Term Rewriting System R:
[x]
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Termination of R to be shown. 



   R
     ↳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
     ↳OC
       →TRS2
         ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

HALF(s(s(x))) -> HALF(x)
LASTBIT(s(s(x))) -> LASTBIT(x)
CONV(s(x)) -> CONV(half(s(x)))
CONV(s(x)) -> HALF(s(x))
CONV(s(x)) -> LASTBIT(s(x))

Furthermore, R contains three SCCs. 


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳Usable Rules (Innermost)
           →DP Problem 2
             ↳UsableRules
           →DP Problem 3
             ↳UsableRules


Dependency Pair:

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


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))


Strategy:

innermost




As we are in the innermost case, we can delete all 8 non-usable-rules.


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 4
                 ↳Size-Change Principle
           →DP Problem 2
             ↳UsableRules
           →DP Problem 3
             ↳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
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳Usable Rules (Innermost)
           →DP Problem 3
             ↳UsableRules


Dependency Pair:

LASTBIT(s(s(x))) -> LASTBIT(x)


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))


Strategy:

innermost




As we are in the innermost case, we can delete all 8 non-usable-rules.


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 5
                 ↳Size-Change Principle
           →DP Problem 3
             ↳UsableRules


Dependency Pair:

LASTBIT(s(s(x))) -> LASTBIT(x)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. LASTBIT(s(s(x))) -> LASTBIT(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
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
           →DP Problem 3
             ↳Usable Rules (Innermost)


Dependency Pair:

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


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))


Strategy:

innermost




As we are in the innermost case, we can delete all 5 non-usable-rules.


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
           →DP Problem 3
             ↳UsableRules
             ...
               →DP Problem 6
                 ↳Modular Removal of Rules


Dependency Pair:

CONV(s(x)) -> CONV(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(0)=  0  
  POL(CONV(x1))=  1 + x1  
  POL(s(x1))=  1 + x1  
  POL(half(x1))=  x1  

We have the following set D of usable symbols: {0, CONV, 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
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
           →DP Problem 3
             ↳UsableRules
             ...
               →DP Problem 7
                 ↳Modular Removal of Rules


Dependency Pair:

CONV(s(x)) -> CONV(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(CONV(x1))=  x1  
  POL(s(x1))=  x1  
  POL(half(x1))=  x1  

We have the following set D of usable symbols: {CONV, 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
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
           →DP Problem 3
             ↳UsableRules
             ...
               →DP Problem 8
                 ↳Dependency Graph


Dependency Pair:

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


Rule:

none


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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