Term Rewriting System R:
[x, y]
double(0) -> 0
double(s(x)) -> s(s(double(x)))
double(x) -> +(x, x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(s(x), y) -> s(+(x, y))

Termination of R to be shown. 



   R
     ↳Removing Redundant Rules



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

double(0) -> 0
+(x, 0) -> x

where the Polynomial interpretation:
  POL(0)=  1  
  POL(s(x1))=  x1  
  POL(+(x1, x2))=  x1 + x2  
  POL(double(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:

double(x) -> +(x, x)

where the Polynomial interpretation:
  POL(s(x1))=  x1  
  POL(+(x1, x2))=  x1 + x2  
  POL(double(x1))=  1 + 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
         ↳RRRPolo
           →TRS3
             ↳Removing Redundant Rules



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

+(s(x), y) -> s(+(x, y))

where the Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(+(x1, x2))=  2·x1 + x2  
  POL(double(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
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →TRS4
                 ↳Removing Redundant Rules



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

+(x, s(y)) -> s(+(x, y))

where the Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(+(x1, x2))=  x1 + 2·x2  
  POL(double(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
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →TRS5
                 ↳Overlay and local confluence Check



The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost. 


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →TRS6
                 ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

DOUBLE(s(x)) -> DOUBLE(x)

Furthermore, R contains one SCC. 


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


Dependency Pair:

DOUBLE(s(x)) -> DOUBLE(x)


Rule:


double(s(x)) -> s(s(double(x)))


Strategy:

innermost




As we are in the innermost case, we can delete all 1 non-usable-rules.


   R
     ↳RRRPolo
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳RRRPolo
             ...
               →DP Problem 2
                 ↳Size-Change Principle


Dependency Pair:

DOUBLE(s(x)) -> DOUBLE(x)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. DOUBLE(s(x)) -> DOUBLE(x)
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:
s(x1) -> s(x1)

We obtain no new DP problems.

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