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

Innermost Termination of R to be shown. 



   R
     ↳Removing Redundant Rules



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

app(bits, 0) -> 0

where the Polynomial interpretation:
  POL(0)=  0  
  POL(bits)=  1  
  POL(s)=  0  
  POL(half)=  0  
  POL(app(x1, x2))=  x1 + x2  
was used.

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


   R
     ↳RRRPolo
       →TRS2
         ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(half, app(s, app(s, x))) -> APP(s, app(half, x))
APP(half, app(s, app(s, x))) -> APP(half, x)
APP(bits, app(s, x)) -> APP(s, app(bits, app(half, app(s, x))))
APP(bits, app(s, x)) -> APP(bits, app(half, app(s, x)))
APP(bits, app(s, x)) -> APP(half, app(s, x))

Furthermore, R contains two SCCs. 


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


Dependency Pair:

APP(half, app(s, app(s, x))) -> APP(half, x)


Rules:


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


Strategy:

innermost




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


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 3
                 ↳A-Transformation
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

APP(half, app(s, app(s, x))) -> APP(half, x)


Rule:

none


Strategy:

innermost




We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 4
                 ↳Size-Change Principle
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




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


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


Dependency Pair:

APP(bits, app(s, x)) -> APP(bits, app(half, app(s, x)))


Rules:


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


Strategy:

innermost




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


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 5
                 ↳A-Transformation


Dependency Pair:

APP(bits, app(s, x)) -> APP(bits, app(half, app(s, x)))


Rules:


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


Strategy:

innermost




We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.


   R
     ↳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)))


Rules:


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


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


The result of this processor delivers one new DP problem.



   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 7
                 ↳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.



   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 8
                 ↳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:
0:00 minutes