Term Rewriting System R:
[X, L, X1, X2]
aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AINCR(cons(X, L)) -> MARK(X)
AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))
AADX(cons(X, L)) -> MARK(X)
ANATS -> AADX(azeros)
ANATS -> AZEROS
AHEAD(cons(X, L)) -> MARK(X)
ATAIL(cons(X, L)) -> MARK(L)
MARK(incr(X)) -> AINCR(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> AADX(mark(X))
MARK(adx(X)) -> MARK(X)
MARK(nats) -> ANATS
MARK(zeros) -> AZEROS
MARK(head(X)) -> AHEAD(mark(X))
MARK(head(X)) -> MARK(X)
MARK(tail(X)) -> ATAIL(mark(X))
MARK(tail(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
ATAIL(cons(X, L)) -> MARK(L)
MARK(tail(X)) -> ATAIL(mark(X))
MARK(head(X)) -> MARK(X)
AHEAD(cons(X, L)) -> MARK(X)
MARK(head(X)) -> AHEAD(mark(X))
ANATS -> AADX(azeros)
MARK(nats) -> ANATS
MARK(adx(X)) -> MARK(X)
AADX(cons(X, L)) -> MARK(X)
AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))
MARK(adx(X)) -> AADX(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(incr(X)) -> AINCR(mark(X))
AINCR(cons(X, L)) -> MARK(X)


Rules:


aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0





The following dependency pairs can be strictly oriented:

MARK(tail(X)) -> MARK(X)
MARK(tail(X)) -> ATAIL(mark(X))


Additionally, the following usable rules using the Ce-refinement can be oriented:

mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
anats -> aadx(azeros)
anats -> nats
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
azeros -> cons(0, zeros)
azeros -> zeros


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a__nats)=  0  
  POL(MARK(x1))=  x1  
  POL(adx(x1))=  x1  
  POL(A__NATS)=  0  
  POL(a__zeros)=  0  
  POL(A__ADX(x1))=  x1  
  POL(tail(x1))=  1 + x1  
  POL(incr(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(A__TAIL(x1))=  x1  
  POL(a__adx(x1))=  x1  
  POL(A__INCR(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  0  
  POL(nil)=  0  
  POL(a__tail(x1))=  1 + x1  
  POL(s(x1))=  x1  
  POL(head(x1))=  x1  
  POL(zeros)=  0  
  POL(a__head(x1))=  x1  
  POL(A__HEAD(x1))=  x1  
  POL(a__incr(x1))=  x1  

resulting in one new DP problem. 



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


Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
ATAIL(cons(X, L)) -> MARK(L)
MARK(head(X)) -> MARK(X)
AHEAD(cons(X, L)) -> MARK(X)
MARK(head(X)) -> AHEAD(mark(X))
ANATS -> AADX(azeros)
MARK(nats) -> ANATS
MARK(adx(X)) -> MARK(X)
AADX(cons(X, L)) -> MARK(X)
AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))
MARK(adx(X)) -> AADX(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(incr(X)) -> AINCR(mark(X))
AINCR(cons(X, L)) -> MARK(X)


Rules:


aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0





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


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(head(X)) -> MARK(X)
AHEAD(cons(X, L)) -> MARK(X)
MARK(head(X)) -> AHEAD(mark(X))
ANATS -> AADX(azeros)
MARK(nats) -> ANATS
MARK(adx(X)) -> MARK(X)
AADX(cons(X, L)) -> MARK(X)
AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))
MARK(adx(X)) -> AADX(mark(X))
MARK(incr(X)) -> MARK(X)
AINCR(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> AINCR(mark(X))
MARK(s(X)) -> MARK(X)


Rules:


aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0





The following dependency pairs can be strictly oriented:

MARK(head(X)) -> MARK(X)
MARK(head(X)) -> AHEAD(mark(X))


Additionally, the following usable rules using the Ce-refinement can be oriented:

mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
anats -> aadx(azeros)
anats -> nats
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
azeros -> cons(0, zeros)
azeros -> zeros


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a__nats)=  0  
  POL(MARK(x1))=  x1  
  POL(adx(x1))=  x1  
  POL(A__NATS)=  0  
  POL(a__zeros)=  0  
  POL(A__ADX(x1))=  x1  
  POL(tail(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__adx(x1))=  x1  
  POL(A__INCR(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  0  
  POL(nil)=  0  
  POL(a__tail(x1))=  x1  
  POL(s(x1))=  x1  
  POL(head(x1))=  1 + x1  
  POL(zeros)=  0  
  POL(a__head(x1))=  1 + x1  
  POL(A__HEAD(x1))=  x1  
  POL(a__incr(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
AHEAD(cons(X, L)) -> MARK(X)
ANATS -> AADX(azeros)
MARK(nats) -> ANATS
MARK(adx(X)) -> MARK(X)
AADX(cons(X, L)) -> MARK(X)
AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))
MARK(adx(X)) -> AADX(mark(X))
MARK(incr(X)) -> MARK(X)
AINCR(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> AINCR(mark(X))
MARK(s(X)) -> MARK(X)


Rules:


aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0





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


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 5
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(s(X)) -> MARK(X)
ANATS -> AADX(azeros)
MARK(nats) -> ANATS
MARK(adx(X)) -> MARK(X)
AADX(cons(X, L)) -> MARK(X)
AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))
MARK(adx(X)) -> AADX(mark(X))
MARK(incr(X)) -> MARK(X)
AINCR(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> AINCR(mark(X))
MARK(cons(X1, X2)) -> MARK(X1)


Rules:


aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0





The following dependency pair can be strictly oriented:

ANATS -> AADX(azeros)


Additionally, the following usable rules using the Ce-refinement can be oriented:

azeros -> cons(0, zeros)
azeros -> zeros
mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
anats -> aadx(azeros)
anats -> nats
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a__nats)=  1  
  POL(MARK(x1))=  x1  
  POL(adx(x1))=  x1  
  POL(A__NATS)=  1  
  POL(a__zeros)=  0  
  POL(A__ADX(x1))=  x1  
  POL(tail(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__adx(x1))=  x1  
  POL(A__INCR(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  1  
  POL(nil)=  0  
  POL(a__tail(x1))=  x1  
  POL(s(x1))=  x1  
  POL(head(x1))=  x1  
  POL(zeros)=  0  
  POL(a__head(x1))=  x1  
  POL(a__incr(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 6
                 ↳Dependency Graph


Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(nats) -> ANATS
MARK(adx(X)) -> MARK(X)
AADX(cons(X, L)) -> MARK(X)
AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))
MARK(adx(X)) -> AADX(mark(X))
MARK(incr(X)) -> MARK(X)
AINCR(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> AINCR(mark(X))
MARK(cons(X1, X2)) -> MARK(X1)


Rules:


aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0





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


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 7
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(adx(X)) -> MARK(X)
AADX(cons(X, L)) -> MARK(X)
AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))
MARK(adx(X)) -> AADX(mark(X))
MARK(incr(X)) -> MARK(X)
AINCR(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> AINCR(mark(X))
MARK(s(X)) -> MARK(X)


Rules:


aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0





The following dependency pairs can be strictly oriented:

MARK(adx(X)) -> MARK(X)
AADX(cons(X, L)) -> MARK(X)


Additionally, the following usable rules using the Ce-refinement can be oriented:

mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
anats -> aadx(azeros)
anats -> nats
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
azeros -> cons(0, zeros)
azeros -> zeros


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a__nats)=  1  
  POL(MARK(x1))=  x1  
  POL(adx(x1))=  1 + x1  
  POL(a__zeros)=  0  
  POL(A__ADX(x1))=  1 + x1  
  POL(tail(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__adx(x1))=  1 + x1  
  POL(A__INCR(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  1  
  POL(nil)=  0  
  POL(a__tail(x1))=  x1  
  POL(s(x1))=  x1  
  POL(head(x1))=  x1  
  POL(zeros)=  0  
  POL(a__head(x1))=  x1  
  POL(a__incr(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 8
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))
MARK(adx(X)) -> AADX(mark(X))
MARK(incr(X)) -> MARK(X)
AINCR(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> AINCR(mark(X))
MARK(s(X)) -> MARK(X)


Rules:


aincr(nil) -> nil
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(X) -> incr(X)
aadx(nil) -> nil
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
aadx(X) -> adx(X)
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
ahead(cons(X, L)) -> mark(X)
ahead(X) -> head(X)
atail(cons(X, L)) -> mark(L)
atail(X) -> tail(X)
mark(incr(X)) -> aincr(mark(X))
mark(adx(X)) -> aadx(mark(X))
mark(nats) -> anats
mark(zeros) -> azeros
mark(head(X)) -> ahead(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0




Termination of R could not be shown.
Duration:
0:01 minutes