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
     ↳Removing Redundant Rules



Removing the following rules from R which fullfill a polynomial ordering:

aadx(nil) -> nil

where the Polynomial interpretation:
  POL(a__nats)=  0  
  POL(adx(x1))=  2·x1  
  POL(a__zeros)=  0  
  POL(tail(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__adx(x1))=  2·x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  0  
  POL(nil)=  1  
  POL(a__tail(x1))=  x1  
  POL(s(x1))=  x1  
  POL(zeros)=  0  
  POL(head(x1))=  x1  
  POL(a__head(x1))=  x1  
  POL(a__incr(x1))=  x1  
was used.

Not all Rules of R can be deleted, so we still have to regard a part of R. 


   R
     ↳RRRPolo
       →TRS2
         ↳Removing Redundant Rules



Removing the following rules from R which fullfill a polynomial ordering:

anats -> aadx(azeros)

where the Polynomial interpretation:
  POL(a__nats)=  1  
  POL(adx(x1))=  x1  
  POL(a__zeros)=  0  
  POL(tail(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__adx(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(a__tail(x1))=  x1  
  POL(zeros)=  0  
  POL(head(x1))=  x1  
  POL(a__head(x1))=  x1  
  POL(a__incr(x1))=  x1  
was used.

Not all Rules of R can be deleted, so we still have to regard a part of R. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳Removing Redundant Rules



Removing the following rules from R which fullfill a polynomial ordering:

ahead(cons(X, L)) -> mark(X)

where the Polynomial interpretation:
  POL(a__nats)=  0  
  POL(adx(x1))=  x1  
  POL(a__zeros)=  0  
  POL(tail(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__adx(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  0  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(a__tail(x1))=  x1  
  POL(zeros)=  0  
  POL(head(x1))=  1 + x1  
  POL(a__head(x1))=  1 + x1  
  POL(a__incr(x1))=  x1  
was used.

Not all Rules of R can be deleted, so we still have to regard a part of R. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →TRS4
                 ↳Removing Redundant Rules



Removing the following rules from R which fullfill a polynomial ordering:

atail(cons(X, L)) -> mark(L)

where the Polynomial interpretation:
  POL(a__nats)=  0  
  POL(adx(x1))=  x1  
  POL(a__zeros)=  0  
  POL(tail(x1))=  1 + x1  
  POL(incr(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__adx(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  0  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(a__tail(x1))=  1 + x1  
  POL(zeros)=  0  
  POL(head(x1))=  x1  
  POL(a__head(x1))=  x1  
  POL(a__incr(x1))=  x1  
was used.

Not all Rules of R can be deleted, so we still have to regard a part of R. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →TRS5
                 ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 1
                 ↳Modular Removal of Rules


Dependency Pairs:

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(tail(X)) -> 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)
MARK(head(X)) -> MARK(X)
MARK(s(X)) -> MARK(X)


Rules:


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





We have the following set of usable rules:

mark(nats) -> anats
mark(s(X)) -> s(mark(X))
mark(head(X)) -> ahead(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(incr(X)) -> aincr(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(zeros) -> azeros
mark(adx(X)) -> aadx(mark(X))
mark(0) -> 0
anats -> nats
ahead(X) -> head(X)
aincr(X) -> incr(X)
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(nil) -> nil
atail(X) -> tail(X)
azeros -> cons(0, zeros)
azeros -> zeros
aadx(X) -> adx(X)
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(a__nats)=  0  
  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)=  0  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(a__tail(x1))=  x1  
  POL(head(x1))=  x1  
  POL(zeros)=  0  
  POL(a__head(x1))=  x1  
  POL(a__incr(x1))=  x1  

We have the following set D of usable symbols: {anats, MARK, adx, azeros, AADX, tail, incr, mark, aadx, AINCR, 0, cons, nats, nil, s, atail, zeros, head, ahead, aincr}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

No Rules can be deleted.

The result of this processor delivers one new DP problem.



   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 2
                 ↳Modular Removal of Rules


Dependency Pairs:

AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))
MARK(adx(X)) -> AADX(mark(X))
MARK(tail(X)) -> 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)
MARK(head(X)) -> MARK(X)
MARK(s(X)) -> MARK(X)


Rules:


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





We have the following set of usable rules:

mark(nats) -> anats
mark(s(X)) -> s(mark(X))
mark(head(X)) -> ahead(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(incr(X)) -> aincr(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(zeros) -> azeros
mark(adx(X)) -> aadx(mark(X))
mark(0) -> 0
anats -> nats
ahead(X) -> head(X)
aincr(X) -> incr(X)
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(nil) -> nil
atail(X) -> tail(X)
azeros -> cons(0, zeros)
azeros -> zeros
aadx(X) -> adx(X)
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(a__nats)=  0  
  POL(adx(x1))=  x1  
  POL(MARK(x1))=  x1  
  POL(a__zeros)=  0  
  POL(A__ADX(x1))=  x1  
  POL(tail(x1))=  1 + 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(s(x1))=  x1  
  POL(a__tail(x1))=  1 + x1  
  POL(head(x1))=  x1  
  POL(zeros)=  0  
  POL(a__head(x1))=  x1  
  POL(a__incr(x1))=  x1  

We have the following set D of usable symbols: {anats, MARK, adx, azeros, AADX, tail, incr, mark, aadx, AINCR, 0, cons, nats, nil, s, atail, zeros, head, ahead, aincr}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

MARK(tail(X)) -> MARK(X)

No Rules can be deleted.

The result of this processor delivers one new DP problem.



   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 3
                 ↳Modular Removal of Rules


Dependency Pairs:

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)
MARK(head(X)) -> MARK(X)
MARK(s(X)) -> MARK(X)


Rules:


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





We have the following set of usable rules:

mark(nats) -> anats
mark(s(X)) -> s(mark(X))
mark(head(X)) -> ahead(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(incr(X)) -> aincr(mark(X))
mark(tail(X)) -> atail(mark(X))
mark(zeros) -> azeros
mark(adx(X)) -> aadx(mark(X))
mark(0) -> 0
anats -> nats
ahead(X) -> head(X)
aincr(X) -> incr(X)
aincr(cons(X, L)) -> cons(s(mark(X)), incr(L))
aincr(nil) -> nil
atail(X) -> tail(X)
azeros -> cons(0, zeros)
azeros -> zeros
aadx(X) -> adx(X)
aadx(cons(X, L)) -> aincr(cons(mark(X), adx(L)))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(a__nats)=  0  
  POL(adx(x1))=  x1  
  POL(MARK(x1))=  x1  
  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(s(x1))=  x1  
  POL(a__tail(x1))=  x1  
  POL(head(x1))=  1 + x1  
  POL(zeros)=  0  
  POL(a__head(x1))=  1 + x1  
  POL(a__incr(x1))=  x1  

We have the following set D of usable symbols: {anats, MARK, adx, azeros, AADX, tail, incr, mark, aadx, AINCR, 0, cons, nats, nil, s, atail, zeros, head, ahead, aincr}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

MARK(head(X)) -> MARK(X)

No Rules can be deleted.

The result of this processor delivers one new DP problem.



   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 4
                 ↳Negative Polynomial Order


Dependency Pairs:

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)
MARK(s(X)) -> MARK(X)


Rules:


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





The following Dependency Pair can be strictly oriented using the given order.

AADX(cons(X, L)) -> AINCR(cons(mark(X), adx(L)))


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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


Used ordering:
Polynomial Order with Interpretation:

POL( AADX(x1) ) = x1 + 1

POL( cons(x1, x2) ) = x1

POL( AINCR(x1) ) = x1

POL( mark(x1) ) = x1

POL( MARK(x1) ) = x1

POL( s(x1) ) = x1

POL( incr(x1) ) = x1

POL( adx(x1) ) = x1 + 1

POL( nats ) = 0

POL( anats ) = 0

POL( head(x1) ) = 0

POL( ahead(x1) ) = 0

POL( nil ) = 0

POL( aincr(x1) ) = x1

POL( tail(x1) ) = 0

POL( atail(x1) ) = 0

POL( zeros ) = 0

POL( azeros ) = 0

POL( aadx(x1) ) = x1 + 1

POL( 0 ) = 0


This results in one new DP problem. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pairs:

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)
MARK(s(X)) -> MARK(X)


Rules:


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





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


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 6
                 ↳Negative Polynomial Order


Dependency Pairs:

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


Rules:


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





The following Dependency Pairs can be strictly oriented using the given order.

AINCR(cons(X, L)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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


Used ordering:
Polynomial Order with Interpretation:

POL( AINCR(x1) ) = x1

POL( cons(x1, x2) ) = x1 + 1

POL( MARK(x1) ) = x1

POL( s(x1) ) = x1

POL( incr(x1) ) = x1

POL( mark(x1) ) = x1

POL( nats ) = 0

POL( anats ) = 0

POL( head(x1) ) = 0

POL( ahead(x1) ) = 0

POL( nil ) = 0

POL( aincr(x1) ) = x1

POL( tail(x1) ) = 0

POL( atail(x1) ) = 0

POL( zeros ) = 1

POL( azeros ) = 1

POL( adx(x1) ) = x1

POL( aadx(x1) ) = x1

POL( 0 ) = 0


This results in one new DP problem. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pairs:

MARK(incr(X)) -> AINCR(mark(X))
MARK(s(X)) -> MARK(X)
MARK(incr(X)) -> MARK(X)


Rules:


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





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


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 8
                 ↳Size-Change Principle


Dependency Pairs:

MARK(incr(X)) -> MARK(X)
MARK(s(X)) -> MARK(X)


Rules:


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





We number the DPs as follows:
  1. MARK(incr(X)) -> MARK(X)
  2. MARK(s(X)) -> MARK(X)
and get the following Size-Change Graph(s):
{2, 1} , {2, 1}
1>1

which lead(s) to this/these maximal multigraph(s):
{2, 1} , {2, 1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
incr(x1) -> incr(x1)
s(x1) -> s(x1)

We obtain no new DP problems.

Termination of R successfully shown.
Duration:
0:06 minutes