Term Rewriting System R:
[x, y]
natsactive -> addactive(zerosactive)
natsactive -> nats
hdactive(x) -> hd(x)
hdactive(cons(x, y)) -> mark(x)
zerosactive -> cons(0, zeros)
zerosactive -> zeros
tlactive(x) -> tl(x)
tlactive(cons(x, y)) -> mark(y)
incractive(cons(x, y)) -> cons(s(x), incr(y))
incractive(x) -> incr(x)
mark(nats) -> natsactive
mark(zeros) -> zerosactive
mark(incr(x)) -> incractive(mark(x))
mark(add(x)) -> addactive(mark(x))
mark(hd(x)) -> hdactive(mark(x))
mark(tl(x)) -> tlactive(mark(x))
mark(0) -> 0
mark(s(x)) -> s(x)
mark(cons(x, y)) -> cons(x, y)
addactive(cons(x, y)) -> incractive(cons(x, add(y)))
addactive(x) -> add(x)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

NATSACTIVE -> ADDACTIVE(zerosactive)
NATSACTIVE -> ZEROSACTIVE
HDACTIVE(cons(x, y)) -> MARK(x)
TLACTIVE(cons(x, y)) -> MARK(y)
MARK(nats) -> NATSACTIVE
MARK(zeros) -> ZEROSACTIVE
MARK(incr(x)) -> INCRACTIVE(mark(x))
MARK(incr(x)) -> MARK(x)
MARK(add(x)) -> ADDACTIVE(mark(x))
MARK(add(x)) -> MARK(x)
MARK(hd(x)) -> HDACTIVE(mark(x))
MARK(hd(x)) -> MARK(x)
MARK(tl(x)) -> TLACTIVE(mark(x))
MARK(tl(x)) -> MARK(x)
ADDACTIVE(cons(x, y)) -> INCRACTIVE(cons(x, add(y)))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

MARK(tl(x)) -> MARK(x)
TLACTIVE(cons(x, y)) -> MARK(y)
MARK(tl(x)) -> TLACTIVE(mark(x))
MARK(hd(x)) -> MARK(x)
MARK(hd(x)) -> HDACTIVE(mark(x))
MARK(add(x)) -> MARK(x)
MARK(incr(x)) -> MARK(x)
HDACTIVE(cons(x, y)) -> MARK(x)


Rules:


natsactive -> addactive(zerosactive)
natsactive -> nats
hdactive(x) -> hd(x)
hdactive(cons(x, y)) -> mark(x)
zerosactive -> cons(0, zeros)
zerosactive -> zeros
tlactive(x) -> tl(x)
tlactive(cons(x, y)) -> mark(y)
incractive(cons(x, y)) -> cons(s(x), incr(y))
incractive(x) -> incr(x)
mark(nats) -> natsactive
mark(zeros) -> zerosactive
mark(incr(x)) -> incractive(mark(x))
mark(add(x)) -> addactive(mark(x))
mark(hd(x)) -> hdactive(mark(x))
mark(tl(x)) -> tlactive(mark(x))
mark(0) -> 0
mark(s(x)) -> s(x)
mark(cons(x, y)) -> cons(x, y)
addactive(cons(x, y)) -> incractive(cons(x, add(y)))
addactive(x) -> add(x)





The following dependency pairs can be strictly oriented:

MARK(tl(x)) -> MARK(x)
MARK(tl(x)) -> TLACTIVE(mark(x))


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

hdactive(x) -> hd(x)
hdactive(cons(x, y)) -> mark(x)
mark(nats) -> natsactive
mark(zeros) -> zerosactive
mark(incr(x)) -> incractive(mark(x))
mark(add(x)) -> addactive(mark(x))
mark(hd(x)) -> hdactive(mark(x))
mark(tl(x)) -> tlactive(mark(x))
mark(0) -> 0
mark(s(x)) -> s(x)
mark(cons(x, y)) -> cons(x, y)
tlactive(x) -> tl(x)
tlactive(cons(x, y)) -> mark(y)
natsactive -> addactive(zerosactive)
natsactive -> nats
zerosactive -> cons(0, zeros)
zerosactive -> zeros
incractive(cons(x, y)) -> cons(s(x), incr(y))
incractive(x) -> incr(x)
addactive(cons(x, y)) -> incractive(cons(x, add(y)))
addactive(x) -> add(x)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(zeros_active)=  0  
  POL(hd_active(x1))=  x1  
  POL(incr_active(x1))=  x1  
  POL(MARK(x1))=  x1  
  POL(TL_ACTIVE(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(tl(x1))=  1 + x1  
  POL(HD_ACTIVE(x1))=  x1  
  POL(add(x1))=  x1  
  POL(add_active(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(hd(x1))=  x1  
  POL(nats)=  0  
  POL(s(x1))=  0  
  POL(zeros)=  0  
  POL(nats_active)=  0  
  POL(tl_active(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pairs:

TLACTIVE(cons(x, y)) -> MARK(y)
MARK(hd(x)) -> MARK(x)
MARK(hd(x)) -> HDACTIVE(mark(x))
MARK(add(x)) -> MARK(x)
MARK(incr(x)) -> MARK(x)
HDACTIVE(cons(x, y)) -> MARK(x)


Rules:


natsactive -> addactive(zerosactive)
natsactive -> nats
hdactive(x) -> hd(x)
hdactive(cons(x, y)) -> mark(x)
zerosactive -> cons(0, zeros)
zerosactive -> zeros
tlactive(x) -> tl(x)
tlactive(cons(x, y)) -> mark(y)
incractive(cons(x, y)) -> cons(s(x), incr(y))
incractive(x) -> incr(x)
mark(nats) -> natsactive
mark(zeros) -> zerosactive
mark(incr(x)) -> incractive(mark(x))
mark(add(x)) -> addactive(mark(x))
mark(hd(x)) -> hdactive(mark(x))
mark(tl(x)) -> tlactive(mark(x))
mark(0) -> 0
mark(s(x)) -> s(x)
mark(cons(x, y)) -> cons(x, y)
addactive(cons(x, y)) -> incractive(cons(x, add(y)))
addactive(x) -> add(x)





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(hd(x)) -> MARK(x)
HDACTIVE(cons(x, y)) -> MARK(x)
MARK(hd(x)) -> HDACTIVE(mark(x))
MARK(add(x)) -> MARK(x)
MARK(incr(x)) -> MARK(x)


Rules:


natsactive -> addactive(zerosactive)
natsactive -> nats
hdactive(x) -> hd(x)
hdactive(cons(x, y)) -> mark(x)
zerosactive -> cons(0, zeros)
zerosactive -> zeros
tlactive(x) -> tl(x)
tlactive(cons(x, y)) -> mark(y)
incractive(cons(x, y)) -> cons(s(x), incr(y))
incractive(x) -> incr(x)
mark(nats) -> natsactive
mark(zeros) -> zerosactive
mark(incr(x)) -> incractive(mark(x))
mark(add(x)) -> addactive(mark(x))
mark(hd(x)) -> hdactive(mark(x))
mark(tl(x)) -> tlactive(mark(x))
mark(0) -> 0
mark(s(x)) -> s(x)
mark(cons(x, y)) -> cons(x, y)
addactive(cons(x, y)) -> incractive(cons(x, add(y)))
addactive(x) -> add(x)





The following dependency pairs can be strictly oriented:

MARK(hd(x)) -> MARK(x)
MARK(hd(x)) -> HDACTIVE(mark(x))


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

hdactive(x) -> hd(x)
hdactive(cons(x, y)) -> mark(x)
mark(nats) -> natsactive
mark(zeros) -> zerosactive
mark(incr(x)) -> incractive(mark(x))
mark(add(x)) -> addactive(mark(x))
mark(hd(x)) -> hdactive(mark(x))
mark(tl(x)) -> tlactive(mark(x))
mark(0) -> 0
mark(s(x)) -> s(x)
mark(cons(x, y)) -> cons(x, y)
tlactive(x) -> tl(x)
tlactive(cons(x, y)) -> mark(y)
natsactive -> addactive(zerosactive)
natsactive -> nats
zerosactive -> cons(0, zeros)
zerosactive -> zeros
incractive(cons(x, y)) -> cons(s(x), incr(y))
incractive(x) -> incr(x)
addactive(cons(x, y)) -> incractive(cons(x, add(y)))
addactive(x) -> add(x)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(zeros_active)=  0  
  POL(hd_active(x1))=  1 + x1  
  POL(incr_active(x1))=  x1  
  POL(MARK(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(tl(x1))=  x1  
  POL(HD_ACTIVE(x1))=  x1  
  POL(add(x1))=  x1  
  POL(add_active(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(hd(x1))=  1 + x1  
  POL(nats)=  0  
  POL(s(x1))=  0  
  POL(zeros)=  0  
  POL(nats_active)=  0  
  POL(tl_active(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:

HDACTIVE(cons(x, y)) -> MARK(x)
MARK(add(x)) -> MARK(x)
MARK(incr(x)) -> MARK(x)


Rules:


natsactive -> addactive(zerosactive)
natsactive -> nats
hdactive(x) -> hd(x)
hdactive(cons(x, y)) -> mark(x)
zerosactive -> cons(0, zeros)
zerosactive -> zeros
tlactive(x) -> tl(x)
tlactive(cons(x, y)) -> mark(y)
incractive(cons(x, y)) -> cons(s(x), incr(y))
incractive(x) -> incr(x)
mark(nats) -> natsactive
mark(zeros) -> zerosactive
mark(incr(x)) -> incractive(mark(x))
mark(add(x)) -> addactive(mark(x))
mark(hd(x)) -> hdactive(mark(x))
mark(tl(x)) -> tlactive(mark(x))
mark(0) -> 0
mark(s(x)) -> s(x)
mark(cons(x, y)) -> cons(x, y)
addactive(cons(x, y)) -> incractive(cons(x, add(y)))
addactive(x) -> add(x)





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(add(x)) -> MARK(x)
MARK(incr(x)) -> MARK(x)


Rules:


natsactive -> addactive(zerosactive)
natsactive -> nats
hdactive(x) -> hd(x)
hdactive(cons(x, y)) -> mark(x)
zerosactive -> cons(0, zeros)
zerosactive -> zeros
tlactive(x) -> tl(x)
tlactive(cons(x, y)) -> mark(y)
incractive(cons(x, y)) -> cons(s(x), incr(y))
incractive(x) -> incr(x)
mark(nats) -> natsactive
mark(zeros) -> zerosactive
mark(incr(x)) -> incractive(mark(x))
mark(add(x)) -> addactive(mark(x))
mark(hd(x)) -> hdactive(mark(x))
mark(tl(x)) -> tlactive(mark(x))
mark(0) -> 0
mark(s(x)) -> s(x)
mark(cons(x, y)) -> cons(x, y)
addactive(cons(x, y)) -> incractive(cons(x, add(y)))
addactive(x) -> add(x)





The following dependency pair can be strictly oriented:

MARK(add(x)) -> MARK(x)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(add(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:

MARK(incr(x)) -> MARK(x)


Rules:


natsactive -> addactive(zerosactive)
natsactive -> nats
hdactive(x) -> hd(x)
hdactive(cons(x, y)) -> mark(x)
zerosactive -> cons(0, zeros)
zerosactive -> zeros
tlactive(x) -> tl(x)
tlactive(cons(x, y)) -> mark(y)
incractive(cons(x, y)) -> cons(s(x), incr(y))
incractive(x) -> incr(x)
mark(nats) -> natsactive
mark(zeros) -> zerosactive
mark(incr(x)) -> incractive(mark(x))
mark(add(x)) -> addactive(mark(x))
mark(hd(x)) -> hdactive(mark(x))
mark(tl(x)) -> tlactive(mark(x))
mark(0) -> 0
mark(s(x)) -> s(x)
mark(cons(x, y)) -> cons(x, y)
addactive(cons(x, y)) -> incractive(cons(x, add(y)))
addactive(x) -> add(x)





The following dependency pair can be strictly oriented:

MARK(incr(x)) -> MARK(x)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(incr(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


natsactive -> addactive(zerosactive)
natsactive -> nats
hdactive(x) -> hd(x)
hdactive(cons(x, y)) -> mark(x)
zerosactive -> cons(0, zeros)
zerosactive -> zeros
tlactive(x) -> tl(x)
tlactive(cons(x, y)) -> mark(y)
incractive(cons(x, y)) -> cons(s(x), incr(y))
incractive(x) -> incr(x)
mark(nats) -> natsactive
mark(zeros) -> zerosactive
mark(incr(x)) -> incractive(mark(x))
mark(add(x)) -> addactive(mark(x))
mark(hd(x)) -> hdactive(mark(x))
mark(tl(x)) -> tlactive(mark(x))
mark(0) -> 0
mark(s(x)) -> s(x)
mark(cons(x, y)) -> cons(x, y)
addactive(cons(x, y)) -> incractive(cons(x, add(y)))
addactive(x) -> add(x)





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes