Term Rewriting System R:
[X, L]
incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Innermost Termination of R to be shown. 



   R
     ↳Removing Redundant Rules



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

adx(nil) -> nil
head(cons(X, L)) -> X

where the Polynomial interpretation:
  POL(adx(x1))=  2·x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  0  
  POL(tail(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(nil)=  1  
  POL(s(x1))=  x1  
  POL(zeros)=  0  
  POL(head(x1))=  2 + 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:

tail(cons(X, L)) -> L

where the Polynomial interpretation:
  POL(adx(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  0  
  POL(tail(x1))=  1 + x1  
  POL(incr(x1))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(zeros)=  0  
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:

nats -> adx(zeros)

where the Polynomial interpretation:
  POL(adx(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nats)=  1  
  POL(incr(x1))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(zeros)=  0  
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
                 ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

INCR(cons(X, L)) -> INCR(L)
ADX(cons(X, L)) -> INCR(cons(X, adx(L)))
ADX(cons(X, L)) -> ADX(L)
ZEROS -> ZEROS

Furthermore, R contains three SCCs. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 1
                 ↳Usable Rules (Innermost)


Dependency Pair:

INCR(cons(X, L)) -> INCR(L)


Rules:


incr(cons(X, L)) -> cons(s(X), incr(L))
incr(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
zeros -> cons(0, zeros)


Strategy:

innermost




As we are in the innermost case, we can delete all 4 non-usable-rules.


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 4
                 ↳Size-Change Principle


Dependency Pair:

INCR(cons(X, L)) -> INCR(L)


Rule:

none


Strategy:

innermost




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

which lead(s) to this/these maximal multigraph(s):
{1} , {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:
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 2
                 ↳Usable Rules (Innermost)


Dependency Pair:

ZEROS -> ZEROS


Rules:


incr(cons(X, L)) -> cons(s(X), incr(L))
incr(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
zeros -> cons(0, zeros)


Strategy:

innermost




As we are in the innermost case, we can delete all 4 non-usable-rules.


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 5
                 ↳Non Termination


Dependency Pair:

ZEROS -> ZEROS


Rule:

none


Strategy:

innermost




Found an infinite P-chain over R:
P =

ZEROS -> ZEROS

R = none

s = ZEROS
evaluates to t =ZEROS

Thus, s starts an infinite chain.

Innermost Non-Termination of R could be shown.
Duration:
0:04 minutes