Term Rewriting System R:
[x]
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
bits(0) -> 0
bits(s(x)) -> s(bits(half(s(x))))

Innermost Termination of R to be shown. 



   TRS
     ↳Removing Redundant Rules



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

bits(0) -> 0

where the Polynomial interpretation:
  POL(0)=  0  
  POL(bits(x1))=  1 + x1  
  POL(s(x1))=  x1  
  POL(half(x1))=  x1  
was used.

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


   TRS
     ↳RRRPolo
       →TRS2
         ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

HALF(s(s(x))) -> HALF(x)
BITS(s(x)) -> BITS(half(s(x)))
BITS(s(x)) -> HALF(s(x))

Furthermore, R contains two SCCs. 


   TRS
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳Usable Rules (Innermost)
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

HALF(s(s(x))) -> HALF(x)


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
bits(s(x)) -> s(bits(half(s(x))))


Strategy:

innermost




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


   TRS
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 3
                 ↳Modular Removal of Rules
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

HALF(s(s(x))) -> HALF(x)


Rule:

none


Strategy:

innermost




We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(HALF(x1))=  x1  
  POL(s(x1))=  x1  

We have the following set D of usable symbols: {HALF}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

HALF(s(s(x))) -> HALF(x)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved. 



   TRS
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳Usable Rules (Innermost)


Dependency Pair:

BITS(s(x)) -> BITS(half(s(x)))


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
bits(s(x)) -> s(bits(half(s(x))))


Strategy:

innermost




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


   TRS
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 4
                 ↳Modular Removal of Rules


Dependency Pair:

BITS(s(x)) -> BITS(half(s(x)))


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))


Strategy:

innermost




We have the following set of usable rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(BITS(x1))=  x1  
  POL(0)=  1  
  POL(s(x1))=  2·x1  
  POL(half(x1))=  x1  

We have the following set D of usable symbols: {BITS, 0, s, half}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

half(s(0)) -> 0


The result of this processor delivers one new DP problem.



   TRS
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 5
                 ↳Modular Removal of Rules


Dependency Pair:

BITS(s(x)) -> BITS(half(s(x)))


Rules:


half(0) -> 0
half(s(s(x))) -> s(half(x))


Strategy:

innermost




We have the following set of usable rules:

half(0) -> 0
half(s(s(x))) -> s(half(x))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(BITS(x1))=  1 + x1  
  POL(0)=  0  
  POL(s(x1))=  1 + x1  
  POL(half(x1))=  x1  

We have the following set D of usable symbols: {BITS, 0, s, half}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

half(s(s(x))) -> s(half(x))


The result of this processor delivers one new DP problem.



   TRS
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 6
                 ↳Modular Removal of Rules


Dependency Pair:

BITS(s(x)) -> BITS(half(s(x)))


Rule:


half(0) -> 0


Strategy:

innermost




We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(BITS(x1))=  x1  
  POL(s(x1))=  x1  
  POL(half(x1))=  x1  

We have the following set D of usable symbols: {BITS, s, half}
No Dependency Pairs can be deleted.
1 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



   TRS
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pair:

BITS(s(x)) -> BITS(half(s(x)))


Rule:

none


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes