK(X1)


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(MARK(x1))=  x1  
  POL(cons(x1, x2))=  1 + x1  
  POL(sqr(x1))=  x1  
  POL(recip(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 9
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(sqr(X)) -> MARK(X)
MARK(recip(X)) -> MARK(X)


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




The following dependency pair can be strictly oriented:

MARK(sqr(X)) -> MARK(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(MARK(x1))=  x1  
  POL(sqr(x1))=  1 + x1  
  POL(recip(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 10
                 ↳Polynomial Ordering


Dependency Pair:

MARK(recip(X)) -> MARK(X)


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




The following dependency pair can be strictly oriented:

MARK(recip(X)) -> MARK(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(MARK(x1))=  x1  
  POL(recip(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 11
                 ↳Dependency Graph


Dependency Pair:


Rules:


aterms(N) -> cons(recip(asqr(mark(N))), terms(s(N)))
aterms(X) -> terms(X)
asqr(0) -> 0
asqr(s(X)) -> s(add(sqr(X), dbl(X)))
asqr(X) -> sqr(X)
adbl(0) -> 0
adbl(s(X)) -> s(s(dbl(X)))
adbl(X) -> dbl(X)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
mark(terms(X)) -> aterms(mark(X))
mark(sqr(X)) -> asqr(mark(X))
mark(add(X1, X2)) -> aadd(mark(X1), mark(X2))
mark(dbl(X)) -> adbl(mark(X))
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(recip(X)) -> recip(mark(X))
mark(s(X)) -> s(X)
mark(0) -> 0
mark(nil) -> nil


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:16 minutes